Mathematics

, ...-... .... """ .. -. HOW TO type or mathematICs SOLVE _1eIY_" WORn == PROBLEM :::. rMrHEMMICS IrIlD II fJQJtI'I DAVID WAYNE, Ph. D. How to Solve Word Problems in Mathematics David S. Wayne, Ph.D. McGraw-Hill New York San Francisco Washington, D.C. Auckland Bogota Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto Library of Congress Cataloging-in-Publication Data applied for. McGraw-Hill ~ A Division of The McGraw·RiU Companies Copyright © 2001 by The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 1 2 3 4 5 6 7 8 9 0 DOC/DOC 0 9 8 7 6 5 4 3 2 1 0 ISBN 0-07-136272-X The editing supervisor was Maureen B. Walker and the production supervisor was Tina Cameron. It was set in Stone Serif by PRD Group. Printed and bound by R. R. Donnelley & Sons Company. McGraw-Hill books are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs. For more information, please write to the Director of Special Sales, Professional Publishing, McGraw-Hill, Two Penn Plaza, New York, NY 10121-2298. Or contact your local bookstore. o This book is printed on recycled, acid-free paper containing a minimum of 50% recycled, de-inked fiber. Contents Preface v Chapter 1-Measurement, Estimation, and Using Formulas 1 Chapter 2-Using Algebraic Equations to Solve Problems 29 Chapter 3-Word Problems Involving Ratio, Proportion, and Percentage 65 Chapter 4-Word Problems Involving Geometry and Trigonometry 91 Chapter 5-Word Problems Involving Statistics, Counting, and Probability 123 Chapter 6-Miscellaneous Problem Drill 147 Appendix-A Brief Review of Solving Equations 155 iii Preface In nearly every part of the United States, students in grades 6 to 12 are undergoing statewide assessments in mathematics that differ significantly from tests that were taken prior to the late 1990s. These current tests are based on revised standards that were created to address a noticeable nationwide weakness in applying the principles and computational skills in math- ematics to realistic, relevant, and practical situations. Most state curriculum documents list as a key component math- ematical reasoning and modeling. These are reflected in the assessments in the form of word problems, and, in most cases, students are required to not only produce a correct answer but also to carefully explain their work. Additionally, many of these assessment tests cover multiple grade levels and require students to recall facts and procedures that may not be part of their current year's course. This book is designed to enhance the student's ability to recall those facts and procedures that are necessary for success in solving word problems. The problems offer opportunities to review the important areas of measurement, estimation, for- mula application, algebraic representation, interpreting geo- metric situations, data analysis, and probability. The solutions include strategies for decoding problems, determining what skills and concepts need to be applied, and how to present a solution that shows complete understanding. This book is not v designed as a textbook that presents information to be learned for the first time; however, there is sufficient explanation of key concepts to enhance the student's understanding. There is also an appendix which contains a review of solving linear and quadratic equations. Each chapter in the book focuses on word problems that have a common element. Chapter 1 presents problems involving measurement and the application of simple formulas. The information and con- cepts used in these word problems form the core of most prob- lems that students face, and these problems are solved by sim- ple arithmetic and reasoning. Chapter 2 helps the students solve more complex prob- lems using algebraic skills and modeling. This chapter provides review and practice in solving linear and quadratic equations. Chapter 3 focuses on problems for which students need to recognize the use of ratios and proportions and apply them correctly. Chapter 4 presents the student with word problems that involve geometric relationships and ways to model these situations. Chapter 5 presents examples of interpreting data and using the basic principles of probability and statistics to re- spond to questions about the data. Chapter 6 is a collection of problems that are not cate- gorized and provides additional practice for students in deter- mining how these problems can best be solved. Whether used in a classroom setting with guidance by a teacher or at home as a supplement to class, this book will provide students with opportunities to better prepare them for being successful in applying mathematics. I'd like to take this opportunity to thank my wife, Phyllis; my daughter, Gayle; my friends and mentors, Carl Goodman and Bert Linder; and many other family members, colleagues, and friends. Without their support and encouragement, this endeavor would not have been possible. vi Chapter I Measurement, Estimation, and Using Formulas You have probably found solving word problems to be the most difficult part of any math course. In most cases, these involve applications of well-known formulas and an under- standing of the relationships between quantities. The key idea in solving these problems is to identify the way in which the quantities are being measured and to recognize which formula to apply. It is also helpful to be able to estimate an answer so that you can determine whether your worked-out solution is reasonable. The examples that follow will show you how to use some of the most common formulas and measurements that commonly appear in word problems. All the formulas and measurements are listed at the end of this chapter for reference. When measurements and calculations result in numbers that are not whole and have long decimal parts, it makes sense to round answers to one, two, or three decimal places depend- ing on the accuracy you need in order to solve the problem. Sometimes a problem will clearly tell you how many decimal places to use. For example, "Find the length to the nearest foot II requires you to give your answer rounded to the nearest whole number. Your calculations should be performed using at least one decimal place to ensure that your answer is rounded cor- rectly. In general, you should use at least one more decimal place in your calculations than the answer requires. If the problem I does not specify how you should state your answer, you should use a reasonable amount of decimal places so that your final answer will be meaningful and practical. For example, when a problem involves money, you could use two decimal places since we rarely calculate with fractions of a penny. Therefore, it is important to look for how an answer is to be given and to think about what would be a reasonable answer. Distance Measuring distance is the same as measuring length. In the English (Le., British and U.S. Customary) system, the units of measuring distance include inches, feet, yards, and miles. In the metric system [Le., International System of Units (SI)], the units include millimeters, centimeters, meters, and kilometers. It is important to know how one unit of measurement can be converted to another. An easy way to convert measurements is to divide the number of smaller units by the amount of those units that are in the larger unit. For example, the number of feet (ft) is equal to the number of inches (in) divided by 12, or the number of meters (m) is equal to the number of millimeters (mm) divided by 1000. For conversions that aren't as simple, another method will be shown. An important measurement of distance is perimeter, which is the distance around a region. This is used in the example below. Example I A rectangular room measures 216 in by 150 in. The interior designer wants to place a strip of molding around the room near the ceiling that costs $2.25 per foot. How much will it cost to buy the molding? Solution I The answer lies in finding the perimeter of the room first in inches and then in feet. The perimeter is the sum of all the edges of the shape. In the case of a rectangle Perimeter = 2 x length + 2 x width Width Length Therefore, the perimeter of our room is 2 x 216 + 2 x 150 = 732 in. We convert this to feet by dividing by 12, and we have 61 ft. The cost of the molding would be 61 ft x 2.25 d ~ l l a r s = $137.25 oot Note that the units in this product are written as fractions. This is a good device to help determine whether the items being multiplied are the correct ones and are in the correct order. The units can be multiplied as if they were fractions and, after "canceling" similar units that appear in the numerators and denominators, the resulting fraction should be the units you wish your answer to have. Thus, for this problem, (feet/I) x (dollars/foot) = (dollars/I) = dollars. Example 2 While planning a trip around Italy, Jim wanted to travel from Rome to Milan to Venice and back to Rome. On a map he found that the distances were approximately 300 kilometers (km) from Rome to Milan, 150 km from Milan to Venice, and 225 km from Venice to Rome. Jim was not comfortable with these measurements and wanted to know the distance in miles. To the nearest mile, what is the total distance of the trip? Solution 1 The conversion for miles to kilometers in the reference table states that 1 mile (mi) is 1.61 km. If we think of this as a fraction, we have Imi 1.61 km = 0.62mi/km For our problem, we need to add the distances in kilome- ters, 300 + 150 + 225 = 675, and then multiply it by 0.62. The answer is miles 675km x 0.62 I = 418.5mi = 419mi ki ometer Remember that an estimate is always useful to have first to be sure that your answer makes sense. Since the conversion factor is greater than 0.5, you would know that your answer would have to be more than l/Z of 300 + 150 + 225 or l/Z of 675 = 337.5 mi and you would feel comfortable with the answer of 419. Distance traveled can also be measured indirectly by knowing the speed at which the travel was occurring and the time taken for the journey. The distance is computed by the simple formula distance = rate x time. Example 3 A car has been traveling at an average speed of 55 mph (miles per hour) for 7 hours. How many miles has the car traveled? Solution 3 Note the phrase "average speed." The car may have trav- eled at somewhat different speeds throughout the 7 hours. However, for the purposes of this problem, you can assume that the car has been traveling at that average speed through- out the journey. (When making this assumption, we usually say that the car has traveled at a constant speed.) The question II How many miles ... traveled?" is really ask- ing "What is the total distance traveled?" Since you know the rate and the time, you can use the formula directly: Distance = 55 mph x 7 hours = 385 mi Notice that mph is really the ratio miles/hour, and we are once again multiplying units to arrive at our answer in miles. Area Area is the measurement of the two-dimensional space within a region such as the space on the floor of a room, the space on a canvas of a picture, or the space on which a house is built. 4 Note that the answer to an area problem uses square units. This leads to conversions different from those in measuring one-dimensional length. 1 square foot (ft 2 ) = 12 in x 12 in = 144 square inches (in 2 ) This means that within a square that measures 1 ft x 1 ft, you can fit 144 squares that measure 1 in by 1 in. 1 ft 1 in D 1 ft 1 in 1 ft2 1 in 2 1 square meter (m 2 ) = 100 centimeters (cm) x 100 cm = 10,000 square centimeters (cm 2 ) This means that within a square that measures 1 m by 1 m, you can fit 10,000 squares that measure 1 cm by 1 cm. When measuring area, it is important to make sure that the two measurements that are being multiplied are in the same units. This would require making conversions as we did when measuring length. Example 4 A rectangular plot of land is roped off for planting. The mea- surements are 8 ft 4 in by 5 ft 8 in. Each plant requires 1 square yard (yd 2 ) of space for planting. How many square yards have been roped off? How many plants can be planted in this plot of land? Solution 4 Be sure to draw a rectangle and label the sides with the given measurements! Drawing diagrams is an important good habit. s 5ftBin 8ft4in In this problem it is important to convert all measure- ments to the same units. Since we don't plant fractions of a plant, we will be looking for a whole-number answer. There- fore, we should convert the inches parts of our numbers to their equivalent fractional or decimal parts of a foot. Thus, for example, 4 in is one-third of a foot (1/3 ft) or 0.33, and 8 in is two-thirds of a foot (2/3 ft) or 0.66. We can also round our measurements to one or two decimal places since our answer must be in whole numbers. The measurements become 8.33 ft by 5.67 ft. The area is 47.23 square feet or ft2. Since 1 square yard (yd 2 ) is 3 ft by 3 ft, there are 9 ft 2 in 1 yd 2 . There are approximately 1 yd 2 2 8.33 ft x 5.67 ft x --2 = 5.25 yd 9 ft in this plot of land. The answer to the second question is a little trickier. If each plant truly reqUires a square shape in which to be planted, then we have to look at a diagram to see that there is enough room for only two plants! 5.67 ft Plant 8.33 ft i- r---- r- Plant cs This points out that when we measure area, we are not ac- tually counting squares. The quantity that we call the area sug- gests that if we could rearrange the space, we could construct that number of squares. The preceding diagram shows that there are only two actual squares in this plot of land that are 3 ft by 3 ft. The rest of the space could create an additional 3.25 squares if it were possible. Circles are popular in word problems of this type. The distance around the circle is referred to as its circumference (rather than perimeter), and the space within the circle is its area. The values of these measurements involve the number 7r (Greek lowercase pi), which doesn't have an exact value, but we approximate it with either 3.14 or with the fraction 22/7. (Technical work, such as engineering, often requires better ap- proximations. Supercomputers have calculated 7r to billions of decimal places beyond 3.14.) The diameter and radius of the circle are also involved in the calculation. The formulas are circumference = 7r x diameter or 2 x 7r x radius, because the diameter is equal to twice the radius, and area = 7r X radius 2 . Example 5 An interior designer wants to cover the floor of a room with a carpet that has a two-color design. The room is square, measur- ing 12 ft on each side. The designer wants a red circle, whose center is the center of the room, that leaves 2 ft to each of the four walls; the remainder of the carpet should be white (Le., there should be a 2-ft white-carpet border around the red-carpet circle in the center of the room). The red carpet costs $12 per square foot, and the white carpet costs $10 per square foot. How much will the carpeting cost? Solution S Once again, it is useful to estimate. The entire room is 12 ft by 12 ft, which creates an area of 144 ft2. Using the two prices, the cost should be between 144 ft2 x $10 / ft2 = $1440 and 144 ft2 x $12/ ft2 = $1728. (If this were a multiple-choice question, you might find that only one of the choices falls 7 into this range. That would have to be the correct choice, and your work would be done!) A diagram is required to find the actual answer. 1-=2+-----=-..,,---+2=-1 12 ft Since we want to leave 2 ft to the wall, the diameter of the circle must be 8 ft and, therefore, its radius must be 4 ft. We use the formula for the area of a circle, area = 7r X radius 2 , and find that the red circular area would be approximately 3.14 x 4 2 = 50.24 ft2. The cost of the red carpet would be 50.24 ft2 x $12/ ft2 = $602.88. The remaining area in the room would be 144 ft2_ 50.24 ft 2 = 93.76 ft2, and the white carpet would cost 93.76 ft2 x$10/ft 2 = $937.60. The total cost would be $1540.48. There is a measurement of land area that doesn't directly state that there are square units, but it is implied. This is re- ferred to as an acre (1 acre = 43,560 ft2). Example 6 A farmer has a farm liz mi x 11/4 mi. How many acres is this farm? How many different crops can the farmer plant, if each crop requires l/Z acre? Solution 6 The farm has 0.5 x 1.25 = 0.625 mi 2 . We need to convert this to acres. The conversion given above relates acres to square feet. We can set up a multiplication of fractions using the cor- rect units in the numerators and denominators to arrive at the 8 number of acres on the farm. 0.625 miz 5280 Z ftz 1 acre I x ml. z x z = 400 acres 43,560ft The farmer could plant 400 -;- l/Z = 800 crops. Volume Volume is a measure of the amount of material a three- dimensional object can hold. The most common objects for which we use formulas are rectangular boxes, circular cones, and spheres. The formulas are given in the reference table at the end of the chapter under the subheading "Measuring Volume." Note that the units of volume are cubic units since we are multiplying three measurements. A cube is a rectangular box whose length, width, and height are all equal. 1 ft 3 = 12in x 12in x 12in = 1728in 3 This means that into a box that measures 1 ft by 1 ft by 1 ft, we can fit 1728 boxes that measure 1 in by 1 in by 1 in. 1 ft 1 ft Similarly, 1 cubic meter (m 3 ) = 100 cm x 100 cm x 100 cm = 1,000,000 cubic centimeters (cm 3 ). This means that into a box that measures 1 m by 1 m by 1 m, we can fit 1 million boxes that measure 1 cm by 1 cm by 1 cm. Example 7 Paper is shredded into rectangular bins that measure 24 in by 12 in by 12 in. These bins are emptied into larger cubical bins that are 4 ft on each side. How many smaller bins can be emptied into each cubical bin? Solution 7 Since we are interested in knowing what it takes to fill something, we need to measure volume. If we can determine the volume of each smaller bin and the volume of the cubical bin, we can divide the volume of the cubical bin by the volume of each smaller bin to arrive at our answer. The volume of each smaller bin is 24 in x 12 in x 12 in or 2 ft x 1 ft x 1 ft = 2 ft 3 . The volume of the cubical bin is 4 ft x 4 ft x 4 ft = 64 ft 3 . Therefore, we can empty 64 -:- 2 = 32 smaller bins into the larger. We often use different units to measure the volume of a liqUid. These don't seem to be cubic units, but they really are. For example, in the metric system we use liters to measure vol- ume. We use the following conversion formula to understand how liters are related to cubic units. 1 milliliter( mL) = 1 cubic centimeter (cm 3 or cc) This means that lliter (L) = 1000 cm 3 • The English system of measurement uses the gallon as its standard, and 1 gallon (gal) = 231 in 3 • Gallons are subdivided as follows: 1 gal = 4 quarts (qt), 1 qt = 2 pints (pt), 1 pt = 2 cups, and 1 cup = 4 fluid ounces (oz). Example 8 A rectangular fishtank is 30 in long, 15 in wide, and 20 in high. Gravel is laid evenly at the bottom of the tank to a height of 2 in, and the tank should be filled to a point that is 3 in from the top of the tank. How much water can the tank hold to the nearest tenth of a gallon? 10 Solution 8 20 in 2 in I ~ __ --,," 15 in After the gravel is laid, the height available for water is 20 - (2 + 3) = 15 in. Therefore, the tank can hold 30 in x 15 in x 15 in = 6750 in 3 of water. Our answer, however, should be in liquid measurements. 6750 in 3 1 gal 1 x 3 = 29.22 = 29.2gal 23 lin Example 9 A balloon when expanded is in the shape of a sphere with a diameter of 24 cm. What is the maximum capacity of the balloon to the nearest liter? Solution 9 Capacity refers to the amount an object can hold; there- fore, we need to measure its volume. The reference table at the end of this chapter shows that the formula for the volume of a sphere is given in terms of its radius, which is one-half of the diameter, which in this case would be 12 cm. Using the formula for the volume of a sphere and 3.14 for Jr, we calcu- late the volume to be 4/3 x 3.14 x 12 3 = 7234.56cm 3 . This is equivalent to 7234.56 milliliters (mL) and, dividing by 1000, to 7.23456 L. Our answer would be 7 L. II Weight Weight is also measured in both the English and metric sys- tems in different ways. In the English system, we use the ounce as the smallest standard unit and the pound (lb; 1 lb = 16 oz) and ton (2000 lb) for larger weights. (The ounce used for weight is usually referred to as a dry ounce and is different from the fluid ounce used when measuring volume.) In the metric system, weight is sometimes referred to as mass. Here, we use the gram (g) as the smallest standard unit and the kilogram (kg; 1 kg = 1000 g) and metric ton (1000 kg) for larger weights. The conversion between systems most often used is that 1 kg is equal to approximately 2.2 lb. Also, 1 kg is the exact weight of 1 L of water. Example 10 In order to prepare for an international wrestling event, Jim had to bring his weight to under 80 kg. If he currently weighs 200 lb and begins a program in which he loses 2.S lb per week, in how many weeks will he reach the required weight? Solution 10 In order to work through this problem, we need to con- vert the information to either pounds or kilograms. Since the rate of losing weight is given in pounds per week, we ought to convert the 80 kg to pounds. Using the "fraction" scheme of conversion, we have 80kg x 2.21b = 176lb 1 1kg Jim would have to lose 200 - 176 = 24 lb. This would take 24lb -:- 2.S lb/week = 9.6 or 10 weeks to reach the eligible weight. Example II The ingredients label on a box of cereal indicates that one serving of cereal contains 24 g of carbohydrates. In order to understand this better, Rachel wants to convert this to ounces. To the nearest tenth, how many ounces of carbohydrates will she be eating, if she has two servings of cereal? 12 Solution II We can use the "fraction" scheme to answer this question with one calculation: 24 g 1 kg 2.21b 160z 2 servings ------''---- x x -- x -- x -------''- 1 serving 1000 g 1 kg lIb 1 = 1.6896 oz = 1.7 oz Temperature Other situations that arise in word problems involve tempera- ture. In the world today we measure temperature in two scales, Fahrenheit and Celsius (or Centigrade). The Celsius scale uses the freezing and boiling pOints of water to create a range from 0° to 100°e. The conversion to Fahrenheit is given by the for- mula F = 1.8 C + 32, where F represents degrees Fahrenheit and C represents degrees Celsius. Example 12 Find the freezing and boiling pOints of water in degrees Fahrenheit. Solution 12 This is a simple application of the formula. Freezing - O°C ---+ F = 1.8 x 0 + 32 = 32°F Boiling - 100°C ---+ F = 1.8 x 100 + 32 = 212°F Example 13 The human body maintains a normal temperature of 98.6°F. If a person developed a fever with a 3° temperature increase above normal, what would be the Celsius reading on a ther- mometer, to the nearest tenth? Solution 13 The new Fahrenheit temperature would be 101.6°F. The formula can still be used, except that we would have to solve 13 an algebraic equation to determine C. 101.6 = 1.8C + 32 (add -32 to both sides of the equation) -+1.8C = 69.6 (divide both sides of the equation by 1.8) 69.6 ° -+ C = IT = 38.66 ... C Shopping (which is 38.7°C to the nearest tenth) Many word problems involve finding the amount one has to spend on a purchase when sales tax and/or discounts are ap- plied. Usually we are given an initial purchase price of an item and a percentage that indicates the discount and/or tax. The solution to most of these problems is found by finding the dol- lar amount of those percentages and adding or subtracting. A formula one can use for finding a discount price is Discounted price = original price - discount x original price The discount needs to be in decimal form in order to perform the multiplication correctly. A formula for finding the cost with sales tax included is Total cost = price + tax x price Again, the tax should be in decimal form. Example 14 Jim wanted to buy several compact disks (CDs). The music store was having a sale that reduced the prices by 15 percent. He selected three CDs that cost $12.99 and one that cost $9.99. How much would this purchase cost if sales tax is 8 1 / 2 percent? Solution 14 Before figuring in the percentages, the total cost of the four CDs needs to be determined. This is (3 x $12.99) + $9.99 = $48.96. Using the first formula to find the new price 14 with the discount, we have New price = $48.96 - (0.15 x $48.96) = $41.62 Note that this price could also have been calculated by finding 85 percent of the price or multiplying $48.96 by 0.85. Using the second formula, we can determine the total cost, Total cost = $41.62 + ($41.62 x 0.085) = $45.16 Note that this could have been calculated by finding 108.5 percent of the price or multiplying $41.62 by 1.085. Interest Another common application of mathematics involves finan- ces. People are interested in determining how much money will result when interest is applied. Interest is simply an amount of money that is added to the original amount and calculated by a percentage of that amount. There are many complicated ways to calculate interest, but the simplest is found using what is called the simple interest formula: Interest earned = p x r x t where p is the original amount called the principal; r is the interest rate as a fraction, decimal, or percent; and t is the num- ber of years that the investment is allowed to grow. The total current amount is found by adding the principal investment and the interest earned. Total amount = principal + interest earned Example IS Samantha has been working hard as a baby-sitter and has $540 collected. She decides to invest this money in a fund that pre- dicts an interest rate of 8 to 11 percent. What could she expect to happen to her original investment after one year? 15 Solution IS You could first estimate an answer by supposing that the rate was 10 percent. This would result in Samantha receiving $54 in interest for an estimate of having a total of $594 at the end of the year. An actual solution would be as follows. At the least, Samantha would receive 8 percent interest. This would result in 1= $540 x 0.08 x 1 = $43.20. At the most, Samantha would receive 11 percent interest. This would result in 1= $540 x 0.11 x 1 = $59.40. At the end of one year, using total amount = p + I, she would have saved between $583.20 and $599.40. Note that the percentage needed to be changed to a dec- imal in order to make the calculation. If you didn't do this, the answer would be unreasonable, as the result would be 100 times larger! The problems that follow will give you plenty of oppor- tunity to use the formulas and ideas that you just studied. Be sure to draw diagrams to help you see the situation clearly, and take a look at the reference table at the end of this chapter to remind yourself of the formulas. Additional Problems I. Gayle would like to fence her garden, which is in the shape of a triangle with two sides measuring 10 1 /2 ft and the other side measuring 8 ft. The material she chose to use costs $4.25 per meter. How much will she spend to fence her gardenl 2. Gayle's triangular garden from Prob. I needs a plastic ground cover in order to keep weeds from growing. She measures the distance from the 8-ft side to the opposite vertex of the triangle to be 6 ft. How much plastic should she usel 3. The length around a wooden rectangular picture frame is 50 in. There is a 2-in border on all sides of the picture within the frame. If the longer side of the picture is I I in, what is the length of the shorter side of the framel 4. The owner of a corner store wants to paint the upper portion of a side exterior wall that is 16 ft tall to make the store more noticeable. The length of the wall is 25 ft. He buys a gallon of yellow 16 paint to do the job and is told that each gallon can cover 400 ft2. What portion of the wall will be yellow? 5. A circular swimming pool is being installed in a neighbor's yard. The county requires that the pool be 6 ft from the end of the property. If the yard is 30 ft by 26 ft and the pool is to be centered in the center of the yard, what is the largest area that can be used for the pool? 6. Melissa is planning to have the front of her property landscaped. Her front yard is rectangular and measures 12 m by 14 m. She plans to put down sod over the entire yard. except for a brick circular design in the middle that has a radius of lOft. To the nearest dollar, how much will Melissa need to spend to cover the entire yard if sod sells for $6 per square foot? 7. How high would the sides of a rectangular water tank have to be if the tank base has an area of 2 m 2 and the tank must hold 5520 L of water? 8. An ice-cream parlor owner charges $1.95 for an ice-cream cone. The cone is first packed in. and then a scoop of ice-cream sits like a ball on top. The top of the cone has a diameter of 2 in. and the cone is 6 in tall. The scoop also has a diameter of 2 in. How much profit is the owner making. if the ice-cream costs her $30 per gallon? 9. An airplane traveling from San Fransisco to Dallas makes a stopover in Phoenix. The average speed of the plane was 320 mph. If it took 2 hours to reach Phoenix and another 2 hours and 45 minutes to reach Dallas, how many miles has the plane flown? 10. On the first day of a cross-country automobile trip with her family, Jennifer covered a distance of 680 mi in 9 1 / 2 hours. The speed limit on the highway she drove is 65 mph. How much longer or shorter time would it have taken Jennifer if she drove at the speed limit? 1 I. The current school record for the 100-m dash is 9 seconds. At what speed, in miles per hour calculated to the nearest tenth. would a runner have to run in order to beat the speed record for the 100-m dash? 12. In a laboratory. it was found that a compound boils at a temperature of 42°C. The heating apparatus can raise the temperature by 2°F per minute. In how long a period can the compound be brought to boil if the heating process begins at 68°F? 13. While planning a winter vacation for his family, Frank investigates temperatures in different parts of the world. The information he finds on the Internet is given in degrees Celsius. He wants to travel to a place where the temperature would be between 75 and 85°F. 17 What range of temperatures should he enter into the search 14. A store is having a "half-price sale"; that is, if you buy one item at full price, you can buy a similar item for half-price. If the sales tax is 6 percent, how much would it cost to buy two pairs of jeans, each costing 15. An owner of a clothing store attempting to clear out her old inventory marks down the outfits she wishes to sell by 10 percent each week. Lindsey has $90 to spend on an outfit. She sees that it is currently marked at $135. How many weeks should she wait before she can afford the outfit, if sales tax is 8 3 /4 percend 16. Mr. Miller invested $15004 years ago in a fund and hasn't made any withdrawals. If the average annual rate of interest over the 4 years was 12 percent, how much money is currently in the 17. A certain credit card charges 18 percent annual interest on the unpaid monthly balance. Mike purchased a computer for $1300, which already includes sales tax, using this credit card. If he pays $200 every month, how long will it take Mike to payoff his debd What was his total cost for the 18. Gary found that his attic needed more insulation to prevent loss of heat. He has to be careful about how much weight he adds to the attic. Insulation comes in rolls that weigh 7 Ib 4 oz. Each roll will insulate 40 ft2. If his attic is rectangular and measures 15 ft by 18 ft, how many rolls of insulation will be needed and how much weight, to the nearest pound, will this insulation add to the attic? 19. Mrs. Alexander, who lives in New York, would like to purchase three pieces of furniture made in England that she saw in a catalog. The cost of shipping by air across the Atlantic Ocean is $365 per 100 lb. In addition, there is a 3 percent international tariff. If the three pieces weigh 16, 32, and 5 I kg, how much money would it cost to import these pieces of furniture from 20. European Airways allows each passenger to check two suitcases that measure 50 cm by 30 cm by 72 cm. The airline assumes that luggage weighs 75 kg per cubic meter. What is the maximum weight, in pounds, of luggage that the plane would be carrying if there are 200 passengers on Solutions to Additional Problems I. The total perimeter of the garden is 10 1 /2 ft + 10 1 /2 ft + 8 ft = 29 feet. In order to calculate the cost, Gayle needs to convert this to 18 meters. Using the conversion scheme of creating fractions. we have 29ft 12in 2.54cm 1m -- x -- x x = 8.84 m I I ft I in 100 cm The total cost is 8.84 m x $4.25/m = $37.57 8ft 2. The distance from the 8-ft side to the opposite vertex is the height of the triangle (draw this line in the diagram from the previous problem and label it 6 ft). The base is the I 01/2-ft side. Using the formula for the area of a triangle from the reference table. A = 112 X base x height. we have A = 0.5 x 10.5 x 6 = 31.5 ft2. 3. The perimeter of the frame is given as 50 in. (Note how the perimeter is disguised as the amount of wood used to make the frame.) If the longer side of the picture is II in and there is a 2-in border on either side. then the longer side of the frame has to be 15 in. The two longer sides of the frame. therefore. use 30 in of the wood and the formula for the perimeter of a rectangle. P = 2 x length + 2 x width. tells us that each of the shorter sides has to be half of the remaining 20 inches. The shorter side of the frame is lOin. 10 in 12 in 15 in 11 in 12 in 19 4. The gallon of paint would cover 400 ft'. Assuming that we would have a rectangular area painted yellow at the top of the wall, we could use the 25 feet to represent the base of the rectangle. The formula for the area of a rectangle, A = base x height, tells us that 400 = 25 x height or height = 400 -7 25 = 8 ft. This is exactly one-half of the wall. 25 ft 8 ft Yellow area 16ft - - - - - - - - - - - - - - ~ 5. Since the pool has to be 6 ft from the property line, the diameter of the pool can be no more than 26 - 12 = 14 ft. Therefore, the radius of the largest possible circular area is 7 ft. Using the formula for the area of a circle, A = IT X radius', we have A = 3.14 X 7' = 153.86 ft'. 6. To solve this problem, we need to calculate the area of the entire rectangular yard and subtract the area of the circle. Since the price of the sod is given in dollars per square foot, we need to convert the measurements of the yard to feet. These conversions are: 12m 100cm I in 1ft --x x x - = 39.4ft = 39ft I 1m 2.54cm 12in 14m 100cm I in 1ft --x x x - = 45.9ft = 46ft I 1m 2.54 em 12in Using the formula for the area of a rectangle, A = length x width, the area of the entire yard is 39 x 46 = 1794 ft'. Using the formula for the area of a circle, A = IT x radius', the circular design has an area of A = 3.14 X 10' = 314 ft'. Therefore, Melissa needs to buy 1794 - 314 = 1480 ft' of sod. This will cost her 1480 ft' x $6/ft' = $8880. 7. We know the volume of the tank to be 5520 L. Since we are being asked to find a height, we have to be working in measurements of length. We therefore need to convert liters into cubic meters in order to work with the same units. From the reference table, we 20 see that I m 3 is equal to I kiloliter (kL) or 1000 L. Therefore. the tank will hold 5520 + 1000 = 5.52 m 3 • We need to calculate the volume of the tank. The formula for the volume of a rectangular three-dimensional shape is V = area of the base x height. Using this. we have 5.52 m 3 = 2 m 2 x height (in meters). Therefore. the height is 5.52 + 2 = 2.76 m. Notice how. in this last equation. we have to have equal units on each side. m 3 = m 3 . This is a good way to be sure that you are applying formulas correctly. 8. We need to know exactly how much ice cream is used in the ice-cream cone. Since gallons is mentioned in the problem. we know that we will be calculating volume. Using the formulas for the volumes of a cone and of a sphere from the formula reference table. we have to identify the radius and the height of the cone and the radius of the scoop. In both cases. the diameter is 2 in; therefore. the radius of each is I in. We apply the formulas Volume of cone = 1/3 x 3.14 X 12 x 6 = 6.28 in 3 and Volume of scoop = 4/3 x 3.14 X 1 3 = 4.18 in 3 • and the total volume is 10.46 in 3 • However. we need to convert this to gallons since the cost is given in dollars per gallon. The reference table tells us that there are 277.42 in 3 in I gal. Therefore. the number of gallons is 10.46 + 277.42 = 0.0377 or 0.04 gal. The ice-cream parlor owner's cost in making the cone is 0.04 gal x $30/gallon = $1.20. The owner is making a profit of $1 .95 - $1 .20 = $0.75 on the cone. 9. The total time taken for the flight is 4 hours and 45 minutes. To use this in the formula for distance traveled. we have to convert this to a decimal number in hours. We can use (45 minutesll) x (I hourI 60 minutes) = 0.75 hour. Our time is. therefore. 4.75 hours. Applying the formula. we have Distance traveled = 320 mph x 4.75 hours = 1520 mi. 10. Using the speed limit of 65 mph in the formula for distance traveled. we have 680 mi = 65 mph x the number of hours taken. Therefore. the number of hours is 680 + 65 = 10.46. or approximately 10 1 / 2 hours. It would have taken Jennifer I hour more. If needed. you could also calculate her rate using the formula 680 = average speed x 9.5 or average speed = 680 +9.5 = 71. 57 or 72 mph. which means that she was traveling faster than the speed limit allowed. I I. The formula for distance traveled is applied to determine the record speed in meters per second: I 00 m = average speed x 9 21 seconds or speed = 100 -;- 9 = I I . I I m/second. This has to be converted to miles per hour, and we can use the "fraction" scheme for doing this: I I. I I m 60 seconds 60 minutes I km I mi ---x x x x--- I second I minute I hour 1000 m 1.61 km = 24.84 h mi our To the nearest tenth, a runner would have to run faster than 24.9 mph to beat the record. 12. To find the time taken, we need to convert the Celsius temperature to Fahrenheit. We can use the formula F = 9/5 X C + 32 to get F = 1.8 x 42 + 32 = I 07.6°F or 108°F (where F = degrees Fahrenheit and C = degrees Celsius, as in Example 12). Therefore, the temperature has to rise 108 - 68 = 40°F. At 2° per minute, it would take 20 minutes for the compound to boil. 13. By applying the formula C = 5/9 x (F - 32) to 75°F and to 85°F, we have a lower temperature of 5/9 x (75 - 32) = 23.9°C and a higher temperature of 5/9 x (85 - 32) = 29.4°C. 14. The discount has to be applied first. That is, one of the pairs of jeans will sell for the full price of $24.99 while the other will sell for one-half of that, or $12.50. Therefore, the 6 percent sales tax has to be applied to the sum of the prices, $37.49. The total cost of the purchase is found from the formula Total cost = price + tax x price, and we compute $37.49 + .06 x $37.49 = $39.74. 15. In this problem it is important to remember that the $90 that Lindsey has to spend must include the 8 3 /4 percent sales tax. We can find the different costs of the outfit as the weeks pass by making successive computations and keeping a list. Using the formula Discounted price = original price - discount x original price, the price after the first week would be $135 - $135 x O. 10 = $121 .50. After the second week, it would be $121.50 - $121.50 x 0.10 = $109.35. After the third week, the price would be 22 $109.35 - $109.35 x 0.10 = $98.42. After the fourth week, the price would be below $90; that is, $98.42 - $98.42x O. 10 = $88.58. We have to compute the price with tax to see if Lindsey's $90 would be enough. The total price would be $88.58 + 0.0875 x $88.58 = $96.33, which is more than she could spend. The price after the fifth week would be $88.58 - $88.58 x .10 = $79.72. The total price with tax would be $79.72+0.0875 x $79.72 = $86.70. Therefore, Lindsey would have to wait 5 weeks and hope that the outfit hasn't been sold. 16. Each year there is a new amount that is earning interest. (This is usually called compounding interest.) Using the formulas Interest earned = principal x rate of interest x time invested and Current amount = principal + interest earned, we can determine how much Mr. Miller has at the end of each year. At end of year Interest earned New amount 1 $1500xO.12=$180 $1500+ $180 = $1680 $1680+ $201.60 = $1881.60 $1881.60+ $225.79 = $2107.39 $2107.39 + $252.89 = $2360.28 2 $1680 x 0.12 = $201.60 3 $1881.60 x 0.12 = $225.79 4 $2107.39 x 0.12 = $252.89 A table like this is a good tool for keeping track of calculations and gives you a way to look back on your work in an organized fashion. 17. This problem is similar to one asking for a total amount earned with interest. That is, a new amount is owed each month after adding on interest on the unpaid balance and $200 deduction. A table is useful to follow the month-by-month situation. Since the interest is added monthly, we need to use 18 percent -;- 12 = 1.5 percent to calculate the added interest each month. Before beginning the calculations, an easy estimate should be made. Since $1300 -;- $200 = 6.5, we would expect that with interest adding to the amount, it should take 7 or perhaps 8 months to payoff the debt. At end of month Interest added Total debt New balance $1300 x O.oJ5 = $19 50 $1300 + $19 50 = $1319 50 $1319.50 - $200 = $1119 50 2 $111950 x 0015 = $16 79 $111950 + $1679 = $1136 29 $1136 29 - $200 = $936 29 3 $93629 x 0015 = $1404 $93629 + $14 04 = $950 33 $95033 - $200 = $75033 4 $750.33 x 0015 = $1125 $75033+$1125 = $76158 $761 58 - $200 = $561 58 5 $56158 x 0015 = $8 42 $561.58 + $8 42 = $57000 $57000 - $200 = $370.00 6 $370.00 x 0015 = $5 55 $37000 + $5 55 = $375 55 $375.55 - $200 = $175 55 7 $17555 x 0015 = $263 $17555 + $2 63 = $178.18 Pay $178 18 to complete debt 23 We see that our estimate was a good one. The total cost of the computer is found by adding the interest amounts to the original price of $1300; that is. $1300 + $7S. IS = $137S. IS. 18. First we need to determine how many rolls of insulation are needed. The area of the attic is 15 ft x IS ft = 270 ft2. Therefore. we need 270 -:- 40 = 6.75 or 7 rolls. For the second part of the question. we can use the exact number 6.75 since we do not have to use the remaining portion of the roll. First. we need to convert 7 Ib 4 oz into a decimal number in pounds. Since there are 16 oz in one pound. 4 oz is 1/4 or 0.25 lb. Therefore. each roll weighs 7.25 lb. The weight is therefore 6.75 rolls x7.25 Ib/roll = 4S.9375 or 49 lb. 19. The answer is found by breaking the problem down into smaller parts. 20. 24 Step I Find the total weight in kilograms and convert it to pounds. The total weight is 16 + 32 + 51 = 99 kg. Converting to pounds. we have 99 kg x 2.21b/kg = 217.Slb. Step 2 Find the cost per pound. The shipping rate is given with respect to 100 lb. Dividing by 100 gives the cost per pound. $3.65 per pound. Step 3 Find the shipping charge. The formula from the reference table Step 4 Step 5 Step I Step 2 Step 3 Step 4 Step 5 Total dollar amount of item = quantity of item x dollar amount of one item is applied to get 217.Slb x $3.65/1b = $794.97. Compute the tax to add. $794.97 x 0.03 = $23.85. The total charge is $794.97 + $23.S5 = $8IS.S2. The dimensions of the suitcase should be converted to meters in order to use the weight rate. Dividing each measurement by 100 gives us 0.5. 0.3. and 0.72 m. The suitcase has a volume of 0.5 m x 0.3 m x 0.72 m = O.IOS m 3 • The weight of one suitcase would be .IOS m 3 x 75 kg/m 3 = S. I kg. There could be up to 200 x 2 = 400 suitcases on the plane for a total weight of 400 x S. I = 3240 kg. Multiplying by 2.2 would give us the number of pounds; That is. 3240 kg x2.2lb/kg = 712Slb. Table of Common Measurements and Conversions Length or Distance in Linear Units Straight-line distance between 2 points: 1 Unit English system 1 inch (in) Basic unit 1 foot (ft) 12 in 1 yard (yd) 36 in 1 mile (mi) 5280 ft Metric system 1 millimeter (mm) 1/1000 m 1 centimeter (cm) 1/100 m 1 decimeter (dm) l/to m 1 meter (m) Basic unit 1 kilometer (km) 1000 m Conversions between systems 1 in 2.54 cm 39 in 1 m 1 mi 1.61 km Area or Space in Square Units 1 unit S"", ocmp'" by • ",""'m,""=" 'q.= 1 ,,;, D 1 "l""'" ,,' English system 1 square inch (in 2 ) Basic unit 1 square foot (ft2) 144 in 2 1 square yard (yd 2 ) 9 ft 2 1 acre 43,560 ft 2 Metric system 1 square meter (m 2 ) Basic unit 1 hectare (ha) to,OOO m 2 Conversions between systems to.8 ft2 1 m 2 2.47 acres 1 ha Volume or Capacity in Cubic Units Space within a three-dimensional cube: @ 1 cubic unit 1 unit English system 1 ounce (oz)* Basic unit 1 cup 40z" 1 pint (pt) 2 cups = 8 oz 1 quart (qt) 2 pt = 32 oz 1 gallon (gal) 4 qt or 277.42 in 3 bnperud (UK) ~ I l U o n / The US hqmd. ~ I l U o n I' 231 cubic mche, English system 1 ounce (oz)t Basic unit 1 pound (lb) 160z t 1 ton 2000lb "These are referred to as fluid ounces. tThese are referred to as dry ounces. 1 unit Metric system 1 millilter (mL) 1 cm 3 or cc 1 unit Conversions between systems 1.06 quarts 1 liter 1 liter (L) 1000 mL or 1 dm 3 1 kiloliter (kL) 1000 L or 1 m 3 Weight or Mass Metric system 1 gram (g) Basic unit 1 kilogram (kg)t 1000 g 1 metric ton (t) 1000 kg Conversions between systems 2.21b 1 kg lOne kilogram is equal to the weight of one liter of water. 25 Table of Common Formulas Seen in Word Problems Measuring Length Perimeter of a rectangle = 2 x length + 2 x width or 2 x base + 2 x height: Ii Width or Length or base Circumference of a circle = 2 x 7r x radius (7r is approximately 3.14) Distance traveled = average rate x time traveled Measuring Area Area of a rectangle = length x width or base x height "<e. of ,Ido', D .. do A<e. of. p",U,log<.m = 00 .. x lrelgh" J / Base Area of a triangle = liz x base x height: Base Area of .n ",ull",," trl.ngio=,/3/ 4 x ,Id" , D Side Upper base Area of a trapezoid = average of the bases /l \ x height or l/Z x (lower base + upper base) x height: /l Height Lower base Area of a circle=7r x radius 2 (7r is approximately 3.14): Continue 26 Surface area of a three-dimensional sphere = 4 x Jr x radius 2 : Radius Measuring Volume Volume of a rectangular solid = length x width x height or area of the base x height: Height Base Length Volume of a cube = side 3 : Side Volume of a sphere = 4/3 x radius 3 Radius Volume of a circular cylinder = Jr X radius 2 x height: Height Volume of a circular cone = 113 x Jr x radius 2 x height: Money-Related Formulas Total dollar amount of item = quantity of item x dollar amount of one item Total cost = price + tax rate x price Discounted price = original price - discount x original price Interest earned = principal x rate of interest x time invested Current amount = principal + interest earned Width 27 Chapter 2 Using Algebraic Equations to Solve Problems In the previous chapter, you studied problems that were solved by making basic arithmetic computations. In this chapter, you will learn how to apply algebra to solve problems. In most cases, this will require modeling the situation with an equation in which what you wish to find will be expressed by a letter representing an unknown quantity. We call this the variable of the equation. The equations will come from applying ba- sic formulas and well-known concepts. (You can review basic tools of algebra by visiting the Appendix in the back of the book.) It is important to carefully identify the unknown variable in the problem and determine the relationship between it and the other information in the problem. It is also important that your answer really does solve the problem! There are many places to make mistakes in these problems, such as creating the wrong equation, solving the equation incorrectly, or making an arithmetic mistake. Problems Involving Numbers The simplest of all word problems are those that involve find- ing numbers. Most of these problems will tell you how two or more numbers are related. The key will be to identify all the relationships between the numbers that are mentioned in the problem. Example I The larger of two numbers is equal to 14 more than 3 times the smaller number. Find the smaller number if their difference is 38. Solution I The first step of any word problem is to find a way to use a variable to represent the unknowns in the problem. Since the problem specifically tells us how the larger and smaller num- bers are related, we can use one variable. It is also a good idea to select a letter for the variable that reminds us of what we're looking for. The letter n is always a good choice for a number. The first sentence of this problem tells us that if we know the smaller number, we can compute the larger one. It makes sense, therefore, to use the variable to represent the smaller one. We should also start our work with a statement of how we are assigning variables. This is often called the "Let" statement. Let the smaller number = n We are now ready to start translating the first sentence into an algebraic equation. A good idea is to first rewrite the problem leaving most of the words and using mathematical symbols for adding, subtracting, multiplying, dividing, and equality when it is easily recognized. This would give us Larger number = 14 more than 3 x smaller number Now we have to take a look at the phrase on the right. We must realize that to have more than something means to add to that something. Larger number = 3 x smaller number + 14 We can now substitute our variable expressions for the smaller and larger parts. Remember to drop the multiplication symbol when multiplying by a variable. Larger number = 3n + 14 30 Both numbers in the problem are now represented by variables, and we are ready to form the equation that we will use to solve the problem. This will come from the other rela- tionship given in the problem, "their difference is 38." Again we should first translate the operation symbols and write the remaining words. We should realize that finding a difference means to subtract, and if the difference is positive, we are tak- ing the smaller from the larger. We get Larger number - smaller number = 38 Substituting our variables gives us 3n+ 14 - n = 38 Simplifying and solving for n, we get 2n+ 14 = 38 2n= 24 n= 12 Therefore, the smaller number is 12, and the larger number is 3 x 12 + 14 = 36 + 14 = SO. To check, we need only verify that their difference is 38. Clearly, SO - 12 = 38"( Sometimes one of the relationships regarding the num- bers is given indirectly; that is, it might be hidden in a key phrase. Example 2 Separate 70 into two parts so that 5 times the smaller part is equal to 14 more than twice the larger part. Solution :1 The problem is really asking for two different numbers whose sum is 70. We can assign a variable, n, to the smaller part. The larger part must be what is left when n is taken away from 70, 70 - n. The "Let" statement is Let n = smaller part 70 - n = larger part 31 Translating only the operations gives us 5 x smaller part = 14 more than twice the larger part Now we have to take a look at the phrases on the right: specif- ically, "more than" and "twice." We must realize that to have more than something means to add to that something. 5 x smaller part = twice the larger part + 14 Now we use the fact that to have twice something means to multiply that something by 2. 5 x smaller part = 2 x larger part + 14 Now we are ready to substitute our variable expressions for the smaller and larger parts. Remember to drop the multi- plication symbol and to use parentheses when we need to mul- tiply by an expression with more than one term. 5n = 2(70 - n) + 14 Solving in the usual way, we get 5n = 140 - 2n + 14 Add 2n to both sides to get 7n = 154 n=22 The smaller part is 22, and the larger part is 70 - 22 = 48. The check is easy. Simply follow the numbers and words of the problem, 5 times the smaller is 5 x 22 = 110; 14 more than twice the larger is 2 x 48 + 14 = 96 + 14 = 110"'! Alternative Solution 2 Another way to solve the problem is to use two different vari- ables to represent both the parts of 70. When we use two 32 variables, we need two equations. (See the Appendix for a re- view of how to work with equations with two variables.) Let x = the smaller number Let y = the larger number The first equation comes from the fact that the sum of the numbers is 70. x+ Y= 70 The second equation comes from the way these parts are re- lated. The translation should still be performed as outlined above, but now we use the variables x and y to substitute for the smaller part and for the larger part. Sx = 2y+ 14 We work with this pair by solving the first equation for y y= 70-x and substituting in the second equation Sx = 2(70 - x) + 14 Note that we have the exact same equation to solve (even though the variable here is x instead of n) and we should arrive at the solution, x = 22 and y = 48. The next example shows you another way to express one of the relationships indirectly. Example 3 In this example,S less than 6 times the largest of three con- secutive positive integers is equal to the square of the smaller minus 2 times the middle integer. Find all three integers. Solution :I It is clear that we are dealing with three different num- bers. The key phrase in this problem is consecutive positive inte- gers. To understand this, let's look at each word. The numbers 33 have to be integers. This is the set of negative and positive whole numbers and zero { ... , -3, -2, -1,0,1,2,3, ... }; that is, we won't accept any answers that are fractions, decimals, or roots. Since the numbers are positive, we are looking only for numbers from the set {I, 2, 3 ... }. The most important word in the phrase is consecutive. This means that the three numbers are all one apart. We can then represent our numbers with the same variable, n. Let n = the smallest integer n + 1 = the middle integer n + 2 = the largest integer N ow we can use our translation scheme. First only the symbols 5 less than 6 x largest = square of the smallest - 2 times the middle Less than something means to take away from that something 6 x largest - 5 = square of the smallest - 2 x middle Square of means to raise the number to the second power 6 x largest - 5 = smallest 2 - 2 x middle Now we can substitute our variables and remove the multipli- cation symbols. 6(n + 2) - 5 = n 2 - 2(n + 1) This is a quadratic equation since we have an equation with a power of 2. (See the Appendix for a review of solv- ing quadratic equations.) It must be simplified and have ° on one side in order to solve it by factoring. 6n + 12 - 5 = n 2 - 2n - 2 34 To both sides of the equation, we subtract n 2 , add 2n, and add 2, to arrive at We can now multiply every term in the equation by -1 to get n 2 - 8n- 9 = 0 We can now factor and solve for n. (n - 9)(n + 1) = 0 Therefore n=9 or n= -1 Remember, we are looking for positive integers, so we can reject -1 as an answer. Since n was the smallest, the three integers are 9, 10, and II. To check, we simply use the numbers and follow the words of the problem: 5 less than 6 times the largest is 6 x 11 - 5 = 66 - 5 = 6I. The square of the smallest minus 2 times the middle is 9 2 - 2 x 10 = 81 - 20 = 61..,1 (Not all consecutive integer problems will give rise to quadratic equations as you will see in the supplementary exercises.) Problems Involving Age Problems that involve the ages of two or more people are very similar to number problems. (After all, an age is a number.) Since we need to have two relationships, these problems may cause you to consider ages at the present time and at another time in the past or future. Example 4 Sandy is 4 times as old as Robby is now. In 10 years, Sandy will be 6 years older than Robby. How old are they now? 35 Solution 4 Since we are told how to find Sandy's age if we know Robby's age, we should use a variable for Robby's age and de- termine the symbolic representation of Sandy's age with the same variable. Following the procedure outlined in the previ- ous problems, we have Let Robby's age = x Sandy's age = 4 x Robby's age Sandy's age = 4x "In 10 years" means that we will add 10 to each of their ages. We ought to state what their ages will be. In 10 years, Robby's age will be x + 10 and Sandy's age will be 4x + 10 We should now translate the second sentence using these fu- ture ages and substitute the future variables Sandy's age = Robby's age + 6 4x + 10 = x + 10 + 6 Simplifying and solving for x, we get 4x + 10 = x + 16 3x = 6 x=2 Therefore, Robby's current age is 2 and Sandy's current age is 8. To check, we should simply follow the words of the prob- lem. Sandy's current age is 4 times Robby's age or 8 = 4 x 2 . ../ In 10 years, Sandy will be 18 and Robby will be 12. Clearly, Sandy will be 6 years older than Robby . ../ Example 5 Jessica is 27 years old and Melissa is 21 years old. How many years ago was Jessica twice as old as Melissa? 36 Solution S In this problem, the variable is the number of years in the past, and we have to subtract this amount from their current ages. Let x = the number of years in the past Therefore, x years ago, jessica was 27 - x years old and Melissa was 21 - x years old. The relationship x years ago was jessica's age = 2 x Melissa's age Substituting the past ages, we get the equation for the problem. 27 - x = 2(21 - x) Simplifying and solving for x, we get 27 - x = 42 - 2x x = 15 years ago To check, we have to compute the girls' ages 15 years ago. jessica was 27 - 15 = 12 years old and Melissa was 21 - 15 = 6 years old. Clearly, jessica was twice as old as Melissa"'! In Chap. 1 we solved problems that make use of many of the basic formulas that arise in mathematics. Those problems merely required that we identify the appropriate formula, sub- stitute known values, and solve for the missing value. Now we will examine situations where those formulas are used as the basis for the equations needed to solve more complex prob- lems. In each problem, we will see that the key to the problem is finding out which of the quantities mentioned are meant to be equal or which quantities are added to get a total amount. Problems Involving Motion In problems that involve motion, there are usually one or two objects moving along a straight line at a constant rate (speed) for a period of time. The two objects can be moving toward each other, moving away from each other, or moving in the 37 same direction as in a race. In any situation, there is a total distance traveled by the objects. The governing relationship for these problems is Distance = rate x time Example 6 Two cars are traveling on the same highway toward each other from towns that are 150 mi apart. One car is traveling at 60 mph, while the other is traveling at SO mph. If they started at the same time, how much time, to the nearest minute, will have passed at the moment when they pass each other? Solution 6 In order to use the formula, we have to decide what is given and what we have to find by examining each of the parts. Drawing a picture can help. The car traveling at 60 mph will cover more of the distance than the car traveling at SO mph. Fast car Slow car -------------------------------x---------------- ~ - - - - - - - - - - - - - - - - - 1 5 0 m i - - - - - - - - - - - - - - - - - - ~ The picture shows that the total distance traveled is a known amount and that it must be equal to the distances traveled by each car. This is the basis of the model for this problem. Total distance = distance traveled by faster car + distance traveled by slower car We know the rates of each car, and the unknown is the time. Since the cars started at the same time and pass each other at the same time, the time they've both traveled is the same. Therefore, we can use one variable for this time, t. As before, it is a good idea to state the use of the variable before any work is begun. Let t = time traveled by both cars 38 Using the relationship Distance = rate x time for each car, our model becomes 150 = 60t + SOt 150 = 110t 150 t= 110 t = 1.363636 ... Remember that this answer must be in hours since the rates were given in miles per hour. The problem asks for the answer to be given to the nearest minute. The decimal part of the hour can be converted to minutes by multiplying .37 x 60 = 22.2 minutes and, when rounded, the answer is 1 hour and 22 minutes. To check the answer, we need to compute the distances traveled by each car. To be accurate, we will use the actual fraction 15/11 for the time. The faster car traveled 15/11 x 60 = 900/ 11 , and the slower car traveled 15/ 11 x SO = 750/ 11 • By adding these fractions, we find that together they traveled 1650/ 11 = 150 mLy" Example 7 Allison and Rachel were competing in a race. Allison had a 10- minute head start and was running at 3 km per hour. Rachel ran at 5 km per hour. In how many minutes did Rachel catch up with Allison? Solution 7 In this problem, the key factor is that at the moment that Rachel catches up to Allison, they will have traveled the same distance. The model for the problem is Distance traveled by Allison = distance traveled by Rachel Since we are asked to find Rachel's time, we will represent it by t. Note that the units of t must be hours, since the rates are given in hours. Allison ran 10 minutes more, which will be represented by t + 1/6, since 10 minutes is one-sixth of an hour. Let t = Rachel's time t + 1/6 = Allison's time Using Distance = rate x time, we have 3 (t+ 1/6) = 5t 3t+ 1/2 = 5t 1/2 = 2t or t = 15 minutes To check, we compute the distance traveled by each girl us- ing t = 1/4 for Rachel and t = 1/4 + 1/6 = 5/12 for Allison. Rachel's distance was 5 x 1/4 = 1.25 km, and Allison traveled 3 x 5/ 12 = 1.25 km as well. Note that we could have converted the rates to kilometers per minute by dividing each rate by 60. This would allow t to represent minutes, and we could use 10 minutes in the equation instead of 1/6 hour. (:0) (t+ 10) = (:0) t 3t 30 5t 60 + 60 = 60 3t+ 30 = 5t 30 = 2t 15 = t The relationship used in these problems can be rewritten to express the rate as a quotient: 40 R distance traveled ate = ------- time This form of the relationship may be necessary to use in prob- lems when the rates aren't explicitly given. Problems Involving Work There are other situations which follow the same principle using a rate. For example, when someone works at a job, we can use Number of jobs completed = rate of work x time or, equivalently Example 8 R number of jobs completed ate = ---------'--- time If Joe can paint five rooms in 8 hours and Samantha can paint three rooms in 6 hours, how long will it take them to paint seven rooms working together? Solution 8 Using the rate formula, we see that Joe's rate of work is 5/S rooms per hour and Samantha's rate of work is I/Z room per hour. This problem is actually similar to Example 6, in that they both work for the same amount of time and together they complete the given total amount. Let t = the time to complete the job together The model is Total number of rooms painted = rooms painted by Joe + rooms painted by Samantha 7 = 5/st + 1M = 5/S t + 4/st 7 = 9/st 56 = 9t 41 Example 9 It takes Suzanne 10 hours longer than Laurie to take their store's inventory. If it takes 3 3 / 4 hours for them to do the job together, how long would it take Suzanne to do it alone? Solution V Since Laurie takes less time to complete the job, we will let her time be represented by t. If we can find this time, then the answer to the problem will be t + 10. The number of jobs considered in this problem is 1. There- fore, Laurie's rate is l/t and Suzanne's rate is l/(t + 10). The model for the problem is similar to Example 8, where the to- tal number of jobs completed is the sum of the amount of work done by each girl in the 2 hours. 1 job = amount done by Suzanne + amount done by Allison Using 15/ 4 for 3 3 / 4 , the amount done by Suzanne is l/t+ 10 x 15/ 4 , and the amount done by Laurie is Ift x 15/ 4 , Substituting into the preceding model, we have 1 __ 1_ x 15 ~ x 15 _ 15 (_1_ ~ ) - t+ 10 4 + t 4 - 4 t+ 10 + t 1 1 1 -=--+- 15/ 4 t + 10 t (Note that this line tells us that the combined rate is equal to the sum of the individual rates. If you remember this fact, you can use it as a model when the problem involves only completing 1 job.) Simplify the fraction on the left to get the fractional equation 411 15 = t+ 10 + t Multiplying both sides of the equation by the least common 42 denominator,15t(t + 10), we get 4t(t + 10) = 15t + 15(t + 10) Simplifying, we get a quadratic equation Subtracting 30t and 150 from both sides gives us Dividing all terms by 2 simplifies the equation to We can now factor and solve 4t 2 + 40t = 30t + 150 4t 2 + lOt - 150 = 0 (2t + 15)(t - 5) = 0 2t+ 15 =Oort - 5 =0 t=-7.5ort=5 Since the problem requires physical time, we reject the negative answer and we see that t = 5 is Laurie's time. There- fore, Suzanne can complete the job in 15 hours working alone. Problems Involving Mixing Quantities A situation similar to those in the problems above is finding out what occurs when we mix different types of objects. In these problems there is also a rate involved, but it is not a rate involving time, as we will see. Example 10 At the chocolate factory, John's job is to experiment with mix- ing different amounts of milk and chocolate to find different flavors. If he mixes 2 quarts (qt) of type A, which contains 15 percent chocolate, with 1 qt of type B, which contains 10 per- cent chocolate, he creates type C. What percent of type C is chocolate? Solution 10 In this situation, the rate is the percentage of chocolate. For each type, we have the relationship Amount of chocolate in the type = percent of chocolate x amount of the type 43 where the percentage is taking the form of the rate as we saw in previous problems. In type A, we have 0.15 x 2 = 0.3 qt of chocolate In type B, we have 0.10 x 1 = 0.1 qt of chocolate Together, in the mixture type C, we have 0.4 qt of choco- late. Since the mixture has exactly 3 qt of liquid, we calculate the percentage that is chocolate. To do this, we rewrite the relationship (amount of chocolate)/(amount of type) = per- centage of chocolate. Since there are 0.1 + 0.3 = 0.4 qt, the answer is easily found to be 0.4/3 = 0.1333 ... or 13.3%. Another type of mixture problem involves the price of a mixture of two items where each sells for different individual prices. The underlying formula is also a kind of rate relation- ship since price is usually given as dollars per pound, cents per pound, or dollars per kilogram. For instance Cost . amount =pnce or cost = price x amount Example II Nancy's Nut Stand sells cashews for $5/lb and honey-roasted peanuts for $2/lb pound. How many pounds of each type of nut should Nancy use to make 8 lb of a mixture that will sell for $3/lb? Solution II The problem asks us to find the amount of both kinds of nuts. We can solve the problem using only one variable since we know that the total amount must be 8 lb. Let p = number of pounds of cashews 8 - P = number of pounds of honey-roasted peanuts The model for the problem is that the total cost for the mixture should be the same as the sum of the costs for each individual kind of nut: Cost of mixture = cost of cashews + cost of honey-roasted peanuts 44 and using the relationship Cost = price x amount, we have (3)(8) = Sp + 2(8 - p) 24 = 5 P + 16 - 2 P 8= 3p P = 8/3 or 22/3 The mixture should have 22/3 lb of cashews and 8 - 22/3 = 5 1 / 3 lb of honey-roasted peanuts. Alternative Solution 11 Another way to solve the problem is to use two variables and get two equations to work with. (See the Appendix for a review of how to work with equations with two variables.) Letx = amount of cashews in mixture Let y = amount of honey-roasted nuts in mixture The first equation is the total amount of the mixture x+y=8 The second equation comes from the cost relationship Cost = price x amount, and the model Cost of cashews + cost of honey-roasted peanuts = cost of mixture. Sx + 2y = (3)(8) = 24 The two equations together form a simultaneous system of equations and can be solved easily by the following process. Multiply the first equation by 2 and subtract it from the first Sx + 2y = 24 2x + 2y = 16 3x =8 x = 8/3 = 22/3 45 Use this value of x in the first equation and solve for y: 22/3 + y = 8 Y = 51/3 It's okay if you are not familiar with working with two equations. As we saw, all of the problems can be solved with one variable and one equation. However, you will be a better problem solver by having several ways to solve a problem in your arsenal. Problems Involving Money One other kind of problem that is easily solved by an alge- braic equation involves combinations of money of different denominations. These are problems similar to the previous ones because the amount of money that a coin is worth is a sort of rate. For example, a quarter is worth 25 cents per coin and a dime is worth 10 cents per coin. The amount of money we have is found by the relationship Amount of money = value of coin x number of coins Here, the value of the coin is taking the form of the rate. Example 12 Joey has $2.35 in nickels, dimes, and quarters. If he has two less quarters than dimes and one more nickel than dimes, how much of each coin does he have? Solution 12 The key to the problem is the way in which the numbers of quarters and nickels are related to the number of dimes. We can find the amounts of quarters and nickels if we know the amount of dimes. Therefore, we will assign the variable to this amount. Let x = number of dimes 46 Since the number of quarters is two less than this amount, we have x - 2 = number of quarters Since the number of nickels is one more than the number of dimes, we have x + 1 = number of nickels The governing relationship is Total amount of money = amount in quarters + amount in dimes + amount in nickels Using the Amount = rate of coin x number of coins, we have 2.35 = .25(x - 2) + O.lOx + 0.05(x + 1) Since we are given the total amount in dollars, we need to use decimals to express the rate of each coin as a decimal fraction of a dollar. If we multiply every term by 100, we can clear the decimals 235 = 25(x - 2) + lOx + 5(x + 1) Simplifying and solving for x, we have 235 = 25x - 50 + lOx + 5x + 5 235 = 40x - 45 280 = 40x 7=x Therefore, there are seven dimes, five quarters, and eight nick- els. This is easily checked: 70¢ + $1.25 + 40¢ is indeed $2.35.1' 47 Summarizing what we have done with all our word prob- lems, we saw that we need to follow some basic steps: Step 1 Read the problem carefully and identify the unknowns. Step 2 Identify the formula or key concept that is needed in the problem and rewrite the problem, keeping most of the words but using basic arithmetic symbols for operations and equality. Step 3 Assign a variable to one of the unknowns with a "Let" statement and determine the symbolic representations of all the other knowns with this variable using the relationships in the problem. Step 4 Substitute the representations of the unknowns into the mathematical statement you created in step 2. Step 5 Use algebra to solve the equation(s) and determine the value of the unknowns you represented by a variable. Step 6 Check your answer to see if it really does solve the prob- lem and satisfy all the necessary conditions. It is worth remembering that in many problems, there are at least two relationships among the unknowns. If only two unknown quantities are asked for, the mathematical model of the problem is sometimes more easily seen when using two variables and creating a system of two equatiOns to solve. Additional Problems I. The sum of two numbers is 62. The larger number is as much greater than 34 as the smaller number is greater than 10. What are the two numbersl 2. Separate the number 17 into two parts so that the sum of their squares is equal to the square of one more than the larger part. 48 3. The difference between two numbers is 12. If 2 is added to 7 times the smaller, the result is the same as when 2 is subtracted from 3 times the larger. Find the numbers. 4. The sum of three consecutive integers is I I greater than twice the largest of the three integers. What are theyl 5. Find three consecutive odd integers whose sum is 75. 6. Find four consecutive even integers such that the sum of the squares of the two largest numbers is 34 more than the sum of the other two. 7. A father is now 5 times as old as his son. In 15 years he will be only twice as old as his son. How old are they nowl 8. Sarah's mom is 42 years old, and her aunt is 48 years old. How many years ago was Sarah's aunt 3 times as old as her moml 9. A cyclist traveling 20 mph leaves 'h hour ahead of a car traveling at 50 mph. If they go in the same direction, how long will it take for the car to pass the cyclisd 10. In the early days of building railroads in the United States, the first transcontinental railroad was started from two points at the ends of the 2000-mi distance over which tracks had to be laid. Two teams at different ends of the country started at the same time. The team in the east moved forward at 2 mi per day over mostly flat land, and the team from the west moved at 1.5 mi per day over mountainous terrain. How long did it take for the teams to meet and the last spike to be nailed in, and how many miles of track did each team layl 1 I. On an international flight across the Atlantic Ocean, a certain type of plane travels at a speed that is 220 km/hour more than twice the speed of another. The faster plane can fly 7000 km in the same time that the slower plane can fly 2700 km. At which speeds do these planes fly in miles per hourl 12. Mr. Begly's office has an old copying machine and a new one. The older machine takes 6 hours to make all copies of the financial report, while the new machine takes only 4 hours. If both machines are used, how long will it take for the job to be completedl 13. The candystand at the multiplex has 100 Ib of a popular candy selling at $1.20/lb. The manager notices a different candy worth $2.00/lb that isn't selling well. He decides to form a mixture of both types of candy to help clear his inventory of the more expensive type. How many pounds of the expensive candy should he mix with the 100 Ib in order to produce a mixture worth $1.75/lbl 14. At her specialty coffee store, Melissa has two kinds of coffee she'd like to mix to make a blend. One sells for $1.60/lb and the other, $2.10/lb. How many pounds of each kind must she use to make 49 75 Ib of mixed coffee that she can sell for $1.90/Ib and make the same profit? 15. joey has $5.05 in 34 coins on his dresser. If the coins are quarters and dimes, how many of each kind are there? 16. The triplets john, jen, and joey gave their father all their coins toward a present for their mother. john had only quarters to give. jen had only dimes and gave 4 times as many coins as john gave. joey had only nickels and gave seven more than twice the number of coins that jen gave. If the total amount of the contribution was $22.40, how much did each child contribute? 17. Mrs. Stone likes to diversify her investments. She invested part of $5000 in one fund that had a return of 9% interest and the rest into a different fund that earned I I %. If her total annual income from these investments was $487, how much does she invest at each rate? 18. Three people each received part of $139 so that the first received twice what the second received and the third's portion exceeded the sum of the other two by one dollar. How much money did each person receive? 19. A farmhand was hired for a 60-day period and would live on the ranch. For each day that he worked he would receive $45, but for each day that he did not work he would pay $10 for his room and board. At the end of the 60 days, he received $2260. How many days did he work? Solutions to Additional Problems I. Using one variable, we have the following. Let x = the smaller number; therefore, 62 - x is the larger number. In Prob. I, the expression "as much greater than" indicates that we want the difference between the numbers. 50 The larger number - 34 = the smaller number - 10 (62 - x) - 34 = x - 10 28 - x = x - 10 38 = 2x x = 19 The smaller number is 19, and the larger number is 62 - 19 = 43. Check: 43 - 34 = 9 and 19 - 10 = 9 . ../ An alternative solution using two variables is as follows. Let x = smaller number and let y = larger number. The equations would be x + y = 62 and y - 34 = x - 10. The second equation could be rewritten as -x + y = 24. Adding the first equation to this new form of the second equation gives us 2y = 86 and y = 43. Substituting into the first equation, we have x + 43 = 62 and x = 19. 2. Both "parts" of the number 17 can be represented using one variable. We should use the single variable to represent the larger part, since another piece of the puzzle relates to it. Let x = the larger part and 17 - x = smaller part and x + I = one more than the larger part. Recognizing the sum of their squares to mean larger part 2 + smaller part 2 , we have the quadratic equation Squaring the binomials, we have x 2 + 289 - 34x + x 2 = x 2 + 2x + I Combining terms gives us 2X2 - 34x + 289 = x 2 + 2x + I Subtracting terms on the right from both sides of the equation produces x 2 - 36x + 288 = 0 We can factor this and solve (x - 12)(x - 24) = 0 x - 12 = 0 or x - 24 = 0 x = 12 or x = 24 Since the larger part of 17 must be less than 17, we reject the answer of 24. The larger part is 12 and the smaller part is 5. Check: 12+5= 17. 122+5 2 = 144+25= 169 and (12 + If = 13 2 = 169.../ 51 3. The problem is best understood and modeled when two variables are used for larger and smaller numbers. The model is Larger number - smaller number = 12 7 x smaller + 2 = 3 x larger - 2 Let x = larger number and let y = smaller number. The two equations are x - y = 12 and 7y + 2 = 3x - 2. We can rewrite the second equation as -3x + 7y = -4, and we will be able to eliminate one of the variables by adding equations if we multiply every term in the first equation by 7: 7x - 7y = 84. Adding the equations gives us 4x = 80 and x = 20. Using the original first equation to find y, we have 20 - y = 12 and y = 8. The numbers are 20 and 8. Check: The difference of the numbers is 20 - 8 = 12. The other relationships are 7 x 8 + 2 = 56 + 2 = 58 3 x 20 - 2 = 60 - 2 = 58\""" 4. Consecutive integers imply that the three numbers are one apart. S2 Therefore The model is Let x = first (smallest) number x + I = second (middle) number x + 2 = third (largest) number First number + second number + third number = 2 x third number + I I (Be sure not to confuse the phrase "is I I greater than" with" II is greater than," which translates into an inequality" I I > "!) Then x + (x + I) + (x + 2) = 2(x + 2) + I I. This linear equation can be solved in the usual way. (At this point, see if you can identify what steps are being taken.) 3x + 3 = 2x + 4 + I I 3x + 3 = 2x + 15 x = 12 The numbers are 12, 13, and 14. Check: 12, 13, and 14 are consecutive integers. 12 + 13 + 14 = 39 and 2 x 14 + II = 28 + I I = 39.1' 5. By definition, an odd number is an integer (fractions are not considered to be odd or even). To be consecutive odd numbers the numbers must differ by 2. Therefore Let x = first (smallest) odd number x + 2 = second (middle) odd number x + 4 = third (largest) odd number The model is simply First odd number + second odd number + third odd number = 75 The equation is therefore x + (x + 2) + (x + 4) = 75 3x + 6 = 75 3x = 69 x = 23 The three consecutive odd numbers are 23, 25, and 27. Check: 23 + 25 + 27 = 75.1' 6. Consecutive even integers must be 2 apart. Therefore. Let x = first (smallest) even integer x + 2 = second even integer x + 4 = third even integer x + 6 = fourth (largest) even integer The model for the problem is (Third integer)2 + (fourth integer)2 = first integer + second integer + 36 53 The equation is therefore (x + 4)2 + (x + 6)2 = X + (x + 2) + 34 x 2 + ax + 16 + x 2 + 12x + 36 = 2x + 36 2x2 + 20x + 52 = 2x + 36 2x2 + lax + 16 = 0 x 2 + 9x + a = 0 (x + I)(x + a) = 0 x + I = 0 or x + a = 0 x = -lor x =-a Since we want even integers, we reject x = - I and our smallest even integer is -a. The four even integers are -a, -6, -4, and -2. Check: (_4)2 + (_2)2 = 16 + 4 = 20 and -a + -6+ 34= 20 . .( Don't be disturbed by having negative numbers as answers. There is no reason to assume that all number problems involve only positive integers! 7. The two relationships are clearly stated in the problem. One relationship can serve as the basis for assigning a variable. Since the father's age is given in terms of the son's age S4 Let x = son's age now 5x = father's age now The second relationship provides us with the model Father's age now + 15 = 2 x (son's age now + 15) 5x + 15 = 2(x + 15) 5x + 15 = 2x + 30 3x = 15 x=5 The son is now 5 and the father is 5 x 5 = 25. Check: In 15 years the son will be 20 and the father will be 40. 40 = 2 x 20 . .( 8. The unknown in this problem is the number of years in the past from now. The model for the problem is Sarah's aunt's age in the past = 3 x Sarah's mom's age in the past Let x = the number of years from then until now. Sarah's aunt's age was 48 - x. Sarah's mom's age was 42 - x. The equation is therefore 48 - x = 3(42 - x) 48 - x = 126 - 3x To both sides we add 3x and subtract 48 to get the simple equation 2x = 78. Dividing both sides by 2 gives us x = 39. Therefore, 39 years ago, Sarah's aunt was 3 times as old as Sarah's mom. Check: 39 years ago, Sarah's aunt was 48 - 39 = 9 and Sarah's mom was 42 - 39 = 3. 9 = 3 x 3."f 9. The "hidden" concept here is that at the moment when the car passes the cyclist, they have traveled the same distance. The model for the problem will be Distance traveled by car = distance traveled by cyclist For each distance, we make use of the formula Distance = rate x time. Since time is the unknown and the cyclist travels 0.5 hour longer than the car, we have Let t = time traveled by car t + 0.5 = time traveled by cyclist The equation we have from the model is 20 x (t + 0.5) = SOt 20t + 10 = SOt 10 = 30t t = 1/3 hour or 20 minutes 55 Check: The cyclist travels for 113 + Ih = 5/6 hours and covers a distance of 20 x 5/6 = 100/6 or 5Of6 mi. The car travels for 113 hour and covers a distance of 50 x 113 = sOh mi./' 10. This is really a "motion" problem when thinking of the teams as very slow moving vehicles. The model for the problem is based on the fact that the total distance is equal to the sum of the distances covered by both teams or Distance covered by team from east + distance covered by team from west = 2000 The "hidden" fact here is that when two objects are moving along the same path toward each other, then at the point when they meet, they have spent the same amount of time traveling! Since the rate is given in miles per day, we let t = the number of days spent by both teams. Then, using Distance = rate x time for each team's distance, we have 2 mi/day x t + 1.5 mi/day x t = 2000 2t + 1.5t = 2000 3.5t = 2000 t = 2000 -:- 3.5 = 571.43 days or approximately 571.43/365 = 1.56 or Ilh years. The team from the east laid approximately 2 x 571.43 = I 142.8 or I 143 mi of track. The team from the west laid approximately 1.5 x 571.43 = 857.1 or 857 miles of track. Check: Since the problem called for the amount of track laid, the only necessary check is that 1143 mi + 857 mi = 2000 mi. I I. Notice that the problem specifies that we give our answer in miles per hour. This adds only one more step in the solution as we can find the rates in kilometers per hour and convert to miles per hour at the end. Working in the metric system does not change the way in which we solve the problem. There are clearly two relationships given in the problem. The first is straightforward and allows us to use one variable 56 to represent both rates. Let r = rate of slower plane 2r + 220 = rate of faster plane The problem gives us the distances for the same amount of time traveled by both planes. From Distance = rate x time, we have the alternative form of Time = distance/rate, and we can derive the model Distance of fast plane Rate of fast plane distance of slow plane rate of slow plane and the equation 7000 2700 2r + 220 r Cross-multiplying, we have 7000r = 2700(2r + 220) Simplifying and solving, we obtain 7000, = 5400, + 594,000 1600, = 594,000 r=371.25 The slower plane is traveling at 371.25 km/hour and the faster plane is traveling at 962.50 km/hour. MUltiplying each of these rates by the conversion factor of 0.62 mi/km, we have that the slower plane is traveling at approximately 230 mph and the faster plane can travel at 597 mph. Check: For the check, we will use the metric answers. The slow plane travels 2700 km in 2700 -;- 371.25 = 7.2727 ... hours. In this time, the fast plane travels 962.5 x 7.2727 ... = 7000 km.1' 12. The model here states that if the machines are working together for the same amount of time, each will do a fraction of the job. 57 Therefore I job = fraction done by old machine + fraction done by new machine The rate of each machine can be found from R number of jobs completed ate = ----..:-.---=--- time The rate of the older machine is '/6 job/hour. and the rate of the new machine is '/4 job/hour. Let t = the number of hours the machines work together. Using the Number of jobs completed = rate of work x time for each machine. we have I = '/6t + '/4t I = s/'2t t = '2/5 or 2.4 hours Check: In 2.4 hours the older machine has done 2.4/6 = 0.4 of the job and the newer machine has done 2.4/4 = 0.6 of the job; thus 0.4 + 0.6 = I job."! 13. The underlying concept here is that the amount of money earned from selling the mixture at $1.75 per pound should equal the total of having sold each of the two candies at their normal price or 58 Dollar amount from mixture = dollar amount from cheaper candy + dollar amount from expensive candy We use the formula Dollar amount = price x amount sold. Let x = the number of pounds of expensive candy and we have that x + 100 be the amount of the mixture. The equation from the model is 1.75 (x + 100) = I .20 x 100 + 2.00x. Simplifying the equation and mUltiplying every term by I 00 gives us 175x + 17,500 = 12,000 + 200x 5500 = 25x x = 220lb Check: When all candy is sold, the mixture would bring in $1.75 x 320 = $560; specifically, 100 Ib of the cheaper candy would bring in $1 .20 x 100 = $120, and 220 Ib of the expensive candy would bring in $2.00 x 220 = $440. $120 + $440 = $560.1' 14. The phrase "make the same profit" indicates that Melissa is seeking to receive the same amount of money from the blend as she would if she sold the same amounts that are in the mixture separately. The model is therefore Dollar amount received from blend = dollar amount from cheaper coffee + dollar amount from the expensive coffee The two amounts of coffee can be represented by the same variable easily. Let x = number of pounds of cheaper coffee in blend 75 - x = number of pounds of expensive coffee in blend Each amount is found by the formula Dollar amount = price x amount used. Therefore, our equation is 1.90 x 75 = 1.60x + 2.10(75 - x) 142.50 = 1.60x + 157.5 - 2.IOx -15 = -0.5x x = 30 Melissa needs to use 30 Ib of the cheaper coffee and 45 Ib of the more expensive coffee. Check: The amount from the blend is $142.50, specifically, $1.60 x 30 = $48.00 and $2.10 x 45 = $94.50. $48.00 + $94.50 = 142.50.1' 15. The two relationships in this problem are Total number of coins = 34 Value of the coins = $5.05 59 Let d = the number of dimes and let q = the number of quarters. The first relationship gives us the equation d + q = 34. The second relationship uses the following concepts. Amount of money = value of coin x number of coins Total amount of money = amount in quarters + amount in dimes The second equation is therefore 0.1 Od + 0.25q = 5.05. To get both equations into a form in which we can add or subtract them to eliminate one of the variables. we multipy the first equation by 25 and the second equation by 100 to get 25d + 25q = 850 10d + 25q = 505 and we subtract to get 15d = 345. Therefore. d = 23. and from the original first equation 23 + q = 34 or q = II Check: 23 dimes + I I quarters = 34 coins and $.10 x 23 + $.25 x II = $2.30 + $2.75 = $5.05,,( (Try solving the problem. using d for the number of dimes and 34 - d for the number of quarters.) 16. The model for the problem is given by 60 Amount from John's quarters + amount from Jen's dimes + amount from Joey's nickels = $22.40 Each amount is calculated by Amount of money = value of coin x number of coins. We should try to express all three numbers of coins with the same variable instead of using three different variables. To do this. we need to study the given relationships carefully. Let x = number of coins (quarters) john gave. jen gave 4 times as many coins as john. so 4x = number of coins (dimes) Jen gave Joey gave 2 times the number of coins Jen gave + 7, so 8x + 7 = number of coins (nickels) Joey gave Using these representations in the model, we have 0.25x + 0.10 x 4x + 0.05(8x + 7) = 22.40 Simplifying and multiplying every term by 100, we have 25x + 40x + 40x + 35 = 2240 105x + 35 = 2240 105x = 2205 x =21 Using this in our "Let" statements, we deduce that John gave 21 quarters, Jen gave 4 x 21 = 84 dimes, and Joey gave 8 x 21 + 7 = 175 nickels. Check: 84 is 4 x 21,/ and 175 is 2 x 84 + 7 . ./ Also $0.25 x 21 + $0.10 x 84 + $0.05 x 175 = $5.25 + $8.40 + $8.75 = $22.40./ 17. This problem is similar to those we've looked at in that the underlying model is Total interest earned = interest earned from 9% fund + interest earned from I I % fund We use the formula Interest earned = principal x rate of interest x time invested and note that the time over which this interest was earned was I year. The unknown in the problem is the amount invested in each 61 fund. Since the total amount is $5000, we let x = the amount invested at 9% and 5000 - x = the amount invested at I I percent. The model gives us the equation 487 = 0.09x + 0.11 (5000 - x) 48,700 = 9x + 55,000 - Ilx 48,700 = 55,000 - 2x 2x = $6300 x = $3150 Mrs. Stone invested $3150 in the 9% fund and the balance, $5000 - $3150 = $1850 in the I I percent fund. Check: The interest from the 9 percent fund was 0.09 x $3150 = $283.50, and the interest from the II percent fund was 0.11 x $1850 = $203.50; then $283.50 +$203.50 = $487.00.1' 1 8. It would become very complicated to try to use three variables to represent the different amounts that each person received. This forces us to try to use only one variable to model the problem. The situation is clearest when we use the second person's amount as the single variable since the first is easily determined when we know this amount. Let x = amount received by second person. Clearly, 2x = amount received by first person. 62 The third person must have received the remaining portion of the $139 or 139 - (x + 2x). 139 - 3x = amount received by third person Note that the phrase "exceeds by some amount" can be translated as "is that amount more than." The model is, therefore Third person's portion = first person's portion + second person's portion + $1 139 - 3x = 2x + x + 1 139 - 3x = 3x + I 138 = 6x x = 23 The second person received $23, the first person received $46, and the third person received $139 - $69 = $70. Check: $23 + $46 + $70 = $139; then $23 + $46 + $1 = $70.1' 19. This problem doesn't belong to any category that was mentioned in this chapter, but it is still basically a "rate x amount" problem and is easily solved with some algebra. The rates are $45 earned per day and $10 paid per day. Having two variables makes the modeling easy. (See the Appendix for a review of how to work with equations having two variables.) Let x = number of days that the farmhand worked Let y = number of days that the farmhand did not work We know that the total number of days is 60. Therefore, the first equation is x+y=60 The basic concept is that Amount earned from working - amount paid for room and board = amount received The second equation is, therefore, 45x - lOy = 2260. Multiplying every term in the first equation by 10 and adding this to the second equation gives lOx + lOy = 600 45x - lOy = 2260 Adding the two equations gives us 55x = 2860, and we find x = 52. The farmhand worked for 52 days and did not work for 8 days. Check: 52 + 8 = 60; then 52 x $45 -8 x $10 = $2340 - $80 = $2260.1' 63 Chapter 3 Word Problems Involving Ratio, Proportion, and Percentage Many situations presented in word problems involve compar- ison of the sizes of groups and objects. The answer sought may not be the number of objects, but the fraction that indi- cates how much of the larger group is the smaller or how two groups are related to the whole. These fractions are referred to as ratios. Sometimes when ratios are given, they are given by the statement a:b. This is equivalent to the fraction a/b. The frac- tion allows us to use arithmetic in the usual way to solve problems. Example I In a math class, there are 32 students. If the ratio of boys to girls is 3:5, how many boys and how many girls are there in the class? Solution I One way to understand this problem is to restate the ra- tio in the following way. For every three boys there are five girls. With this we can create a visual representation of the situation using B for a boy and G for a girl. Writing groups of BBBGGGGG and keeping track of how many we have in a cumulative way gives us: 65 Groups BBBGGGGG BBBGGGGG BBBGGGGG BBBGGGGG Running count 8 16 24 32 Adding the Bs and Gs tells us that there are 12 boys and 20 girls. The check for this answer is to create the fraction of boys/girls and see if it is equal to 3/s. Check: 12/z0 = 0.6 = 3/s . ./ Alternative Solution 1 The method described above was quick, but it would be te- dious if the group were much larger than 32. The problem really involves finding the number of groups and applying a rate. The ratio 3 boys:S girls can be translated into two rates:3 boys/group and 5 girls/group. The model for the situation is Class size = number of boys + number of girls where each number is found by Rate of boys or girls per group x the number of groups (In Chap. 2, there are many problems using similar models.) If we think of the number of groups as the unknown, we can solve the problem algebraically by assigning a variable to this and modeling the situation with an equation. Let x = the number of groups Number of boys = 3x Number of girls = Sx Using these quantities in the model, we have the equation 66 3x + Sx = 32 8x =32 x=4 There are four groups, each consisting of 3 boys and 5 girls. Therefore, there are 3 x 4 = 12 boys and 5 x 4 = 20 girls. Example 2 An advertisement for toothpaste claims that 4 out of 5 den- tists recommend a certain brand. A consumer group wanted to check the accuracy of this claim and surveyed 180 dentists. How many dentists would they expect to not recommend this brand, if the advertisement's claim is accurate? Solution 1. The phrase "4 out of 5" indicates that the ratio of those who recommend to those who do not recommend is 4: 1. Using the picture scheme R for recommend and D for doesn't rec- ommend, we would have to list and count groups of RRRRD. Rather than listing groups, we easily realize that there has to be 180/5 = 36 groups and, therefore', there would be 36 Ds if the claim were accurate. The more structured algebraic approach would be Let x = the number of groups Number of those who recommend = 4x Number of those who do not recommend = x 4x + x = 180 5x = 180 x =36 There would be 4 x 36 = 144 dentists who recommend the brand and 1 x 36 = 36 who do not. Ratios can appear in many of the types of problems we saw in the previous chapter. Number Problems Example 3 Two positive numbers are in the ratio of 5:8. Find the numbers, assuming that the square of the smaller is 15 less than 10 times the larger. 67 Solution :I If we think of multiples of 5 and 8 as "groups," we can use a solution similar to that for Probs. 1 and 2; specifically, we can list pairs of multiples and check the conditions for each pair. Multiple 1 2 3 Smaller 5 10 15 Larger 8 16 24 Smaller 2 25 100 225 10 x larger-IS 80 -15 = 65 160-15 = 145 240 - 15 = 225'( As before, this is an easy solution because the actual numbers came up early in the list. If the actual numbers were larger, listing them in this fashion would be tedious. The algebraic approach would be Let x = "multiple" needed 5x = smaller number 8x = larger number The equation would come from the model Smaller 2 = 10 x larger - 15 and we would have the quadratic equation: (5X)2 = 10(8x) - 15 25x 2 = 80x - 15 25x 2 - 80x + 15 = 0 Dividing every term in the equation by 5 gives us the simpler quadratic 68 5x 2 - 16x + 3 = 0 (5x - l)(x - 3) = 0 5x - 1 = 0 or x - 3 = 0 x = 1/5 or x=3 Since x is a multiple, it must be a whole number. Therefore, we reject the answer of lIs. Our numbers would be the third multiples of 5 and 8 or 15 and 24. Check: 15 2 = 225; then 10 x 24 - 15 = 240 - 15 = 225."f Age Problems Example 4 Jason is Chris' older cousin. The ratio of their ages is 9:5. In 5 years, Chris' age will be 4 more than half of Jason's age. How old are the boys now? Solution 4 The ratio indicates that the current ages of the boys are multiples of 5 and 9. Making a table as follows will lead to the answer quickly. Now In 5 years lIz Jason's Chris' age - liz of Chris Jason Chris Jason age Jason's age 5 10 15 9 11 15 71/z 18 15 23 11 1 1z 27 20 32 16 Alternative Solution 4 Using algebra Let x = the multiple Chris' current age = 5x Jason's current age = 9x The model is -31/z +3 1 /z +4 Chris' age + 5 = lIz Oason's age + 5) + 4 Substitution gives 5x + 5 = 1/z(9x + 5) + 4 5x + 5 = 4.5x + 2.5 + 4 "f 69 5x + 5 = 4.5x + 6.5 0.5x = 1.5 x=3 Chris' age is 5 x 3 = 15, and Jason's age is 9 x 3 = 18. Check: In 5 years their ages would be 20 and 32. Specifically, 20 = liz x 32 + 4 = 16 + 4 = 20.v' Perimeter Example 5 The perimeter of a triangular plot of land is 273 m. A surveyor is making a map of the land and notes that the ratio of the sides is 2:9:10. What is the actual length of each side of the plot? Solution S In solving this problem, the use of a table would be a lengthy task. An algebraic approach gets to the heart of the problem quickly. (Make sure to draw a diagram that reflects the ratios to help visualize the problem accurately.) 2 ~ 10x Let x be the multiple. The sides of the triangular plot are 2x, 9x, and lOx. The model is Sum of all sides = 273 We have the equation 70 2x+9x+10x=273 21x = 273 x = 13 The sides are therefore 2 x 13 = 26 m, 9 x 13 = 117 m, and 10 x 13 = 130 m. Check: 26 + 117 + 130 = 273.1' Mixture Example 6 Carole's candystand sells a popular mixture of two of her can- dies which sell for $2.70/lb and $1.95/lb. The ratio of the ex- pensive candy to the cheaper candy in the mixture is 3:5. If Carole sold her entire supply of the mixed candy during the coming month, she would have revenue of $928.20, which would be the same as if she sold each candy separately. How many pounds of candy does she have in supply, and at what price does she sell it for? Solution 0 This seems like a more complicated problem than the previous ones in that more information is given. The key to the problem is that the revenue would be the same as if she sold each type of candy separately. Sometimes when a problem seems complicated, we should try to use a simple number and work with the information in the problem to help get a better understanding of what is occurring. For example, since the ratio of the candies in the mixture is 3:5, let's assume that 8lb was sold. In other words, we might ask, "What would be the revenue if 3 lb of the expensive candy and 5 lb of the cheaper candy were sold?" Clearly, the answer to this question is 3 x $2.70 + 5 x $1.95 = $17.85. The price of the mixture would be $17.85 -:- 8 lb, giving us $2.23/lb (rounded). We easily see that 8 lb is not the answer we are looking for, but it helped us model the problem. The model is Number of pounds of expensive candy x $2.70 + number of pounds of cheaper candy x $1.95 = $928.20 71 Using algebra: Let x = the multiple Number of pounds of expensive candy = 3x Number of pounds of cheaper candy = 5x Total number of pounds sold = 8x The model gives us the equation 3x x 2.70 + 5x x 1.95 = 928.20 17.85x = 928.20 x = 52 Therefore, 8 x 52 = 416lb ofthe mixture is in supply and sells for $928.20..;- 416 = $2.23/lb. (Note that this is the same price as when 8 lb is sold. This would be the price for any amount sold, since we are maintaining the same ratio in all quantities of the mixture.) Check: The mixture contains 3 x 52 = 156 lb of the expen- sive candy, which would have a revenue of 156 x $2.70 = $421.20, and 5 x 52 = 260 lb of the cheaper candy, which would have a revenue of 260 x $1.95 = $507.00. The total rev- enue is $421.20 + $507.00 = $928.20.1' Proportion Word problems sometimes involve comparing two situations that are numerically or geometrically similar. This often re- quires a scaling of the quantities in a proportional manner. For example, a numerically similar situation could be work- ing with a recipe that lists quantities required to make four servings needs to be scaled to make more servings when more than four are required. A geometrically similar situation could be working with two rectangles that are of different sizes but of the same length:width ratio. When dealing with such situations, we create a mathe- matical statement indicating that the ratios are the same. This is called a proportion and would appear in ratio form as a:b = c:d or in fraction form as alb = cld. 72 From the ratio form, we call a and d the extremes while band c are referred to as the means. From the fraction form, we see that if we multiply both sides of the equation by bd, we arrive at ad = bc. This is sometimes referred to as cross- multiplying in order to solve a proportion. The more mathematical phrase is: In a proportion, the product of the means is equal to the product of the extremes. This will be the governing principle in solving the following problems. Measurement of Quantities Example 7 Among the ingredients in a recipe for a sweet dessert are 21/2 cups of regular sugar and 1/4 cup of brown sugar. The directions indicate that this will serve 8 people. If there will be 12 people to serve, how much of each type of sugar will be needed? Solution 7 The underlying assumption in the problem is that the recipe needs to be scaled proportionally when seeking to make more servings. The first step is to decide which two of the three quantities given in the original recipe are necessary for the ra- tios that will appear in the proportion: 2 liz, 1/ 4 , or 8. It is helpful in solving problems of this kind to rephrase the question so that the problem is better understood: "What happens when the number of people changes to 121" This indicates that our ratios and proportion must involve 8 and 12. This also helps us see that we will need to solve two differ- ent proportions: one for regular (refined white) sugar and the other for brown sugar. The model for the problem is, therefore (a) The amount of regular sugar for 8 people is proportional to the amount of regular sugar for 12 people. (b) The amount of brown sugar for 8 people is proportional to the amount of brown sugar for 12 people. Using (a), we can assign a variable and proceed algebraically. Let x be the amount of regular sugar for 12 people. The model 73 gives us the equation 21/2 8 x 12 Cross-multiplying, we have 8x = 30 and x = 30/8 = 3 3 / 4 cups of regular sugar. Using (b), we can solve in a similar fashion. Let y be the amount of brown sugar for 12 people 1/4 y 8 12 Cross-multiplying, we have 8x = 3 and y = 3/8 cup of brown sugar. Check: One way to check the problem is to see if the ratios of the original amounts to the new amounts is the same as 12:8 or 3/z. For regular sugar, 3 3 / 4 -:- 2 1 /z = 15/ 4 -:- 5/ 2 = 15/ 4 X 2/5 = 30/z 0 = 3/ 2 . ./ For brown sugar, 3/8 -:- 1/4 = 3/8 X 4/1 = 12/8 = 3/z . ./ Measurement of Similar Figures Example 8 On a blueprint for a new office building, the rectangular con- ference room measures 4.5 in by 12 in. The shorter wall of the actual room measures 15 ft. How much carpeting will be needed to cover the floor of the actual room? Solution 8 The underlying assumption is that a blueprint is a scaled drawing of the room; that is, the sides of the room on the blueprint are proportional to the actual sizes or the ratios of the corresponding sides are equal. The model for the problem is (length on blueprint)/(actuallength) = (width on blueprint)/(actual width). Once we find the actual width, we need to compute the area of the floor of the room. Letting w be the actual width, we have 4.5/15 = 12/w. Cross-multiplying gives us 4.5w = 180. Therefore, w = 180/4.5 = 40 ft. The area is 15 x 40 = 600 ft2. In Chap. I, we were concerned about how to keep track of units of measure. Note that in computation of w, we actually 74 have the following calculation, including units: w = (12 in x 15 ft)/4.5 in = 40 ft, which shows that we have made the cor- rect calculation as it gives us our answer in feet when we cancel units. We see that something interesting happened when we compare the areas of the blueprint drawing to the actual draw- ing. The area of the room on the blueprint is 4.5 in x 12 in = 54 in 2 , and the area of the actual room is 600 ft2 = 600 ft 2 x 144 in 2 /1 ft 2 = 86,400 in 2 . The ratios of the two areas is 86,400 in 2 /54 in 2 = 1600. The length of the actual room is 15 feet = 15 ft x 12 in/ft = 180 in, and the ratio of the lengths is 180 in/4.5 in = 40. What we notice is that 1600 = 40 2 . In general, the ratio of the areas is the square of the ratios of the lengths. This fact is useful to remember. Travel From the formula for motion Distance = rate x time we have the following equivalent formula. T . distance lme=--- rate In a problem where the time taken is the same for similar modes of transportation, we can set up a proportion between the distances and rates. Example 9 Test drives of a new automobile indicated that while traveling at two speeds, one of which was 20 mph greater than the other, the higher speed covered a test distance of 1000 ft while the lower speed covered a distance of 800 ft in the same time. What were the tested speeds, and what was the test time to the nearest second? Solution g Since the test times were the same, we can use the ratio of distance/rate for each situation and create a proportion to 75 model the problem: Distance of slower test distance of faster test - - - - ~ - - - - - - - - - - - = - - - - ~ - - - - - - - - - Slower rate faster rate Before beginning, however, we need to convert the distances into miles since the rates are given in miles per hour. 1000 ft 1 mi 1 x 5280 ft = 0.189 = 0.19 mi 800 ft 1 mi -1- x 5280 ft = 0.152 = 0.15 mi Let r be the slower rate and r + 20 be the faster rate. The model gives us the equation 0.15 0.19 - ---=-=- r r +20 Cross-multiplying, we have 0.15(r + 20) = 0.19r. Multiplying by 100, we have 15(r + 20) = 19r, and we can solve this equation in the usual way. 15r + 300 = 19r 300 = 4r r = 75 mph The slower tested speed was 75 mph, while the faster tested speed was 95 mph. Calculating the time tested actually gives us the check on the problem since both distances and speeds should produce the same time. . distance 0.15 Tlme of slower test = = -75 = 0.002 hour rate . distance 0.19 Tlme offaster test = = -95 = 0.002 hour rate To convert to seconds, we use 0.002 hour 60 minutes 60 seconds 7 2 7 d -------- x x = . = secon s 1 1 hour 1 minute 76 Direct Variation Another way in which problems make use of proportions is to refer to a direct variation among changing quantities; that is, when one aspect of a situation changes, another aspect changes in a proportional way or at the same rate. Example 10 A construction firm knows that the number of new houses that it can build in a month varies directly with the cost of labor due to overtime charges. If 12 houses can be built when it pays an average of $14.50 per hour, how much should the company expect to pay its workers, if it needs to build 20 houses? Solution 10 Since we are told that the quantities vary directly, we can immediately set up the model proportion. Number of houses Cost of labor = 12 4.50 Let c be the cost of labor we are seeking. The model gives us 20 12 = c 14.50 Cross-multiplying, we have 12c = 290 and c = 290/12 = $24.17 per hour. Check: The check would be that the ratios would be the same, that is, 12 -;- 14.50 = 0.83 and 20 -;- 24.17 = 0.83. v' (Note that the calculator shows differences to the fourth decimal place. This is due to the rounding we did in the problem.) The ratio 0.83 is referred to as the constant of variation. Example II The force applied to stretch an elastic spring varies directly with the amount that it is stretched. If a force of 15 dynes (the metric measure of force) stretches a spring 2.3 cm, how much force, to the nearest dyne, would be required to stretch the spring 6 cm? 77 Solution II Since we are told that the quantities vary directly, we can immediately set up the model proportion: Amount of force 15 Stretch 2.3 Let f be the required force, and we have f 15 6 2.3 and f = 90/2.3 = 39.13 = 39 dynes. Check: The constant of variation must be the same: 15 -:- 2.3 = 6.5 and 39 -:- 6 = 6.5."! Percentage Problems that involve percentages are really problems involv- ing proportions. When asked for a percentage of a number, you can think of it as asking "How much would we have if we scaled our base amount to be 100?" For example, when asked "What is 35% of 80?" we can translate this to mean "If we have 35 objects out of a set of 100, how much would we have in a similar group of 80?" The prob- lem is then easily modeled by the proportion 35/100 = x/80. This gives us 100x = 35 x 80 = 2800 and x = 28. Percentages appear in many of the problem situations we have seen before. Investment Example 12 When Gary invested $1200 in the Lion Fund, he had $1240 at the end of 3 months. Assuming the same rate of return, what would be the annual rate of return for this fund? Solution 12 In order to determine the annual rate, we have to make use of the given assumption and conclude that over the year 78 he would earn 4 times the interest received, since 3 months is one-fourth of a year. In other words, 4 x $40 = $160. Using the percentage model, we have 160 x = 1200 100 Cross-multiplying gives us 1200x = 16,000 and x = 16,000/ 1200 = 13 1 13%. Check: We can calculate the interest for one year at 13.33 ... percent and, then multiply by 0.25 to determine what the quarterly interest is; that is, $1200 x 0.133333 x .25 = $39.9999, which is approximately $40. (The difference arose because we did not have to use the entire repeating decimal.) ,( Area Example 13 The floor of a social hall that is 45 ft by 82 ft is carpeted except for a circular dance floor in the middle that has a diameter of 20 ft. To the nearest tenth of a percent, what percent of the floor is carpeted? Solution 13 The solution requires finding the area of the entire floor, the area of the circle, and their difference, which is the amount of carpeted space. The percentage model is then used to give us Area of carpeted space x Area of entire room 100 The area of the entire room is 45 x 82 = 3690 ft2. The radius of the dance floor is 10 ft and using 3.14 for rr, the area of the circular dance floor is 3.14 x 10 2 = 314 ft2. Therefore, the carpeted area is 3690 - 314 = 3376 ft2. The model gives us the proportion 3376 x = 3690 100 79 Solving for x, we have x = 3376 x 100/3690 = 91.49 = 91.5%. Check: The check is simple: 3690 x 0.915 = 3376 (rounded).v' More complicated problems involving percentages can be found in mixture problems involving liquids that contain dis- solved ingredients such as chemicals. Example 14 A certain chemical solution used in manufacturing batteries contains water and acid. Initially, the solution is made with SO kg of dry acid and 900 L of water. The solution is mixed and left to sit so that the water evaporates until it becomes a 30% acid solution. How much water, to the nearest liter, has to evaporate for this to be accomplished? Solution 14 First, we need to make sure that our units of measurement are the same. Fortunately, in the metric system, 1 L of water weighs exactly 1 kg. Regarding 30% as 30/100, the model for the problem is the proportion Amount of acid 30 ~ - - - - - - - - - - - - - - - - - - - - ~ ~ = Remaining water + amount of acid 100 Let x be the amount of water that has evaporated. Therefore, 900 - x is the amoun t of remaining water and the total weight, in kilograms, is 900 - x + SO = 950 - x. The model gives us the equation: SO 30 ----- = 950 - x 100 Cross-multiplying, we have 30(950 - x) = 5000. Simplifying and solving, we have 28,500 - 30x = 5000 23.500 = 30x x = 23,500/30 = 783 1 13 Therefore, 783 L of water has to evaporate. 80 Check: The remaining water is 900 - 783 = 117, and the to- tal weight is 50 kg of acid + 117 kg of water = 167 kg, that is, 50/167 = .299 = 29.9%, which is okay since we used a rounded amount."! Summary I. When given a ratio of two or more objects, assign a variable to be the multiple of each number in the ratio and use the specific multiples given by the ratio as the amounts. Model the problem and substitute these variable amounts to derive an equation to solve. Solve for the mUltiple and compute the actual amounts. 2. Identify when a proportion is implied in the problem. This will be the case when phrases such as "in the same ratio as:' "varies directly as," or "is proportional to" appear in the problem. 3. Shapes are similar only when corresponding sides are in proportion. 4. Computing a percentage or using a given percentage will usually be solved by creating a proportion in which the ratio of the actual amounts is equal to x / I 00, where x is the percentage. Additional Problems I. Separate 133 into two parts so that the ratio of the larger to the smaller is 4:3. 2. On a box for a model-airplane kit, labels indicated that the scale used is I :72. If the wingspan of the model is 9 in, how long is the wingspan for the actual plane to the nearest foot? 3. CD Emporium finds that CDs of rock artists outsold CDs of classical music by 5:2 during November. If the profit from a rock CD is $2.50 and the profit from a classical CD is $1.80, how much profit was made if 2345 rock and classical CDs were sold during November? 4. A survey to determine attitudes of students of mathematics was given to 420 sixth-grade students 5 years ago and to the same group now, when they are in eleventh grade. In the sixth grade, the ratio of students who found math easy to those who didn't was 3: I ; now, in the eleventh grade, it is 3:2. How many fewer students find math easy now than before? 5. Glenn High School has 1620 students in which the girl:boy ratio is 5:4. Of this group, 8% of the girls and 5% of the boys went on a field trip. What percent, to the nearest tenth, of the school's population went on this trip? 81 6. The ratio of the longer side of a rectangle to the shorter side is 7:2. The ratio of another rectangle's longer side to its shorter side is 4:3. Given that the longer side of the second rectangle is I I in more than the longer side of the first and the shorter side is 18 in more than the shorter side of the first rectangle, find the ratio of the perimeters of the first rectangle to the second. 7. On a scale model of an outdoor patio at the home design center, there is an equilateral triangle in its center that is to be filled with a specially colored concrete mix costing $15.25 per bag, and each bag can be used to cover 8 ft2. The model is scaled at a ratio of 1:8 compared to the actual size of the patio, and a side of the triangular region is 1.5 feet. How much will it cost to fill the region on the actual 8. The sum of the digits of a two-digit number is 12. If the digits are reversed, a larger number is formed and the ratio of the larger number to the smaller number is 7:4. Find both numbers. 9. The ratio of the ages of a girl, her mother, and the girl's grandmother is 2:5:8. Six years ago the ratio of the girl's age to her mother's age was 4: 13. How old are the three women 10. Carl's crew can complete 7 jobs in the same time that Ben's crew can complete 5 jobs. During the same week, Carl's crew completed 12 more jobs than Ben's crew. How many jobs were completed by both 1 I. Fifty liters of an acid solution contains 18 L of pure acid. How many liters of acid must be added to make a 4% acid 12. The amount of calories in a serving of ice cream varies directly with the amount of sugar the ice cream contains. A company makes a regular chocolate ice cream and a "Lite" version. Regular chocolate ice cream has 240 calories (cal) per serving and contains 64 g of sugar. How many grams of sugar does the Lite chocolate have, if it has only 180 cal per 13. Mrs. Simmons tries to be a shrewd investor and invested her $28,000 in three different stocks in the ratio of 3:4:7. Over the course of a year, the respective returns were 8, 12, and 10%. What was her overall percentage return for the 14. A plane can make the same I OOO-mi trip as an express train in 6 hours' less time. The ratio of their speeds is 5:2. At what speed does each vehicle 82 Solutions to Additional Problems I. Let x be the larger part of 133. Therefore. 133 - x is the smaller part. The problem gives us the exact proportion x : 133 - x = 4 : 3. Using the product of the means is equal to the product of the extremes. we have 4(133 - x) = 3x 532 - 4x = 3x 532 = 7x x = 532/7 = 76 The larger part is 76. and the smaller part is 133 - 76 = 57. Check: 76 + 57 = 133 and 76/57 = 1.333 ... = 11/3 = 'h..f 2. This is a simple problem that will use the following proportion as its model. Wingspan of model = Actual wingspan 72 We can find the actual length in inches and then convert to feet. Let x be the number of inches in the actual wingspan. Therefore. we have 9 in/x = lin. Cross-multiplying. we have x = 9 x 72 = 648 in. 648 in x ~ = 54 ft I 12 in Check: We need to make sure that the ratio of the actual span to the model span is 1:72 or 0.0139. Using the calculated inches. we have 9 -;- 648 = .0139.1' 3. We need to first find the amount of each type of CD sold. The total profit is the sum of the profits from each type. The model for the problem consists of two steps. Step 1 Number of rock CDs + the number of classical CDs = 2.345. Step 2 Total profit = the number of rock CDs x $2.50 + the number of classical CDs x $1.80. 83 Let x = multiple of 2 and 5 2x = number of rock CDs sold in November 5x = number of classical CDs sold in November From the model for step I 2x + 5x = 2345 7x = 2345 x = 2345 -:- 7 = 335 Therefore. 2 x 335 = 670 classical CDs were sold and 5 x 335 = 1675 rock CDs were sold. From the model for step 2 Total profit = 1675 x $2.50 + 670 x $1.80 = $4187.50 + $1206.00 = $5393.50 Check: The check is concerned mostly with the amounts of each type of CD sold. We need to make sure that the ratio is indeed 5:2 or 2.5. specifically. 1675 -:- 670 = 2.5..flt is also necessary to recheck the arithmetic for step 2. 4. We have to compute the actual numbers of students who find math easy and those who do not at each grade level on the basis of the given ratios. For the sixth grade. let x be the multiple. 3x be the number of students who found math easy. and I x be the number of students who did not. 84 3x + x = 420 4x = 420 x = 105 Therefore. at the sixth grade. there were 3 x 105 = 3 15 students who found math easy (and 105 who did not). At the eleventh grade (we should use a different variable so to avoid confusion between our two answers). Let y = multiple 3y = number of students who find math easy 2y = number of students who do not 3y + 2y = 420 5y = 420 y = 84 Therefore, at the eleventh grade there are 3 x 84 = 252 students who find math easy (and 2 x 84 = 168 who do not). The difference 3 15 - 252 = 63 is our answer. Check: The check involves making sure that the numbers we determined are in the ratio given in the problem. The sixth-grade ratio should be 3 : I or 3, that is, 315 ...;- 105 = 3..fThe eleventh-grade ratio should be 3:2 or 1.5, that is, 252...;- 168 = 1.51' 5. The problem must be solved by finding the number of girls and boys who went on the trip, using the percentage model Number of girls on trip + number of boys on trip x Total number of students 100 Realizing this, we see that simply adding 8% + 5% = 13% is most likely not the correct answer. (This is the most common error in a problem like this.) It would be helpful to have an estimate before beginning. The ratio 5:4 indicates that slightly more than half of the students are girls. If the groups were equal, there would be 810 girls and 810 boys. Using 10% as an estimate for 8%, about 80 girls would be on the trip, and 5% of the boys would mean 40 boys on the trip. An estimate of the total number of students on the trip would be 120 and 120 -:- 1620 = 0.074, which is approximately 7 percent. The algebraic scheme for finding the ratios will give us the numbers of girls and boys in the school. Let x = multiple 5x = number of girls 4x = number of boys 5x + 4x = 1620 9x = 1620 x = 180 85 There are 5 x 180 = 900 girls and 4 x ISO = 720 boys. Thus S% of the girls is 900 x O.OS = 72 and 5% of the boys is 720 x 0.05 = 36. Therefore. 72 + 36 = 108 students went on the trip. The model gives us the proportion: I OS/ 1620 = x/I 00 and x = 10.SOO -:-1620 = 6.66 ... = 6.7% rounded to the nearest tenth. Check: The best check for a problem with these many steps is to recheck the arithmetic at each step and to feel sure that the answer is reasonable. The answer is reasonable on the basis of our initial estimate. 6. Since the only information about the first rectangle is the ratio of the sides. we will need to use the information about the second rectangle to help find the sides. Let x be the multiple needed for the first rectangle. 7x be the longer side of the first rectangle. and 2x be the shorter side. Then 7x + I I is the longer side of the second rectangle and 5x + IS is the shorter side of the second rectangle. We can set up the following proportion for the second rectangle. 86 7x + II 4 - 2x + 18 3 Cross-multiplying. we have the equation 3(7x + II) = 4(2x + IS). Simplifying and solving. we obtain 21x + 33 = 8x + 72 13x = 39 x=3 Therefore. the dimensions of the first rectangle are 7 x 3 = 21 in and 2 x 3 = 6 in. Its perimeter is 2 x (21 + 6) = 54 in. The sides of the second rectangle are 21 + I I = 32 in and 6 + IS = 24 in. Its perimeter is 2 x (32 + 24) = I 12 in. The ratio we want is 54: I 12 or 54/1 12. which reduces to 27/56 or 27:56. Alternative Solution After having assigned the variable and obtained representations for the sides. we could have set up our answer in variable form immediately. Specifically. the perimeter of the first rectangle is 2(7x + 2x) = ISx and the perimeter ofthe second rectangle is 2(7x + II + 2x + IS) = ISx + 58. The ratio of the perimeters would be 18x / (18x + 58). Once we found that x = 3. from the proportion. the ratio is quickly seen as 54/1 12. 7. The ratio plays only a small part in this problem. A scale of 1:8 simply means that the actual patio will be 8 times larger than the model. Therefore. the actual side of the triangle will be 8 x 1.5 = 12 ft. Using the formula for the area of an equilateral triangle from the reference table in Chap. I. we find that the area is .J3/4 x 122 or approximately 1.73 -:- 4 x 144 = 62.28 ft2. Since each bag covers 8 ft2. we will need 62.28 -:- 8 = 7.785 bags. which means that we have to buy 8 full bags. The cost is $15.25/bag x 8 bags = $122.00. 8. A simple way to solve the problem is to try all possibilities. The only two-digit numbers whose sum of digits is 12 and which are smaller than the number whose digits are reversed are 39.48. and 57. The ratio 7:4 is to equivalent to 1.75. The ratios to consider are 93/39. 84/48. and 75157. The only one of these equal to 1.75 is 84/48. Therefore. the numbers are 48 and 84. An algebraic solution to problems involving the digits of a number are best solved by using two variables. one for the tens digit and one for the units digit. We know that the original number is the smaller of the numbers. Let u be the units digit and t be the tens digit of the original number. Therefore. lOt + u is the original number and IOu + t is the number when the digits are reversed. The problem gives us two pieces of information to use: (0) t + u = 12 and (b) (IOu + t)/ (lOt + u) = 7/4. Using (b) and cross multiplying. we have 4( I Ou + t) = 7( I Ot + u). Simplifying gives us 40u + 4t = 70t + 7u. From (0) we see that t = 12 - u and when substituted into the equation. we have an equation in one variable to solve 40u + 4( 12 - u) = 70( 12 - u) + 7 u 40u + 48 - 4u = 840 - 70u + 7u 36u + 48 = 840 - 63u 99u = 792 u = 792/99 = 8 (and t must be 12 - 8 = 4) Therefore. the original number is 48 and the larger number is 84. Check: We need to ensure that the ratio of the numbers is 7/4 = 1.75. or 84/48 = 1.75.,( 87 9. An algebraic solution is the fastest way to the answer. Let x be the mUltiple for the current ages, 2x be the girl's age, 5x be the mother's age, and 8x be the grandmother's age. Six years ago, the girl's age was 2x - 6 and the mother's age was 5x - 6. With these, we have the proportion 2x - 6 4 ---= 5x - 6 13 Cross mUltiplying gives us the equation 13 (2x - 6) = 4(5x - 6). Simplifying and solving, we have 26x - 78 = 20x - 24 6x = 54 x=9 Currently, the girl is 2 x 9 = 18 years old, the mother is 5 x 9 = 45 years old, and the grandmother is 8 x 9 = 72 years old. Check: Six years ago, the girl and her mother were 12 and 39 years old, respectively. The quotient ratio 12/39 is reducible to 4113.,( 10. The first sentence of the problem gives us the ratio 88 Jobs completed by Carl's crew 7 = - Jobs completed by Ben's crew 5 Let x be the multiple, 7x be the number of jobs completed by Carl's crew, and 5x be the number of jobs completed by Ben's crew during the week. The second sentence gives us the simple model Number of jobs completed by Carl's crew = number of jobs completed by Ben's crew + 12 Inserting the variables, we have the equation 7x = 5x + 12 and, clearly, x = 6. Therefore, Carl's crew completed 7 x 6 = 42 jobs and Ben's crew completed 5 x 6 = 30 jobs. The total number of jobs was 72. Check: We need to ensure that 42:30 is equivalent to 7:5. We find that 42/30 does reduce to 715.,( II. To better understand the problem, it is helpful to realize that the original solution is 36 percent acid; that is, Amount of acid = .!! = 36 = 36% Amount of acid + water 50 100 Since we are adding more acid, we let x be the amount of acid added. This enables us to create the proportion (IS + x) / (50 + x) = 40/ 100. Cross-multiplying gives us the equation 100(IS + x) = 40(50 + x). Simplifying and solving for x, we obtain Check: ISOO + 100x = 2000 + 40x 60x = 200 IS + 3 1 /3 211/3 64 160 64 3 ---=--=--;--=-x- 50 + 3 1 /3 53 1 /3 3 3 3 160 64 = - = 0.4 = 40%1' 160 12. The phrase "varies directly" indicates that the ratios of calories to amount of sugar are proportional. Therefore, letting g be the number of grams of sugar in the Lite version, we have the proportion 240/64 = ISO/g. Cross mUltiplying gives us 240g = ISO x 64 = 11,520 and g = 11,520 -;- 240 = 4S g of sugar. Check: 240/64 = 3.75 and ISO/4S = 3.75j 13. Note the word respective. This is usually meant as a direction to make a correspondence between the first items in each list, the second items in each list, and so on. In this problem, it means that the smallest investment earned S%, the middle amount earned 12%, and the largest investment earned 10%. Each percentage return must be calculated for the money in that investment. Therefore, we have to solve this problem in several steps. (0) Find the amount of each investment. Let x be the multiple. The three amounts invested are 3x, 4x, and 7x. Therefore, 3x + 4x + 7x = 14x = 2S,OOO and x = 2000. The three amounts are actually $6000, $SOOO, and $14,000. 89 (b) Find the new amounts on the basis of the percentage returns. We can use the formula A = p+ prt, where t = I. For the $6000 investment, she earned 6000 x .08 = $480. For the $8000 investment, she earned 8000 x . 12 = $960. For the $14,000 investment, she earned 14,000 x.1 0 = $1400. The total return was $480 + $960 + $1400 = $2840. (c) Find the overall percentage using 2840/28,000 = x/I 00, that is, x = 2840 x 100 -:- 28,000 = 10.14%. Check: The most important aspect of the problem to check is the amounts of each investment represented in the ratio of 3:4:7. Checking for a ratio involving three or more terms can be done pairwise: $8000/$6000 = 4/3, $14,000/$8000 = 7/4. Both are true. The rest of the check would be to recheck the arithmetic in finding the earnings and overall percentage.,( 14. We can use two variables to set up this problem quickly. Let x be the multiple for the speeds. Since the plane is obviously the faster vehicle, 5x is the speed of the plane and 2x is the speed of the train. Let t be the time of the train and, therefore, t - 6 is the time of the plane. Using Distance = rate x time, we have 2xt = 1000 or xt = 500, and realizing that the distances traveled by each are equal, we have the equation 2xt = 5x(t - 6). Simplifying gives us 90 2xt = 5xt - 30x or 30x = 3xt. Using xt = 500 from above, we have 30x = 1500 and x = 50. Therefore, the train travels at 2 x 50 = 100 mph and the plane travels at 5 x 50 = 250 mph. Check: We need to check the ratios of the speeds and the time taken by each vehicle. The quotient ratio 250/ I 00 can be reduced to 5/2.,( Using Time = distance -;- rate, the time taken by the train is 1000 mi -;- 100 mph = 10 hours, and the time taken by the plane is 1000 mi -;- 250 mph = 4 hours, which is 6 hours less.,( Chapter 4 Word Problems Involving Geometry and Trigonometry Geometry, the study of the relationships among shapes, is a major component of mathematics as it enables us to under- stand the physical world around us. Trigonometry is the study of the relationships within triangles and, therefore, the two areas are closely related. Word problems that involve geom- etry and trigonometry mostly require an understanding of a few basic relationships between lines, angles, and polygons. The key step is to identify the relationship that applies. Many of the basic relationships are listed at the end of the chap- ter in the Table of Fundamental Geometric and Trigonometric Relationships; others are mentioned in the solutions to the problems as they arise. Algebraic equations will arise, and you should refer to Chap. 2 and to the Appendix for review and practice if you have difficulty following the steps in their so- lutions. Diagrams are essential in problems involving shapes. In most cases, the shapes are simple to draw. At other times the problems describe the geometric figure in detail in order to test your ability to visualize the situation. This requires careful reading and following the description. Problems Involving Angles Example I The ratios of the measures of the angles of a triangle are 2:7:9. Find the measure of the largest angle. 91 Solution I The model for the problem comes from the fact that the sum of the measures of the angles of a triangle is 180°. It is nearly obvious that the angle measures 20°, 70°, and 90° solve the problem. To be sure, we make use of the method seen in Chap. 3 to solve ratio problems. A diagram is also helpful in understanding the problem. 2x Let x be the multiple. The angles are, therefore, 2x, 7x, and 9x. The model gives us the equation 2x + 7 x + 9 x = 180 18x = 180 x = 10 The angles are indeed 2 x 10 = 20°, 7 x 10 = 70°, and 9 x 10 = 90°. Example 2 Two cables of equal length are supporting a vertical antenna on a roof. Each is attached to the top of the antenna and to a spot on the rooftop. If the cables form a 40° angle with each other, what are the measures of the angles formed by the cables and the rooftop? Solution :1 A diagram is necessary to visualize the situation and iden- tify the shape involved. 92 Since the cables are of equal length, the triangle has equal sides and is, therefore, isosceles. In an isosceles triangle the angles opposite the congruent sides are congruent. Using this along with the fact that the sum of the measures of the angles of a triangle is 180 0 , we can derive an equation to solve. Let x be the measure of each of these congruent angles. x+x +40 = 180 2x + 40 = 180 2x = 140 and Many word problems involve a pair of parallel lines. The relationships that arise with parallel lines concern the angles formed when a third line is introduced that intersects both parallel lines. This line is called a transversal. The next example demonstrates such a situation. Example 3 The largest of the angles formed by a transversal that crosses two parallel lines is 4S 0 less than twice the smallest of the angles. Find the measures of these angles. Solution 3 All the facts about the angles that are formed when a transversal crosses two parallel lines indicate that only two different angle measures appear. 93 Therefore, the "largest" angle is the larger of these two mea- sures and the" smallest" is the smaller of these measures. These are clearly adjacent angles that lie on the same straight line and are, therefore, supplementary. The model uses this fact and the following relationship in the problem: Measure of larger angle + measure of smaller angle = 180° Measure of larger angle = 2 x measure of smaller angle = 45° Since the problem relates the larger angle to the smaller angle, we will represent the smaller angle by the variable. Let x = smaller angle 2x - 45 = larger angle x + 2x - 45 = 180 3x = 225 x = 75 The smaller angle is 75 and the larger angle is 2 x 75 - 45 = 105°. Check: The angles must be supplementary: 75° + 105° = 180° . .( Problems Involving Line Segments of Polygons Many relationships exist among the line segments found in triangles, quadrilaterals, and other polygons depending on the 94 type of figure in question. These line segments could be sides of the figure, diagonals, altitudes, medians, angle bisectors, and others. The table at the end of the chapter lists many of these relationships. The most common word problems involve similar figures that are found in the diagram that models the problem. Example 4 At some time in the afternoon a 6-ft person casts a shadow that is 10 ft long while a nearby flagpole casts a shadow that is 22 ft long. To the nearest foot, how tall is the flagpole? Solution 4 In order to identify the shapes we are working with, we need to draw a diagram. Some basic assumptions must be made: (1) the person and the flagpole are standing perpen- dicular to level ground and (2) the sun's rays form the same angles with the head of the person and the top of the flagpole. These assumptions are standard for this kind of problem and lead to the following diagram. Person Flagpole h 6ft 10 ft J Person's shadow , J Flagpole's shadow 22 ft Note that two similar right triangles are formed on the basis of the assumptions. Since corresponding sides of similar triangles are in proportion, we have Height of person height of flagpole Length of person's shadow length of flagpole's shadow Let h be the height of the flagpole, and we have 6/10 = h122. Cross-multiplying gives us 10h = 132 and h = 13.2. The height of the flagpole to the nearest foot is 13 ft. 95 The next example demonstrates the need to carefully read the description of the shape involved in the problem. Part of the goal of such a problem is to test your ability to visualize the situation. This requires careful reading and following the description. Example 5 One of the shorter sides of a rectangle is extended 10 units beyond a vertex, and a line segment is drawn from its new endpoint through the rectangle to the vertex opposite the ver- tex extended through. This segment cuts the side it is drawn through into two segments of 15 and 24 units. What is the area of the rectangle? Solution 5 Finding the area of the rectangle requires finding the length and width. The problem tells us that one side is 24 + 15 = 39 units. We need to study a diagram of the problem in order to find the means to find the width. The directions given in the problem must be followed with care in order to arrive at the correct diagram. The ver- tices of the starting rectangle ought to be labeled, and as the directions are followed new labels should mark pOints of in- tersection. A ...--_____ ---,8 24 D L - - - - ~ - - - - - - i C 10 E Suppose that we extend side BC through C to a new endpoint E. The problem tells us that CE = 10 units. The vertex opposite C is A, and we draw segment AE, which cuts side DC at F. The problem tells us that DF is 24 and FC is 15. The diagram makes it clear that there are two 96 similar right triangles, t'}.ADF and t'}.ECF. Since correspond- ing sides of similar triangles are in a proportion, we have AD/ EC = DF /CF or AD/I0 = 24/15. Thus AD = 240 -:- 15 = 16 units. The area of the rectangle is 39 x 16 = 624 square units. Problems Involving Right Triangles The most frequent real-world applications of geometry include situations that involve right triangles. This is due to finding right triangles in abundance in the physical structures of the world around us and the importance of finding perpendicu- lar distances. The two most important tools in right-triangle problems are the Pythagorean Theorem and the trigonometric ratios: sine, cosine, and tangent. Using the Pythagorean Theorem The Pythagorean Theorem is used when we know two sides of the right triangle and need to find the third. It relates the two legs and the hypotenuse of a right triangle in the following way: The square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. Using variables a and b to represent the lengths of the legs and c to represent the length of the hypotenuse, we have the familiar statement a 2 + ~ = c 2 . a b Remember that the hypotenuse is always the side opposite the right angle and is the longest side of the right triangle. In solving problems with the Pythagorean Theorem, you will have to compute square roots. These are seldom going to be whole numbers. Therefore, be sure to give the answers to the degree of accuracy asked for in the problem. 97 Example 6 A 30-ft ladder leans against the side of a building and reaches the flat roof. If the ladder stands at a point 10 ft from the house, how tall is the house to the nearest foot? Solution 15 A simple diagram of the situation clearly reveals that we are working with a right triangle in which we know two sides and have to determine the third. Building h 10 More specifically, we know one leg and the hypotenuse. If we let h be the height of the building, we have the equation h 2 + 10 2 = 30 2 . h 2 + 100 = 900 = 800 h = )800 = 28.28 The height of the building to the nearest foot is 28 ft. Several sets of numbers, referred to as Pythagorean Triples, are useful to remember. Each triple is a set of three whole num- bers that satisfy the Pythagorean Theorem. The four most com- mon triples are 98 3,4,5 5,12, l3 8,15,17 7,24,25 3 2 + 4 2 = 9 + 16 = 25 = 52 52 + 122 = 25 + 144 = 169 = l3 2 8 2 + 15 2 = 64 + 225 = 289 = l7 2 7 2 + 242 = 49 + 576 = 625 = 25 2 If you notice two of the three numbers in a problem, you can save computation time. Be sure that the smaller numbers rep- resent the legs and the largest one represents the hypotenuse! Actually, any triplet that is in the same ratio as any of the above will also satisfy the Pythagorean Theorem. For example 6,8,10 6 2 + 8 2 = 36 + 64 = 100 = 10 2 Look for these scaled triplets as well. Example 7 The sides of a triangle are IS, IS, and 18. What is the length of the altitude to the noncongruent side? Solution 7 The altitude creates two congruent right triangles and, hence, it bisects the side. The following diagram indicates that we are looking for a leg of a right triangle whose other leg is 9 and whose hypotenuse is 15. A multiple of the 3:4:5 triple is 9:12:15. There is no other possibility for the leg but 12. We can also solve the problem algebraically to be con- vinced of this. Let h be the length of the altitude. Therefore, h 2 + 9 2 = 15 2 . h 2 +81 = 225andh 2 = 144. Then h = J144 = 12. The Pythagorean Theorem shows up in other figures where right triangles are formed. Look for this when "drop- ping" altitudes or working with perpendicular lines. Example 8 The parallel sides of an isosceles trapezoid are 8 and 18 units. If the area of the trapezoid is 156 square units, how long are the nonparallel sides? Solution 8 The answer is found by dropping altitudes from the shorter of the parallel sides to the longer. The shape formed in the center is a rectangle, and the right triangles formed on its sides are congruent. Therefore, the longer base is d i v i d ~ d into segments of 5, 8, and 5. Each nonparallel side of the trapezoid is a hypotenuse of the right triangle, in which one leg is 5 units and the other leg is the altitude of the trapezoid. 8 111\ 5 8 5 We can calculate the length of the altitude from the for- mula for the area of a trapezoid. Area = liz x (base 1 + base 2) x h We have 156 = liz x 26 x h = 13h. Therefore, h = 156/13 = 12. We now have a right triangle whose legs are 5 and 12. The hypotenuse must be 13 from the familiar 5,12,13 Pythagorean Triple. Example 9 A floor pattern requires tiles that are in the shape of a rhombus. The manufacturer makes these tiles according to the lengths of the diagonals. How long is each side of the tile, if the diagonals are 10 and 16 in? Solution 9 The solution to this problem lies in the fact that the di- agonals of a rhombus are perpendicular. Since a rhombus is a parallelogram, the diagonals bisect each other. Therefore, the diagonals create four congruent right triangles in the interior, 100 where each side of the rhombus is a hypotenuse of the right triangle. Letting s represent the side of the rhombus, the Pytha- gorean Theorem gives us the computation S2 = S2 + 8 2 = 2S + 64 = 89 and s = .J89. The problem doesn't ask for a specific degree of accuracy. It is best, therefore, to leave the answer in radical form. However, it is a good idea to get its value to see if the answer is reasonable. The square root .J89 is approxi- mately 9.43. This is reasonable since it is larger than either of the other two sides of the right triangle. Using the Trigonometric Ratios Trigonometric ratios are ratios of two sides of the right triangle and are used when we have problems that involve two sides and an acute angle of a right triangle. The ratios are referred to with respect to the angle in question and are called the sine of the angle, the cosine of the angle, and the tangent of the angle. The legs of the right triangle are identified as being either the side opposite the angle or the side adjacent to the angle. The other side is, of course, the hypotenuse. The ratios are specifically S · f I length of opposite side me 0 an ang e = --,----....::...---::------:-=-=--=----- length of hypotenuse (sine is usually abbreviated as sin) C · f I length of adjacent side osme 0 an ang e = ---"------,-----'----- length of hypotenuse (cosine is usually abbreviated as cos) 7'. f I length of opposite side langent 0 an ang e = .. length of adJacent sIde (tangent is usually abbreviated as tan) 101 Letting the legs be represented by a and b and the hypotenuse by c as in the diagram below, B c a A b C we have . A a b a sm =- cosA =- tanA= b c c while . b smB =- a cosB =- b tanB = - c c a Your calculator or a table of trigonometric values can tell you the approximate value of the sine, cosine, or tangent of any acute angle, and it also can tell you the angle if you know the ratio. Finding Sides of a Right Triangle Example 10 The warning on the side of a 2S-ft ladder says that the angle it makes with the ground should be no smaller than 40° before it becomes unsafe. To the nearest foot, what is the height along a wall that this ladder makes at such an angle? Solution 10 The diagram for the problem presents us with a right tri- angle. We should label the vertices of the triangle for reference. 102 The side of the building is the side opposite the angle we are using, and the ladder is the hypotenuse. Therefore, we need to use the sine ratio. The verbal model is S · f 400 height along wall lne 0 = ---=---::------:-."....::...::-::-- length of ladder or using h for the height, sin 40° = h12S. The calculator tells us that sin 40° = 0.6428 (four decimal places is standard). Therefore, we have h = 25 x 0.6428 = 16.07. To the nearest foot, our ladder will reach a height of 16 ft. In word problems involving trigonometry, we often find the phrases "angle of elevation" and "angle of depression." In Example 10, the 40° angle that the ladder makes with the ground would be the angle of elevation of the ladder to the point where it reaches on the wall. In other words, the angle of elevation is the measure of the upward tilt off the horizontal required to view an overhead object. The angle of depression is the measure of the downward tilt off the horizontal to view an object below. Object Horizontal Angle of elevation Horizontal Angle of depression In both cases a right triangle is formed when drawing the height of the viewer or the object, and the trigonometric ratios can be used. It is important to note that the angle of depression lies outside the triangle and is the complement of its adjacent angle. Therefore, it is also equal to the other acute angle of the right triangle. Example II From a point 150 ft from the center of the base of a hill, the angle of elevation to the top of the hill is 28°. A tram from that point carries people to the top. To the nearest foot, how high is the hill, and how long is the tram's cable? 103 Solution II The angle of elevation is the angle that the tram cable makes with the ground. The height of the hill is measured as the perpendicular axis from its top to the center of its base. The following diagram models the problem, and we label the ver- tices for reference. (Remember, you don't have to be an artist; use simple lines to make the right triangle easy to identify!) B Cable A L.... ___ --'_ - - ' ~ - - - - ' L-__ ~ , - - - ~ , C 200 We know the length of the side adjacent to the angle AC = 200. The height is the side opposite the angle BC, and the length of the cable is the hypotenuse AB. In order to find the height, we easily determine that the appropriate trigonometric ratio is tangent since it is the one of the three that involves the opposite and adjacent sides. Letting h be the height, we have tan 28. 0 = BC / AC = h/200. There- fore, h = 200 x tan 28° = 200 x 0.5317 = 106.34 = 106 ft, to the nearest foot. We can find the length of the cable in two different ways, one using trigonometry and the other using the Pythagorean Theorem . • We can use the cosine ratio since it is the one that involves the adjacent side and the hypotenuse. Letting c be the length of the cable, we have cos 28° = AB/ AC = 200/c. When we cross multiply here, we have c x cos 28° = 200. Therefore c = 2007 cos 28° = 200 7 0.8829 = 227 feet, to the nearest foot. • We can use the Pythagorean Theorem to find the hypotenuse of the triangle, since we were given AC and we found Be. Note that we are using a value that we had to determine and we had rounded that answer. To be accurate, we should use the value of BC to more decimal places, 106.34, and work with several decimal places until we round off at the end of 104 the calculations. Letting c be the cable length, we have c 2 = 200 2 + 106.34 2 = 40.000 + 11,308.20 = 51,308.20 Therefore, c = ",51,308.20 = 226.51 ft = 227 ft, to the near- est foot. (If we would have used 106 ft for Be, our answer would have been 226 ft, which is close, but not correct.) Example 12 An observer from the top of a lighthouse that is 40 ft tall and stands on a cliff 80 ft above sea level notices a boat off shore. Boats must be warned if they come too close to the cliff, be- cause of the rocks below the waterline. If the observer notes that the angle of depression is 50', how far from the cliff is the boat? Solution 11 The diagram clearly shows the right triangle that must be used. It also indicates that the height includes the cliff and the tower, that is, 80 + 40 = 120 ft. 40 80 Boat- d The diagram also indicates that the angle within the triangle we need to use measures 40°. The height of the cliff is the adjacent side of the angle, and the distance between the boat and the cliff is its opposite side. Therefore, we should use the tangent ratio. Letting d be the distance, we have tan 40° = d/120 and d = 120 x tan 40 = 120 x 0.8391 = 100.7 = 101 ft, to the nearest foot. lOS Finding Angles of a Right Triangle The previous examples demonstrated how the trigonometric ratios are used to find the lengths of the sides of a right triangle. They are equally as useful when trying to find the acute angles of a right triangle. To accomplish this, you have to know the lengths of two sides. The important step is to identify the po- sition of the given sides relative to the angle you seek to find. Example 13 A young tourist visiting Washington, D.C. is awed by the size of the Washington Monument, which is 555 feet tall. The tourist, who is 5 ft tall, is standing 50 yards (yd) from the mon- ument. At what angle, to the nearest degree, is she elevating her head when she looks at the top of the monument? Solution 11 Two aspects of this problem must be accounted for be- fore we can accurately identify the right triangle. One is the obvious change of units in the distance that the tourist is from the monument. She is standing 50 yd away, which is equal to 150 ft. The second aspect is more subtle. Since we are concerned with the angle of elevation of her head, we must consider the horizontal axis to begin at the height of her head. We can safely say that is 5 ft. Therefore, the vertical height used is 555 - 5 = 550 ft. 550 ft Ground The monument is the side opposite the angle, and the distance is the side adjacent to the angle. We can make use of the tangent ratio to solve the problem. 106 Let x be the measure of the angle. We have tan Xo = 550/150 = 3.6667. Your calculator should have a key press that will find the inverse tangent (sometimes written as tan- 1 or arc tan). This will tell you what angle has the calculated number as its tangent. We could write x = the inverse tangent of 3.6667 or x = tan- 1 3.6667. Enter 3.6667, and press the key to get x = 74.7° or, to the nearest degree, 75°. Special Right Triangles Just as there are some easy triplets of numbers to help with using the Pythagorean Theorem, certain special right triangles are well known and used frequently in problems involving trigonometry. The most easily remembered trigonometric ratio is sin 30° = 0.5 or l/Z. This means that in a right triangle that has a 30° angle, the hypotenuse is twice the size of the side opposite the 30° angle. If these sides of the triangle were 1 and 2, the Pythagorean Theorem would be used to find that the other leg is v'3. 60" 2 We refer to this as the special 30-60-90 right triangle. From the diagram, we have six trigonometric ratios that are easy to remember and have exact radical values. . 1 sm30° = 2 sin 60° = ~ v'3 cos 30° =- 2 1 cos 60° = 2 1 tan 30° = v'3 tan 60° = v'3 and we have the following ratio: leg opposite 30° : leg adjacent 30° : hypotenuse = 1 : v'3 : 2. This makes problems like the following easy to solve. 107 Example 14 The diagonal of a rectangle is B in long and makes a 30° with the longer side. What is the perimeter of the rectangle? Solution 14 The diagram with the vertices labeled shows that we have a right triangle, tiBAD, with hypotenuse BD. Since the an- gle is 30°, it is a 30-60-90 right triangle and we know that AD:AB:BD= 1: J3: 2. 8 o c Therefore, since BD = B in we have AD = 4 in and AB = 4J3 in. The perimeter of the rectangle is 4 + 4 + 4J3 + 4J3 = B + BJ3. This is an exact answer and does not require a decimal approximation unless one is asked for. (Note that J3 is approximately 1.73 if it is needed.) The other special right triangle is the isosceles right trian- gle. Since the acute angles must be equal, they are both 45°. Since the legs are equal, their ratio is 1. Finally, since the legs are the sides opposite and adjacent to a 45° angle, we have that tan 45° = I! The Pythagorean Theorem gives us the hy- potenuse exactly as ../2. The diagram below depicts the special 45-45-90 right triangle. We have the three trigonometric ratios sin 45° = 1/../2, cos 45° = 1/../2, and tan 45° = I, and the sides are in the ra- tio leg:leg:hypotenuse = 1 : 1 : ../2. 108 Example IS What is the exact area of a square whose diagonal is 8 in? Solution IS x The diagonal creates two isosceles right triangles within the square. Using the ratio leg:hypotenuse = l/vIz, if we let x be the side of the square, we have the proportion x/8 = l/vIz. Cross- multiplying gives us xvlz = 8 and x = 8/vIz inches as the side of the square. The area of the square is, therefore, (8/vIz)2 = 64/2 = 32 in 2 . Summary I. Make sure to draw simple but accurate diagrams. 2. Look for right triangles. isosceles triangles. parallel lines. and other shapes that have known relationships. 3. Assign a variable to the unknown length or angle. and create a model equation that uses the identified relationship. 4. Solve the equation for the variable. and be sure to use it to determine all the lengths or angles asked for in the problem. S. Check your arithmetic and algebra. and make sure that the answer is reasonable for the problem. Additional Problems I. In a triangle. two of the angles are in a ratio of 5:3. The third angle is 40 0 less than the smaller of the other two. Find the measure of the largest angle. 2. The acute angles of a right triangle are such that the square of the smaller is 25 0 less than 10 times the larger. Find both angles. 3. Two angles formed on the same leg of a trapezoid are such that one is 30 0 less than 5 times the other. Find the measures of the angles. 109 4. In an isosceles triangle, a segment drawn from the vertex angle to the base cuts the vertex angle into two angles whose ratio is 5:3. If the smaller of these angles is 20° less than the base angle of the isosceles triangle, what are the measures of the angles formed by the line segment? 5. An entrance to a mine is dug horizontally into a hill. At a point 45 m along the entrance, a ventilating shaft is constructed vertically upward and breaks through at a point 80 m along the hillside. How tall is the ventilating shaft, to the nearest meter? 6. A basketball player is standing directly in front of the basket at the foul line, which is 15 ft from the basket. The player passes the ball to another player 8 ft directly to the right. This player takes the shot from that point. If the three-point line is 23 ft 9 inches from the basket, will this shot count for three points if it is made? 7. A kite flying on a 150-ft string is anchored to the ground. At one point the string is making a 70° angle with the ground. How high above the ground is the kite? 8. A surveyor uses a transit to determine the height of the Empire State BUilding in New York City. When pointed at the top of the antenna on the building, the angle of elevation given by the transit is approximately 80° from a point 260 ft away from the building. To the nearest foot, how tall is the Empire State Building including the antenna? 9. In testing the acoustics for a theater, the distance from the stage to the last row in the balcony has to be determined. If the angle of depression from that row to the stage is 28° and the vertical height above the stage is approximately 50 ft, how far, to the nearest foot, does the sound have to travel to reach someone sitting there? 1 O. An outdoor patio along the back of a house has to be constructed on an incline to allow rainwater to drain away from the house. To the nearest tenth of a degree, what is the angle at which the patio inclines, assuming that the length across the patio extends 12 ft from the house and that it is 3.7 in higher at the side adjoining the house than on the other side? 1 I. A 25-ft vertical pole is to be supported by two cable wires, each 30 ft long, on different sides of the pole. To the nearest foot, what is the distance between the points at which the cables are anchored to the ground? What is the angle, to the nearest degree, that these wires make with the ground? 12. Find the area, to the nearest square centimeter, of a regular pentagon whose perimeter is 100 cm. 110 13. On a baseball diamond. the distances between the bases is exactly 90 ft. The catcher has to be able to make an accurate throw from homeplate to second base. If the catcher throws the ball at 80 mph. how many seconds. to the nearest tenth. will it take the ball to reach second base? 14. Find the exact area of a parallelogram whose sides measure lOin and 20 in and the acute angle formed by the sides is 60°. Solutions to Additional Problems I. Let x be the multiple to use with the ratio. Therefore. the angles. in descending size order. are 5x. 3x. and 3x - 40. The fact that the sum of the measures of the angles of a triangle is 180°will give us the equation to solve. 5x + 3x + 3x - 40 = 180 Ilx = 220 x = 20 The largest angle is 5 x 20 = I 00°. Check: The middle angle is 3 x 20 = 60°. and 100°: 60° is indeed 5:3. The remaining angle is 60° - 40° = 20° and 100° + 60° + 20° = 180° . ./ 2. The key point is that the acute angles of a right triangle are complementary. That is. their sum is 90°. I I I The relationship given provides the model (Measure of smaller angle)2 = lOx measure of larger angle - 25 Let x be the measure of the smaller angle and 90 - x be the measure of the larger angle. Using these in the model, we have the quadratic equation x 2 = 10(90 - x) - 25 (which needs to be simplified, and all terms must be brought to one side) x 2 = 900 - lOx - 25 x 2 + lOx - 875 = 0 We seek factors of -875 whose difference is + 10. A bit of trial and error will yield -25 and +35. Therefore (x - 25)(x + 35) = 0 x = 25 or -35 We can reject -35 since angles must have a positive measure. The smaller angle measures 25°, and the larger measures 65°. Check: 25 2 = 625 and lOx 65 - 25 = 650 - 25 = 625.1' 3. The leg of the trapezoid can be regarded as a transversal that, if extended, would cross the parallel sides. I ; The two angles are, hence, interior angles on the same side of the transversal, and they are supplementary. The model is, therefore, 112 the sum of the two angles = 180 0 • Let x = one angle 5x - 30 = the other angle x + 5x - 30 = 180 6x = 210 x = 35 The angles are 35 0 and 5 x 35 - 30 = 175 - 30 = 145 0 • Check: We need to make sure that the angles are supplementary: 35 0 + 145 0 = 180 0 .,( 4. In an isosceles triangle. the base usually refers to the noncongruent side and the angles formed on this side are called the base angles. The third angle is called the vertex angle of the isosceles triangle. The diagram for this picture requires an isosceles triangle with a line segment that is clearly not an altitude. since the altitude to the base would also cut the vertex angle exactly in half. 5y\\X ill Since we are dealing with ratios. we will assign a variable for the multiple as follows. Let x be the multiple. Therefore. 3x is the smaller angle formed by the segment and 5x is the larger one. Therefore. the vertex angle is 8x. To avoid confusion. we can represent the base angle by a different variable. Let y be the measure of the base angle. We can use the sum of the angles of a triangle = 180 0 to model the problem: 8x + y + y = 180 or 8x + 2y = 180. To solve a problem with two variables. we need a second equation involving both of them. This comes from the other piece of information. namely. the smaller angle formed by the segment is 20 0 less than the vertex angle. Therefore. we have the equation 3x = y - 20 or y = 3x + 20. We can substitute for y in the first equation. giving us 8x + 2(3x + 20) = 180. Simplifying and 113 solving for x, we obtain 8x + 6x + 40 = 180 14x +40 = 180 14x = 140 x = 10 Therefore, the angles formed by the segment are 3 x 10 = 30 0 and 5 x 10 = SOc. Check: The base angles would each have to be 20° more than the smaller angle, that is, 50 0 • The sum ofthe angles ofthe isosceles triangle is 50° + 50 0 + 80 0 = 180 0 .,( 5. The phrase "vertically upward" indicates that the shaft is perpendicular to the horizontal (axis) and a right triangle is formed whose legs are the entrance and shaft and whose hypotenuse is the hillside. L f--- 45 m----l Therefore, we can apply the Pythagorean Theorem. Let x be the length of the vertical shaft. We have x'+ 45' = 80' or x' + 2025 = 6400; Thus x' = 4375 and x = J4375 = 66.14. To the nearest meter, the shaft is 66 m. 6. In order for the shot to be worth three points, the distance to the basket has to be greater than or equal to 23 ft 9 in or 23.75 ft (since 9 in is three-fourths of a foot). The use of the word directly enables us to assume that we are dealing with perpendicular distances and, hence, we have the following diagram. A f-r-_---'1"'S"'ft __ -7 c Bft d B 114 The first player is at A, the second player is at B, and the basket is at C. Be is the distance we are trying to determine, and it is the hypotenuse of a right triangle whose legs are AC and AB. We recognize these to be the legs of the 8, IS, 17 Pythagorean Triple, and the hypotenuse is 17, which is too short for a three-point shot. If you don't recognize the triple, you can use the Pythagorean Theorem. Let d be the distance, and we have d 2 = 8 2 + 15 2 = 64+ 225 = 289 and d = J289 = 17 ft. 7. The diagram shows that we have a right triangle where the height of the kite is the side opposite the 70° angle and the string is the hypotenuse. Therefore, we make use of the sine ratio. Let h be the height of the kite. Then sin 70 c = h /150. From a calculator or table, we find sin 70 0 = 0.9397. Therefore, h = 150 x 0.9397 = 140.96. To the nearest foot, the kite is 141 ft above the ground. h 150 It 8. The diagram indicates that we have a right triangle. We see that 260 ft is the length of the side adjacent to the given acute angle and the height of the Empire State Building is the side opposite this angle. Therefore, we need to use the tangent ratio. Let x be the height of the building, and we have tan 80° = x /260 or x = 260 x tan 80° = 260 x 5.6713 = 1474.5 ft. To the nearest foot, the height is 1475 ft. (The actual height is 1472 feet. The difference is due to rounding the angle given by the transit.) 11 ! i .1 . I 80° II ; i 260 It lIS 9. The angle of depression from the last row in the balcony to the stage is the same as the angle of elevation from the stage to that row. Therefore. we have a right triangle where the hypotenuse is the distance we wish to find and the vertical height is the side opposite the known acute angle. Stage We can make use of the sine ratio. Let d be the distance. and we have sin 28° = SOld or d = 50.;- sin 28" = 50.;- 0.4695 = 106.49. The sound will have to travel approximately 107 ft. 10. To find the angle of incline. we will have to find a right triangle in which we know two of the sides. At the house side of the patio we have a vertical height of 3.7 in. and. since this is measured perpendicularly to the ground. we have a right triangle that extends to the other side of the patio. 12 ft or 144 in away. (We must work in the same units when taking ratios of sides.) ':,l ~ __ (_ 12tt=144;n The diagram indicates that we have the sides opposite and adjacent to the angle of incline. which indicates that we need to use the tangent ratio. Let x be the measure of the angle of incline. and we have tan XO = 3.7/144 = 0.0257. Using the inverse tangent. we have x = I .47° or. to the nearest tenth of a degree. 1.5". I I. The problem asks us to find two pieces of information. The model for this situation is easily seen to be an isosceles triangle where the cables are its congruent sides and the pole is its altitude. The altitude of a right triangle forms two congruent right triangles. and. since we know two of its sides, we can use the Pythagorean 116 Theorem to find the third. This side of the right triangle is one-half the distance between the cables on the ground. Cable Let x reprelient the third side of the right triangle. The Pythagorean Theorem gives us x 2 + 25 2 = 30 2 x 2 + 625 = 900 x 2 = 275 x = J275 = 16.58 The distance between the cables is. therefore. 33.16 ft or. to the nearest foot. 33 ft. To find the angle. we could create a ratio with any two of the sides. However. we should use the given information rather than the length we computed. in case we computed incorrectly. The given sides represent the side opposite the angle and the hypotenuse. Therefore. we should use the sine ratio. Let y be the angle. and we have sin yO = 25/30 = 0.8333. The inverse sine gives us y = 56.4° or. to the nearest degree. 56°. 12. The diagram shows us that the interior of the pentagon can be divided into 10 congruent right triangles by drawing from the center all the radii and perpendicular segments to each side. Such a perpendicular segment is called an apothem. The area of the pentagon will be equal to lOx the area of each right triangle. 117 Since all sides are congruent. each side has a measure of 100 cm ...;- 5 = 20 cm. Since the radii form isosceles triangles. the apothem bisects each side. Therefore. we know that each leg of the right triangle is 10 cm. We need to find the length of the apothem in order to find the area of the right triangle. We can use trigonometry. if we can find the acute angle formed by a radius and a side. A basic fact about a regular polygon with n sides is that an exterior angle formed by extending a side is equal to 360/n°. For a pentagon. this is 360...;- 5 = 72°. The interior angle is its supplement. 180° - 72° = 108°. The radius bisects this angle and. therefore. the acute angle in the right triangle is 54°. The apothem is the side opposite this angle and the "half side" of the pentagon is the side adjacent to this angle. Therefore. we can use the tangent ratio. Let a be the length of the apothem. and we have tan 54° = a/IO or a = 10 x tan 54° = 10 x 1.3764 = 13.764. The area of the right triangle is 'h x lOx 13.764 = 68.82 cm 2 . The area of the pentagon is therefore. lOx 68.82 = 688.2 cm 2 or. to the nearest square centimeter. 688 cm 2 . 13. The speed of the throw will come from the distance formula: Distance = rate x time; specifically. Time = distance ...;- rate. The rate is 80 mph. and we will have to convert units. The problem will therefore require several steps. 118 Third base Second base Home plate 90 ft First base (a) Find the distance from home plate to second base. Home plate. first base. and second base form an isosceles right triangle with the distance we seek as the hypotenuse. Since this is a special 45-45-90 right triangle. we know that the hypotenuse is ,.,fi x 90. Since we will be making conversions later. we should keep several decimal places along the way. ,.,fi is approximately 1.4142. and our distance is approximately 127.278 ft. (b) Convert the rate to feet per second. For this we can use the dimension scheme (see Chap. I ) 85 mi 5280 ft I hour 1 minute ---x x x---- 1 hour 1 mi 60 minutes 60 seconds feet = 124.667-- second (c) Calculate the time. Time = distance -;- rate = 127.278 -;- 124.667 = 1.02 seconds or. to the nearest tenth. 1.0 second. 14. The area of a parallelogram is the product of one side (the base) and the altitude drawn to that side (the height). The word exact is an indicator that we will be able to express our answer either as a number that we did not have to round off or as a number involving radicals. In either case. we should be looking for a special right triangle. A ~ t 1 /0 c E The diagram shows that we have a 30-60-90 right triangle and we know that the sides are in the ratio leg opposite the 30° angle: leg opposite the 60 0 angle: hypotenuse = I :.J3 : 2. We know that the hypotenuse AB is lOin. The side opposite the 30 0 angle is. therefore. 5 in. and the side opposite the 60 0 • our altitude. is 5.J3 in. Therefore. the area of the parallelogram is 20 in x 5 .J3 in = 100.J3 in 2 • 119 Table of Fundamental Geometric and Trigonometric Relationships Congruent Similar Supplementary angles Complementary angles Adjacent angles Vertical angles Parallel lines Regular polygon Bisector Altitude Median Definitions Equal in measure and form. Proportional in measure, but identical in form. Two angles whose sum is 180°. Two angles whose sum is 90°. Two angles that share a common side and vertex. Two nonadjacent angles formed by intersecting lines. Two lines that are always the same distance apart and, therefore, will never intersect. A closed figure in which all sides and interior angles are congruent. A line or line segment that divides a side or angle into two congruent sides or angles. A line segment from a vertex of a triangle perpendicular to the opposite side. Its length is the height of the triangle when the opposite side is considered to be its base. A line segment from a vertex of a triangle drawn to the midpoint of the opposite side, therefore, bisecting the opposite side. Relationships between Lines and Angles 1. Vertical angles formed by intersecting lines are congruent. 2. Adjacent angles forming a straight line are supplementary. 3. Adjacent angles forming a right angle are complementary. Continued 120 4. Perpendicular lines form right angles. s. A transversal that intersects a pair of parallel lines forms a. Congruent angles in corresponding positions on each parallel line b. Congruent angles interior to both parallel lines and on alternate sides of the transversal c. Supplementary angles interior to both parallel lines and on the same side of the transversal Relationships Involving Triangles 6. The sum of any two sides of a triangle must be greater than the remaining side. 7. The sum of the angles of a triangle is 180 0 • 8. An exterior angle is equal to the sum of the two nonadjacent interior angles. 9. Two triangles are similar if they have the same three angles. 10. The ratios of corresponding sides of similar triangles are equal. 11. A median divides a triangle into two triangles of equal area. 12. In an isosceles triangle, two sides are congruent and their opposite angles are congruent. 13. In an isosceles triangle, the altitude drawn to the noncongruent side is alsoan angle bisector and a median. 14. In an equilateral triangle, all sides are congruent and each angle measures 60 0 • Relationships Involving Right Triangles 15. The Pythagorean Theorem states that the square of the hypotenuse is equal to the sum of the squares of the legs. 16. The square of the length of an altitude drawn to the hypotenuse of a right triangle is equal to the product of the segments it creates on the hypotenuse. 17. The trigonometic ratios are a Sine of an angle = length of opposite side . length of hypotenuse b Cosine of an angle = length of adjacent side . length of hypotenuse c. Tangent of an angle = : : ~ ~ ! ~ ~ ~ ~ ~ f a ~ ~ i ~ ~ : ~ ~ : Continued 121 Relationships Involving Quadrilaterals 18. The sum of the angles of any quadrilateral is 360 0 • 19. Opposite sides of a parallelogram are parallel and congruent. 20. Opposite angles of a parallelogram are congruent. 21. Consecutive angles of a parallelogram are supplementary. 22. The diagonals of a parallelogram bisect each other. 23. A rectangle, a rhombus, and a square are all parallelograms and have relationships 16 through 19. 24. A rectangle has four right angles, and its diagonals are congruent. 25. A rhombus has four congruent sides, and its diagonals are perpendicular. 26. A square is both a rectangle and a rhombus and has relationships 21 and 22. 27. A trapezoid has only one pair of parallel sides. Relationships Involving Regular Polygons 28. The sum of the interior angles of an n-sided regular polygon is 180 x (n - 2). 29. The measure of each exterior angle of an n-sided regular polygon is 360 -7- n, and the interior angle is its supplement. 30. The segments are drawn from the center of an n-sided regular polygon to the vertices from n congruent isosceles triangles. 31. The ratio of the areas of similar polygons is equal to the square of the ratio of corresponding sides or of the perimeters. Relationships Involving Circles 32. All radii of the same circle or congruent circles are congruent. 33. The diameter is the longest line segment with endpOints on the circle. 34. A radius meets a tangent to a circle at a right angle. 35. Opposite angles of a quadrilateral inscribed in a circle are supplementary. 122 Chapter 5 Word Problems Involving Statistics, Counting, and Probability Many word problems will ask you to analyze data and to make calculations that describe the data. These calculations include familiar statistical ideas such as mean, median, and mode and computing probabilities. A necessary ingredient in all of these is the ability to count objects accurately. In the problems that follow, you will see the most common ways of doing this and the variety of situations that appear most frequently when working with statistics, counting, and probability. Mean, Median, and Mode The words mean and average are synonyms. You have probably computed your test average sometime in your life and know that for a set of scores A sum of scores verage score = number of scores The average score is descriptive of the data in the following way: If all the data were the same, they would all be equal to the mean or average score. For example, if you were in a car traveling at an average speed of SS mph, it would be as if you traveled exactly SS mi every hour of your journey. In reality, you may have traveled different distances during each hour. However, 123 if you divided the total distance by the number of hours of your trip, your computation would be 55. It is useful to remember that the sum of the scores = the average score x the number of scores. Example I Dana's average in math is 82 after taking eight tests. To the nearest whole number, what will her new average be if she earned a 92 on her next test? Solution I Using the relationship described above, the sum of her scores is equal to 82 x 8 = 656. If the new score is a 92, the sum will be 656 + 92 = 748. Therefore, her new average will be 748 -7- 9 = 83.1 or 83. (The best way to check this answer is to calculate the average of eight scores of 82 and one score of 92.) Example 2 Three numbers are in the ratio of 2:4:7. What is the largest of these numbers if their average is 52? Solution 1 The relationship tells us that the sum of the three scores is 52 x 3 = 156. We need to apply the algebra of ratios (see Chap. 3). Let x be the multiple, and we are given the numbers repre- sented by 2x, 4x, and 7x. Therefore, 2x + 4x + 7x = 13x = 156 and x = 12. The largest of the three numbers is 7 x 12 = 84. Check: The other two numbers are 2 x 12 = 24 and 4 x 12 = 48. The average of the three numbers is (24 + 48 + 84) -7- 3 = 156 -7- 3 = 52.1' The median of a set of data is the score that occupies the middle position when the scores are arranged from smallest to largest. In the last example, 48 is the median of the three scores 24, 48, and 84. If the list has an even number of terms, the median is the average of the two in the middle of the list. When the data list is large, it may be presented in the form of a table which indicates the score and the number of 124 times it appears in the list. This number is called the frequency of the score. In Example 2, each score has a frequency of 1. In Example 1, if all the test scores were 82, its frequency would be 8. The mode of the data would be the score with the high- est frequency, that is, the score that appears in the list most often. Example 3 The Fahrenheit temperatures at 8 A.M. were recorded for each day during February in a Canadian city. The data are given in the table below. Find the mean, median, and mode tempera- tures. Solution 3 Temperature _4° _2° 3° 4° 10° Frequency 4 6 9 4 5 The mode is easily seen to be 3° since it has the highest frequency. The median would require listing all the scores in ascending order. By adding the frequencies, we know the total number of scores in the list. In this case, the list has 28 scores. Since this is an even number, we would average the scores in the 14th and 15th positions in the list. -4, -4, -4, -4, -2, -2, -2, -2, -2, -2, 3,3,3,3,3,3,3,3,3,4,4,4,4,5,5,5,5,5 The average of these two scores is the median, 3. Rather than listing all the scores, an easier way to find the median is to count through the frequencies in the table. The -4°s and -2°s occupy the first 10 positions. The 3°s occupy positions 11 to 19, which include the two middle positions. The mean can also be computed from the table. The sum of all the scores is the sum of the products of each score and its frequency. For these data, the sum is -4 x 4 + -2 x 6 + 3 x 9 + 4 x 4 + 10 x 5 = -16 + -12 + 27 + 16 + 50 = 65. If no 125 specific directions are given, it is customary to compute the mean to one more decimal place than that of the orig- inal scores. Therefore, to the nearest tenth, the mean is 65 -;- 28 = 2.3. Example 4 This example demonstrates how the median and the mean can be very different for a set of data. Of James' seven tests this quarter in biology, the median score was 82 and the mean score was 76. If the average of his highest three tests is 90, what is the average of his lowest three scores? Solution 4 Since we know the mean, 76, we can compute the sum of all seven scores; that is, the sum is equal to 76 x 7 = 532. In order to find the average of the three lowest scores, we can let x be their sum. Therefore, x + 82 + 3 x 90 = 532. Multiplying and combining terms, we get x + 352 = 532 and, after sub- tracting 352 from both sides, we see that x = 180. The average of the three lowest scores is 180 -;- 3 = 60. Check: We need to check that the sum of the seven scores we now know is 532: 3 x 60 + 82 + 3 x 90 = 180 + 82 + 270 = 532.-1 Counting Objects Word problems that ask for a count of objects would be very tedious if we had to list all the possibilities to know how many we have. In fact, we saw in the previous examples that count- ing doesn't necessarily have to be done this way if we know how many times a specific object appears in the collection. The following problems ask for a count of the possibilities of making selections from a set of objects. Example 5 To get from class to outside the main entrance of the school, you can use any of three stairways to the first floor, either of two corridors to the main lobby, and either of two doors. How many different paths can you choose from? 126 Solution S This situation can be visualized by a tree diagram or flowchart that diagrams all the possibilities. Stairway 1 / \ C1 C2 ./' '" ./' '" 01 02 01 02 Stairway 2 ./' \ C1 C2 ./' '" ./' '" D1 D2 01 02 Stairway 3 ./' '" C1 C2 ./' '" ./' '" D1 D2 01 D2 (where C1, C2 and 01, D2 represent corridors 1 and 2 and doors 1 and 2, respectively). There are clearly 12 possible ways from the classroom to the outside of the main entrance. Note that the number of branches of the tree is based on the product 3 x 2 x 2 = 12. This is the basis of the Count- ing Principle, which is a popular way to solve such problems. The Counting Principle states that The total number of different selections consisting of picking one item from each of several dif- ferent types is equal to the product of the possibilities for each item selected. Example 6 At the local fast-food restaurant you can choose from eight dif- ferent sandwiches, four different side dishes, and five different beverages. How many different meals can you have consisting of one sandwich, one side order, and one beverage? Solution «5 The Counting Principle leads us to believe that the num- ber of different meals is equal to the product 8 x 4 x 5 = 160. It's that simple! Drawing a tree diagram would take much longer. One way to create a visual for counting problems such as these is to think of filling slots. For the problem described above, we would draw x x Sandwich side order beverage 127 and fill the slots with the number of different choices for each item, namely, 8, 4, and 5. This will help with more complicated problems. Example 7 Student ID numbers all have eight characters. The first two characters are different letters chosen from the letters A through F. The remaining six characters are digits from 1 to 9 which can be repeated. How many different 10 numbers are possible to create? Solution 7 The situation is the same as filling eight slots. The first slot is filled with a letter from a set of six, while the second is filled with a letter from a set of five since we cannot repeat the letter we already used. All the rest are filled with a number from a set of nine each since we can repeat digits. The model is -6- x -5- x -9- x -9- x -9- x -9- x -9- x -9- Letter letter number number number number number number or 6 X 5 X 9 6 = 15,943,230. (A tree diagram would have been impossible to draw!) Sometimes answers such as the previous one will seem incorrect. However, when considering all possibilities, such large numbers will turn up, and they cannot be checked by simple means. Therefore, it is important that you are sure of the way you reasoned through the problem. Example 8 How many ways can four out of seven students be seated in a row of four chairs? How many ways can all seven be seated in a row of seven chairs? Solution 8 Filling slots for the first situation, we have Z X Q x ~ x .1 = 840. For the second situation we continue the product until we reach I, 7 x 6 x 5 x 4 x 3 x 2 x 1. This product is symbolized by 7! and is called "7 factorial": 7! = 5040. 128 Your calculator may have a key for computing factorials, n!. You would enter the number and press the key for the com- putation. Another symbol sometimes used is n P r , where n is the number of objects in the set and r is the number of objects chosen to be arranged. The P stands for the term permutation, which is a synonym for arrangement. The answer to the first question would be symbolized by 7 P4 and the second answer by 7P7, which is 7!. Combinations The previous problems were solved by considering the order in which selections were made. Some situations involve count- ing groups and are not concerned with a specific order. For these problems, we often make use of another symbol, nCr. where C stands for combinations. The formula is nCr =nPr/r!. This makes sense since there have to be less to count if we don't care in what order they appear. For example, the three- letter arrangements abc, acb, bac, bca, cab, and cba are different permutations, but the same combination. Example 9 Mr. Fredricks wants to have a mock election in his social stud- ies class. Students have to elect four people to serve as class officers. If nine students were nominated, how many possible different groups of four could the class elect? Solution 9 If the problem asked for specific officer positions, we would be counting permutations. Since it does not, we will find the number of combinations. We use 9C4 = 9 P 4 = 9 x 8 x 7 x 6 4! 4 x 3 x 2 x 1 Since our answer is going to be a whole number, we should try to reduce before multiplying and eliminate all the factors in the denominator. The fraction reduces nicely to 126, our answer. 129 Sometimes you won't be told explicitly that you are look- ing for combinations. You will often be able to find a way to restate the problem that makes it obvious. Example 10 How many different diagonals can be drawn in a regular oc- tagon (an eight-sided figure with equal sides)? Solution 10 F E G o H c A B If you don't recognize the use of combinations, you can label the vertices A through H and start listing the diagonals sys- tematically: AC, AD, AE, AF, AG, BD, BE, BF, BG, BH, CE, CF, CG, CH, (CA is already in the list), DF, DG, DH, (DA and DB are already in the list), EG, EH, (the rest with E are already in the list), FH (the rest with F are already in the list). We count 20 diagonals. If you think of a diagonal as being a connection between two vertices, the problem is really one of finding all the differ- ent pairs that can be formed from the eight vertices. That is sPz 8 x 7 sCz = 2! = 2 x 1 = 28 However, eight of these pairs are not diagonals, but sides of the octagon. Therefore, the answer is 28 - 8 = 20. Probability Most questions about probability are given as word problems with descriptions of practical situations. Several words are used 130 to help clarify the situations so that a model for probabil- ity problems can be formed. The situation usually describes making selections and is referred to as the experiment. When considering all the ways in which the situation can unfold, there may be several outcomes that are considered to be success- ful in achieving the desired event. The probability is a simple ratio. P b bT f number of successful outcomes ro a 1 lty 0 an event = total number of outcomes This will always be a number ranging from 0 to 1, since the numerator can never be larger than the denominator. Once again, we find ourselves concerned with counting. Therefore, we should expect to make use of the Counting Prin- ciple and, if necessary, the formulas for permutations and com- binations. Example II A bag of candy contains candies of different fruit flavors. Four candies are strawberry-, eight are lemon-, and three are cherry- flavored. If two candies are picked at random without replace- ment, what is the probability that they both will have the same flavor? Solution II To solve this problem, we have to identify the experiment, an outcome, and what constitutes a successful outcome. Experiment: Picking two candies. An outcome: The two flavors picked. A successful outcome: The flavors are the same. We now need to count the total number of outcomes. For this, we can use the Counting Principle. Since we are not replacing the first candy and there are 15 candies in the bag, there are 15 x 14 = 210 different ways to pick 2 of them. To find the number of successful outcomes, we have to add the number of ways we can pick two candies of each flavor. There are 4 x 3 = 12 ways to pick strawberry candies, I J I 8 x 7 = 56 ways to pick lemon candies, and 3 x 2 = 6 ways to pick cherry candies. The number of successful outcomes is, therefore, 12 + 56 + 6 = 74. The probability of picking two candies of the same flavor is 74/210 or, when reduced, 37/105. (Sometimes you may be asked to express the answer as a decimal or percent. This may require rounding. To the nearest thousandth, the answer is 0.352; to the nearest percent, the answer is 35%.) Many probability problems involve familiar objects such as a deck of 52 playing cards, a pair of dice, a circular spinner divided into colored regions, or a two-sided coin. Some points to remember are • A regular deck of 52 playing cards consists of four suits, each having and ace, the numbers 2 through 10, a jack, a queen, and a king. Of these 13 labels, the jack, queen, and king are referred to as picture cards. Jokers are not part of the 52 cards. • A single die is a cube with each face representing a number from 1 to 6. A fair die is one in which there is an equal chance of landing on either side. The top side is considered to be the result of the toss of the die. When a pair of dice is rolled, the sum of the two top sides will be any number from 2 to 12, but there are different probabilities for each of these sums. • A circular spinner is divided into regions by radii drawn from the center. These regions may not necessarily be of equal area. The probability of landing in any region is the ratio of the area of the region to the area of the entire circular region and is directly proportional to the angle formed by the radii that border the region. (The ratio of areas is used for probabilities in any geometric setting.) • A fair coin is one with which there is an equal probability of landing on either side. The sides are usually distinguished by "heads" and "tails," although it could have any kind of markings on either side. The probability of getting either heads or tails when tossing a fair coin is liz. Sometimes we are told that the coin is not fair and there is a bias for one side. 132 Example 12 A card is drawn from an ordinary deck of 52 playing cards and replaced in the deck. A second card is then drawn. What is the probability that both are picture cards? Solution 12 Identifying the components of the problem, we have The experiment: Pick two cards. An outcome: The types of cards picked. A successful outcome: The two cards are picture cards (jacks, queens, or kings not necessarily the same). Since we are replacing the first card, there will still be 52 possibilities for the second pick. Therefore, there are 52 x 52 = 2704 different outcomes. In the deck, there are 12 picture cards and the number of successful outcomes is 12 x 12 = 144. Therefore, the probability is 144/2704. We can also solve the previous problem by a different use of the Counting Principle, which deals with probabilities. Specifically, it states: The probability of an event is equal to the product of the probabilities of being successful at each stage of the experiment. The probability of picking a picture card at each stage is 12/52 or 3/13. Therefore, the probability of picking two picture cards is found simply by 3/13 x 3/13 = 9/169. Note how this helps with expressing the answer in reduced form. Example 13 A fair coin is tossed 5 times. What is the probability of getting five heads? Solution 13 The probability of getting a head on one toss is liz. Using the modified Counting Principle, our answer is simply liz x ) s 5 liz X Ij2 X liz X liz = (liz = 1 '/zS = 1132. Problems involving probabilities sometimes involve data from real-world situations. For these problems, we think of the probability as the fraction or percentage that represents the number of times we are successful. The probability is used as 133 the basis of a prediction or expectation. There is no real difference between this and the previous computations for probability, except that we usually have exact counts to work with. Example 14 The monthly rainfall for a Midwestern (U.S.) city for the past year is given in the table below. How many months would you expect rainfall between 2 and 3 in over a 5-year period? Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Inches 2.9 2.5 3.3 4.2 4.8 4.3 4.1 3.3 3.5 3.1 3.3 2.9 Solution 14 The solution is quite simple. Since 3 months out of 12 meet the condition, the probability of between 2 and 3 in of rain in any month is 3/12 or 25%. Therefore, we expect that over a 5-year period which is 60 months, there will be 25% of 60 or 15 months in which the rainfall is between 2 and 3 in. Summary I. Be sure to identify all situations carefully and determine exactly what it is you are counting. 2. Determine whether you are or are not concerned with the specific order; that is, determine whether you will use permutations or combinations. 3. Describe the experiment, outcomes, and successful outcomes in order to understand the situation. Additional Problems I. So far this season, the average spectator attendance at the nine high school basketball games is 135. Each ticket costs $1.50, and the school expects the total revenue for the year to be $2610. What is the expected average attendance for the remaining three playoff gamesl 2. Six people were raising money for a charity. Jordan and Danny averaged $120, Michelle and Lindsey averaged $105, and Mark and Evan averaged $92. What was the average amount for all sixl 134 3. Five consecutive odd integers are such that the sum of the first three is 60 more than the sum of the last. What is the mean and median of these numbers? 4. Theresa ordered a combo pizza with three different toppings. If there are six toppings to choose from and one of her toppings is mushrooms, how many possible different combo pizzas did she have to choose from? 5. Windstar Airlines flies from Newark to three different cities. From each of these cities, there are six different airlines that will take you nonstop to Dallas. On the return trip, if you do not take the same route, how many different itineraries can you plan starting and returning with Windstar at Newark? 6. Janice is completely unprepared for the quiz on the novel that was read in English class. The quiz consists of 10 multiple-choice questions with four choices per question. She decides to guess at each question. How many different answer sheets could Janice possibly produce? 7. There are 12 runners in a race. How many different outcomes are there for first, second, and third place? 8. The Senate Committee on Armed Services has I I Republicans and 9 Democrats. They decide to form a subcommittee consisting of 4 Republicans and 3 Democrats. How many different subcommittees are possible? 9. During the first 2 years of college at the State University, a student has to take 2 math courses, a science course, and 3 social science courses. How many possible different combinations of courses can a student take if the school offers 6 math courses, 4 science courses, and 10 social science courses for freshmen and sophomores? 10. During the junior year of high school, the English classes must read 3 of 5 novels, 4 of 10 short stories, I of 3 Shakespearian plays, and 5 of 9 poems. What is the likelihood that two different English classes will read the same selections? 1 I. A biased coin is weighted so that the probability of landing heads is 2 in 3. If three coins are tossed simultaneously many times, what percent of the time will they all land tails? 12. In the game of poker, a "flush" is a 5-card hand consisting of cards all of the same suit. From a regular deck of 52 playing cards, what is the probability of picking 5 cards and winding up with a flush? 13. Stacy had to get dressed quickly for school one morning. She reached into her closet for pants and a blouse. She has three pairs 135 of blue pants, two pairs of black pants, and two pairs of brown pants. She also has three white blouses, four blue blouses, and one red blouse. What is the probability that she picked blue pants and a blue 14. The test scores of a class are given in the table below. While going over the test with the class, the teacher randomly picks two test papers out of the pile simultaneously to use as models. What is the probability that these papers will have scores greater than the average of the Scores SS 62 6S 78 83 88 93 100 Frequencies 3 3 S 1 3 9 4 2 15. A circular spinner is divided into four regions of different colors by four radii. If the ratios of the distinct angles formed by the radii are I :3:5: I I, what is the probability of the spinner landing in one of the two smaller 16. On a popular TV game show, the contestant has to spin a wheel. The wheel is divided into 20 regions. All these regions are equal in area except for two which award the player the grand prize. If these regions are each one-fifth the size of the others, to the nearest hundredth, what is the probability that the contestant will win the grand 17. On a circular dartboard whose diameter is 18 in, the bull's-eye in the center has a diameter of 6 in. What is the probability of a thrown dart landing in the 18. On his way out the door to go to school in the morning, Brandon's mother tells him to take three different pieces of fruit from the bin. If there are five bananas, three oranges, and two pears, what is the probability that he will take two bananas and a 19. When emptying her pocket one night, Gayle found that she had eight nickels, five dimes, five quarters, and two pennies. The next 136 day she put these coins in her pocket. At lunchtime, she reached in and took out two coins. What are the chances that she took out less than 35q! 20. In an envelope are 10 straws each measuring a different whole-numbered length from 2 to I I in. When taking three different straws out of the envelope without looking, what is the probability that the three straws could form a Solutions to Additional Problems I. We need to work backward from the expected revenue and compute the expected total attendance for the 12 games. At $1.50 per ticket, the expected total attendance is $2610-;- $1.50 per ticket = 1740 tickets sold. Using the fact that the sum of the attendance figures equals the average attendance times the number of games, we know that the total attendance of the nine games played so far is 9 x 135 = 1215. Therefore, the school needs to sell 1740 - 1215 = 525 more tickets. The average attendance for the remaining three games would be 525 -;- 3 = 175. 2. We need to compute the total amount of money raised by all six people. Jordan and Danny raised $120 x 2 = $240, Michelle and Lindsey raised $1 05 x 2 = $210, and Mark and Evan raised $92 x 2 = $184. The total is $240 + $210 + $184 = $634. Therefore, the average for each person is $634 -;- 6 = $105.67. 3. We need to use some algebra for this. Let x be the first integer. The others are x + 2, x + 4, x + 6, and x + 8. The sum of the first three is x + x + 2 + x + 4 = 3x + 6. Therefore, our equation is 3x + 6 = x + 8 + 60 -+ 2x = 62-+ 3x + 6 = x + 68 -+ x = 31 If the first odd integer is 31, the others are 33, 35, 37, and 39. The mean and median are the same, namely 35. 4. The Counting Principle applies to this problem with three slots to fill. Since we know that one of Theresa's toppings is mushrooms, the first slot has only one choice. The remaining toppings are different. Therefore, the second slot has five choices and the third has four. There are I x 5 x 4 = 20 different combo pizzas that she can create. 137 S. There are four slots to fill in this problem using the Counting Principle. From Newark with Windstar x -- airline to Dallas x -- x airline from Dallas to Newark with Windstar Since you cannot take the same Windstar routes or the same airline to and from Dallas. the number of different itineraries is the product 3 x 6 x 5 x 2 = 180. 6. The Counting Principle is applied with 10 slots. each representing one question with four possible choices for each question. Therefore. the number of different answer sheets is the very large number 4 10 = 1.048.576. (Note that the probability that Janice will answer all the questions correctly is 1/1.048.576 or approximately 0.000000095. a very small chance!) 7. This is a permutation problem since we are interested in the arrangement of the three people who finish in the first three positions. The answer is 12 P3 = 12 x II x 10 = 1320 possible outcomes for the race. 8. The solution to this problem involves both the Counting Principle and combinations. That is. each possible group of Republicans can be joined with any of the possible groups of Democrats. Therefore. the number of different subcommittees will be the product of the two numbers of ways we can pick groups of each. The number of groups of Republicans is C = II P4 = II x 10 x 9 x 8 = 330 I I 4 4! 4 x 3 x 2 x I The number of groups of Democrats is C = 9 P 3 = 9 x 8 x 7 = 84 9 3 3! 3 x 2 x I The number of different subcommittees is 330 x 84 = 27.720. 9. The Counting Principle can be applied with the slots as -----x ------x Math courses Science courses Social science courses There are 138 c _ 6 P 2 _ 6 x 5 _ 15 6 2 - 2! - 2 x I - different ways to choose math courses. There are different ways to choose a science course. There are IOC3= IOP3 = IOx9x8 = 120 3! 3 x 2 x I different ways to choose social science courses. Therefore. there are 15 x 4 x 120 = 7200 different combinations of these courses that a freshman or sophomore can take. 10. The word likelihood is a synonym for probability. The problem needs to be restated in a way that identifies the event. Two classes reading the same selections is the same as already knowing the selection that one class has made and asking. "What is the probability of that being the selection for another c l a s s ~ " The probability is, therefore Number of possible different selections There are sC 3 = 10 different groups of 3 novels. 10C, = 210 different groups of short stories, 3 different Shakespearian plays. and 9C 5 = 126 different groups of poems. Therefore, using the Counting Principle, there are lOx 210 x 3 x 126 = 793,800 different selections to choose from. The probability is I -7- 793,800 = .000001259 or there is almost no likelihood that this will happen! 1 I. A key fact about probability is that the sum of all possible outcomes is I. In other words. if any outcome is considered to be successful, the numerator and denominator of the probability fraction must be equal. Since the only outcomes for tossing a coin are heads and tails and we are given the probability that landing heads is 2/3, the probability of landing tails is 1/3. We use the modified Counting Principle for probabilities and find the probability of three tails on one toss ofthe three coins to be 1/3 x 1/3 x 1/3 = 1/27 or 0.037. Therefore, we can expect this to occur 3.7% of the time. 12. A "flush" is a group of cards of the same suit in no specific arrangement. Therefore. for anyone suit, the number of different 139 flushes is 13C5 = 1287. Since there are four different suits, there are 4 x 1287 = 5 148 different flushes. The total number of different five-card hands is 52C 5 = 2,598,960. Therefore, the probability is 5148 -:- 2,598,960 = 0.00198. (Since this number is nearly 0, there is an extremely small chance that this will occur!) 13. Identifying the components of the problem, we have the experiment-picking pants and picking a blouse (each pick represents a different stage of the experiment); an outcome--picking any of seven pairs of pants and any of eight blouses; and a successful outcome--picking any of three pairs of blue pants and any of four blue blouses. We can use the modified Counting Principle for Probabilities, which states that the probability of a successful outcome is the probability of picking blue pants x the probability of picking a blue blouse. Using combinations, we have the probability of picking blue pants = 3/7 and the probability of picking a blue blouse = 4/8 = 1/2. The answer is, therefore, 3/7x 1/2=3/14. 14. Identifying the components of the problem, we have the experiment-picking two papers without replacement (each pick represents a different stage of the experiment); an outcome--picking any pair of papers; and a successful outcome--picking a pair greater than the average. To find the average score, we have to get the sum of the scores by mUltiplying each score by the number of times it appears in the set of 30 and then add the products. This is easiest to do by adding a third column to the table which will contain the products and adding a row at the bottom that will contain the sums of the second and third columns. 140 Score 55 62 65 78 83 88 93 100 Sums Frequencies 3 3 5 1 3 9 4 2 30 Products 165 186 325 78 249 792 372 200 2367 The average is 2367 -:- 30 = 78.9. The scores of 83,88,93, and 100 are all greater than the average. The frequencies indicate that there are 18 of these scores. Therefore, the probability of the first paper having a score greater than the average is 18/30. Since the second paper is pulled out at the same time as the first, we lose one possibility from those greater than the average and from the total pile. Therefore, its probability would be 17/29. The modified Counting Principle for Probabilities states that the probability of both papers having scores greater than the average = 18/30x 17/29 = 0.35, rounded. Note that this answer is equivalent to (18 x 17)/(2 x I) = (30 x 29)/(2 x I) 18xl7 2xl x - ------ 2 x I 30 x 29 30 x 29 18 x 17 which indicates that the problem could have been solved by finding the ratio of (a) the number of ways to select 2 from the 18 scores that are greater than the average to (b) the number of ways of selecting 2 scores from the entire set of 30. 15. The probability of a spinner landing in a region is the ratio of the area of the region to the area of the circle. The ratio of the areas of the regions is directly proportional to the ratio of these central angles. While it is not necessary to compute the angles, it is helpful in understanding the situation. Using the algebra of ratios, let x be the multiple and let the angles be x, 3x, 5x, and I I x. Their sum is 20x. Since the entire rotation about the center of the circle is 360 0 , we have 20x = 360 and x = 18. Therefore, the two smaller angles are 18 0 and 54 0 • Together this is 72/360 of the 141 rotation. The probability of landing in either of the two smaller regions is, therefore, 72/360 = 2/ 10 or 0.2. 16. In this problem, we are given information about the areas of the region. The probability of a spinner landing in a region is the ratio of the area of the region to the area of the circle. We can use algebra to determine this ratio. Since the smaller regions are one-fifth the size of the others, we let x be the area of each of the smaller regions and Sx be the area of each of the other 18 regions. The total area of the regions we wish to land in is 2x, and the total area of the wheel is sx x 18 + 2x = 92x. Therefore, the probability of landing in either of the smaller regions is 2x /92x = 1/46 or, to the nearest hundredth, 0.02. 17. This problem is similar to Probs. 15 and 16 as it involves finding the ratio of areas. (This, in fact, will be true of any problem involving interior regions of a geometric shape. Such problems are referred to as problems of geometric probability.) The probability of the dart landing in the bull's-eye is the ratio of the area of the bull's-eye to the area of the dartboard. The "bull's-eye" is a circle whose center is that of the larger circular dartboard. Since we know the diameters of each circle, 6 and 18 in, we know that the radii are 3 and 9 in. The area of the bUll's-eye is 9n and the area of the entire dartboard is 81 n. Therefore, the probability is 9n 781 n = 1/9. Note that this is the square ofthe ratios ofthe radii: (3/9)2 = (1/3)2 = 1/9. This will always be true about the ratio of areas of two circles. 18. This problem can be solved using the same idea as in Prob. 8, using combinations. The probability is the ratio of the number of ways Brandon can pick 2 bananas and I pear to the number of ways he can pick any 3 fruits from the collection of 10 fruits. The 142 answer is 10 x 2 120 20 120 I 6 19. When all else fails. a probability problem can be solved by listing all possible outcomes. In this case a table whose column and row headings indicating the different coins can be used. Each cell contains the total value of the two coins. Note that the diagonal contains no values since we cannot use the same coin twice. p p n n n n n n n n d d d d d q q q q q p X 2 6 6 6 6 6 6 6 6 II II II II II 26 26 26 26 26 P 2 X 6 6 6 6 6 6 6 6 II II II II II 26 26 26 26 26 n 6 6 X 10 10 10 10 10 10 10 15 15 15 15 15 30 30 30 30 30 n 6 6 10 X 10 10 10 10 10 10 15 15 15 15 15 30 30 30 30 30 n 6 6 10 10 X 10 10 10 10 10 15 15 15 15 15 30 30 30 30 30 n 6 6 10 10 10 X 10 10 10 10 15 15 15 15 15 30 30 30 30 30 n 6 6 10 10 10 10 X 10 10 10 15 15 15 15 15 30 30 30 30 30 n 6 6 10 10 10 10 10 X 10 10 15 15 15 15 15 30 30 30 30 30 n 6 6 10 10 10 10 10 10 X 10 15 15 15 15 15 30 30 30 30 30 n 6 6 10 10 10 10 10 10 10 X 15 15 15 15 15 30 30 30 30 30 d II II 15 15 15 15 15 15 15 15 X 20 20 20 20 35 35 35 35 35 d II II 15 15 15 15 15 15 15 15 20 X 20 20 20 35 35 35 35 35 d II II 15 15 15 15 15 15 15 15 20 20 X 20 20 35 35 35 35 35 d II II 15 15 15 15 15 15 15 15 20 20 20 X 20 35 35 35 35 35 d II II 15 15 15 15 15 15 15 15 20 20 20 20 X 35 35 35 35 35 qUUmmmmmmmm3535HHHXmmmm qUUmmmmmmmm3535HHHmXmmm qUUmmmmmmmmHH35HHmmXmm qUUmmmmmmmm35H35HHmmm q 26 26 30 30 30 30 30 30 30 30 35 35 35 35 35 50 50 5 0 L - - · ~ - There are 20 x 20 - 20 = 380 filled cells (note that this is 20 x 19). Of these, 70 are not successful; therefore, the probability is 310/380 = 31/38, or the chance that Gayle took out less than 35¢ is approximately 0.82. Identifying the components for this problem, we have the experiment-picking two coins; an outcome-pick any two coins totaling any amount; and a successful outcome-picking two coins whose total value is less than 35¢. The total number of 143 outcomes is c _ 20 P 2 _ 20 x 19 = 190 20 2 - 2! - 2 x I which will be the denominator of the probability ratio. (Note that we are not considering the order in which we pick the coins; that is why this is half of the number we saw in that table.) It should be clear from the problem that there are considerably fewer ways to pick two coins unsuccessfully than successfully. We can find the number of ways to do this and subtract. Any pick of two quarters is unsuccessful and there are SP2 5 x 4 SC2 = - = -- = 10 2! 2 x I ways to do this. Any pick of a quarter and a dime will be unsuccessful, and there are 5 x 5 = 25 ways of doing this. Therefore, our probability is (190 - 35)/190 = 155/190 = 31/38. 20. The experiment-picking three different straws; an outcome--three different lengths; and a successful outcome--the three lengths can represent the sides of a triangle. At first thought, you might say that the probability should be I, thinking that you can form a triangle from any three lengths. This, however, is not true! A basic fact about triangles is that the sum of any two sides must be greater than the third side. Since there is more to consider than just counting the number of different groups of 3, it would be reasonable to find a systematic way to list the combinations and pick out the ones that work. Note that there are loe 3 = 120 possible different triplets that can result and we should list the possibilities in a way that makes it easy to see that we have considered all possibilities. Therefore, we can start a table and include in the cells only those possible third sides which were not considered previously. For example, we can start with supposing that our first two picks are 2 and 3. We can have a side of 4, but anything higher would not be less than 2 + 3. When moving down the column, we go to 2,4. A possible third side is 3, but we already listed the triplet 2,3,4 in the row above. The only other remaining possibility is 5. We move along through the 2s until 2, I I, which would add no additional successful triplets, and then to the next column, where we consider the pairs, including a 3. 144 - ~ VI First Possible First Possible First Possible First Possible First Possible First Possible First Possible First Possible First Possible two third two third two third two third two third two third two third two third two third 2,3 4 2,4 5 3,4 5,6 2,5 6 3,5 6,7 4,5 6,7,8 2,6 7 3,6 7,8 4,6 7,8,9 5,6 7,8,9,10 2,7 8 3,7 8,9 4,7 8,9,10 5,7 8,9,10,11 6,7 8,9,10,11 2,8 9 3,8 9,10 4,8 9,10,11 5,8 9,10,11 6,8 9,10,11 7.8 9,10,11 2,9 10 3.9 10,11 4,9 10,11 5,9 10,11 6,9 10,11 7.9 10,11 8,9 10,11 2,10 11 3,10 11 4,10 11 5,10 11 6,10 11 7,10 11 8,10 11 9,10 11 2,11 X 3,11 X 4,11 X 5,11 X 6,11 X 7,11 X 8,11 X 9,11 X 10,11 X We can now count the number of entries in the "Possible third" column 70. Therefore, the probability of picking three straws that could form a triangle is 70/120 = 7/12 or approximately 0.58. 146 Chapter 6 Miscellaneous Problem Drill This collection of problems will give you an opportunity to use the skills, concepts, and strategies discussed in the previ- ous chapters. If you need help, you can look back at the solu- tions to the examples and additional problems in the chapter mentioned before each set of problems. The answers are given at the end. Good luck! Problems Similar to Those in Chap. I I. Jim and Donna would like to put a fancy molding around their den. Jim likes the molding that costs $2.S0/ft, while Donna prefers the molding that costs $1.7S/ft. If the rectangular den is 15 ft x 12 ft, how much more would Jim be willing to pay for molding than Donna wouldl 2. The distance from homeplate to first base at the local schoolyard is 60 ft, and the entire diamond (which is really a square) is covered with grass except for the pitcher's circle, which is dirt. If the grass covers approximately 3550 ft2, what is the diameter of the pitcher's circle to the nearest footl 3. Laura walked to the store from her house in 20 minutes and then from the store to her friend's house, which was I mi away, in 15 minutes. If she walked at the same rate on both trips, how far is her house from the storel 4. When Danny first arrived in Italy on his trip, he weighed himself and found that he weighed 63 kg. When he returned home from his trip, he weighed himself again and found that he then weighed 142 lb. How many pounds did Danny gain or lose while in ltalyl 147 5. A chef had to prepare soup for 50 people and used the entire capacity of the pot. If each soup bowl holds cup of soup, how many second servings of soup could the chef serve, if the pot has a 3-gal 6. Last year Gabriella invested $3500 in a mutual fund that had an annual return of 12 percent. This year she invested all the principal and interest from that fund into another fund that earned only 6 percent. How much money does she have 7. In November, Josh wanted to buy a football helmet that cost $49.99. The salesperson told him that after the football season in February the helmet would sell for 15 percent less. However, there was another helmet in the store that sold for $42.99. What should Josh do in order to pay the least amount for a helmed Problems Similar to Those in Chap. 1 8. The larger of two numbers is 8 less than 3 times the smaller. Given that the sum of the number:s is 88, find the larger number. 9. Find four consecutive odd integers whose sum is 96. 10. The sum of the squares of the first and last of three consecutive positive integers is 164. Find the three integers. I I. The denominator of a fraction is 3 more than the numerator. If 3 is added to both the numerator and the denominator, the new fraction has a value of 0.7. Find the original fraction. 12. The tens digit of a two-digit number is 3 less than the units digit. The number is 3 more than 4 times the sum of the digits. Find the number. 13. Today Allen is 8 years older than Harvey. In 6 years he will be twice as old as Harvey will be then. How old is Harvey 14. Jeff and Steve left their homes, which are 21 mi apart, at the same time and bicycled along the same straight road toward each other. If Jeff's speed was 4 mph faster than Steve's, how fast was Jeff traveling if they met each other in 45 15. A nuclear reaction projects an atomic particle along a straight line at 900 m per second (m/second). A second particle is projected 5 seconds later along the same path at a speed of 1200 m/second. In how many seconds into its trip will the second particle pass the firsd 16. The prices of two kinds of coffee are $1.75/Ib and $3.40/Ib. The owner of the store creates a 50-lb mixture of the two that sells for 148 $2.25. Assuming that all 50 Ib are sold, how much, to the nearest pound, of the cheaper coffee should she use in the mixture to make sure that she makes the same money as if she sold the coffees separatelyl 17. Amy has twice as many nickels as quarters and two more dimes than nickels. If she has $1.85 in change, how many dimes does she havel 18. Carole is asked to do a job that ordinarily takes her 45 minutes to complete. Melissa usually does the same job in 30 minutes. If they work together, how long will it take for the job to be completedl 19. If two pumps are used to fill a tank at the same time, the tank is filled in 3 hours. The faster pump fills the tank by itself in 5 hours. This morning both pumps were turned on at the same time and left to fill the tank. After I ~ hours, the slower pump stopped working. How much time did it take for the other pump to complete filling the tankl 20. Phyllis invested part of her $10,000 inheritance in a fixed annuity earning 7 percent annually. She invested the rest of her money in various stocks that were expected to earn 8 percent, but actually earned 12 percent over the year. How. much did she invest in both situations if her earnings for the year were $1 050l Problems Similar to Those in Chap. 3 21. A school district boasts that there are three teachers for every 70 students. If the total number of teachers and students in the school is 4015, how many students are in the schooll 22. Separate 144 into two parts so that the ratio of the two numbers is 5:7. 23. The ratio of two positive numbers is 4: I, and the sum of their squares is 153. Find the numbers. 24. The ratio of Gayle's age to Danielle's age is 3:2. In 10 years, Gayle's age will be 22 less than twice Danielle's age. How old is Gayle nowl 25. Lindsey, Kristen, and Nicole ran for student body president. The ratio of votes received respectively were 2:5:6. If 2600 votes were cast, by how many votes did Nicole win the electionl 26. A recipe in a cookbook calls for ~ cup of sugar and 3 cups of flour to make a cake that serves eight people. Joan was expecting more than eight people and used 10 cups of flour. How many cups of sugar did she have to usel 149 27. In June the final exam in math given to all ninth graders is graded by two teacher committees. Committee A usually grades 30 more papers in an hour than committee B. In the time it took for all the papers to be graded, committee A graded 350 papers while committee B graded 280 papers. On the basis of these results, how many papers can each committee grade in one hour, and how many hours did it take for all the papers to be gradedl 28. The chemist knows that the pressure of a gas within a container varies directly with the temperature in the container. If the pressure is measured at 35 Ib/in 2 when the temperature is 18°C, how much higher, to the nearest tenth of a degree, must the temperature be raised so that the pressure reaches 40 Ib/in 2 l 29. In the manufacturing process of a cleaning fluid, 180 kg of a dry base is added to 500 L of water. The mixture is heated so that it becomes a 40 percent basic solution by weight. How much water must be eliminated in this process to achieve the proper mixturel Problems Similar to Those in Chap. 4 30. In a triangle, the measure of the largest angle is 16 0 more than twice the measure of smallest angle and the third angle is 20 0 more than the smallest. What are the measures of all three anglesl 31. The measures of two consecutive angles of a parallelogram are in a ratio of 4:5. Find the measure of the larger angle. 32. The perimeter of an isosceles triangle is 52 cm, and each leg is I 0 cm less than the base. What are the lengths of the three sides of the trianglel 33. At 3 P.M., a 5 Y2-ft-tall person casts a shadow of 8 ft. A nearby tree is 21 ft tall. To the nearest tenth of a foot, how long is the tree's shadow at 3 p.M.l 34. One of the legs of a right triangle is 17 in more than the other. Find the length of the larger leg, given that the hypotenuse is 25 in long. 35. A rigid cable is needed to support a 12-ft antenna on the roof of a building. The cable is to be attached to the top of the antenna and to a point 7 ft from the base of the antenna. To the nearest foot, how much cable must be available to accomplish thisl 36. The parallel sides of an isosceles trapezoid are 16 and 30 cm. If the area of the trapezoid is 184 cm 2 , how long, to the nearest tenth of a centimeter, are the nonparallel sidesl 150 37. In the trapezoid described in the previous problem. what are the measures. to the nearest degree. of the four interior angles of the trapezoid? 38. A 15-ft ladder is leaning against a building at an angle of 39° with the ground. How high. to the nearest tenth of a foot. along the side of the building does the top of the ladder reach? 39. A floor design in the lobby of a building is in the shape of a rhombus. and its diagonals are laid out in red tile. There are as many floors in the building as there are whole degrees in the smaller angle of the rhombus. If the diagonals measure 6 and 8 m. how many floors are in the building? 40. From a point 25 ft from where a kite is directly over the ground. the angle of elevation of the kite is 32°. How long. to the nearest foot. is the string attached to the kite? Problems Similar to Those in Chap. S 41. Jennifer has been improving in math. Her last three test scores were in the ratio of 4:5:7. What was her lowest score if her test average is 72? 42. The weights of some of the members of the wrestling team are 172. 212. 185. 183. 20 I. and 179 lb. Find the median and mean weights of this group. 43. In an experiment. the times of many chemical reactions were recorded and are given in the table below. Find the mean. median. and mode of the reaction times. Time, Number seconds of reactions 23 31 2S 46 28 S2 31 61 3S 10 I S I 44. Mike went to the video store to rent some newly released movies to watch over the weekend. Among the new releases there were four comedies, six action movies, and three dramas. If he wanted one of each, how many different possible groups of three movies could he rent? 45. Among the 12 students in the club, 3 have to be selected to be the officers. How many different groups of 3 students can be chosen to be officers? 46. Four cards are drawn from a standard deck of playing cards without replacing a card after it is drawn. If the first three cards are hearts, what is the probability that the next card drawn is not a heart? 47. "Rolling doubles" means that when two dice are tossed, the up sides are the same. In a certain game, you lose a turn whenever doubles are rolled. How many times should you expect to lose a turn if you toss the dice 78 times during the game? 48. On a circular spinner whose radius is 6 cm, the blue region occupies 911" cm 2 , the green region occupies 1211" cm 2 , and the red and yellow regions equally occupy the rest of the area. What is the probability of the spinner stopping in the red region? 49. A group of seven boys and five girls went to the amusement park. Only four people were able to get on the last remaining car on the roller coaster. To the nearest hundredth, what is the probability that the four people consisted of two boys and two girls? 50. In a bag of cookies, there are four chocolate-chip cookies, eight sugar cookies, and six oatmeal cookies. Steve reaches into the bag without looking and takes out two cookies. What is the probability that they are both chocolate-chip cookies? I. 2. 3. 4. 5. 6. $40.50. 8 ft. I ~ m i . Answers to Chap. 6 Danny gained 3.4 lb. 14 bowls of soup. $4155.20. 7. Josh should wait because the discounted price on the helmet will be $42.50. 8. 64. 152 9. 21. 23. 25. and 27. 10. 8.9. and 10. II. 4f7. 12. 47. 13. 2 years old. 14. 16 mph. 15. 15 seconds. 16. 35 lb. 17. 8 dimes. 18. 18 minutes. 19. 2 ~ hours. 20. $3000 in the annuity and $7000 in the stocks. 21. 3850 students. 22. 60 and 84. 23. 12 and 3. 24. Gayle is 36 years old now. 25. 200 votes; Nicole received 1200 votes. Kristen received 1000 votes. and Lindsey received 400 votes. 26. I ~ cups of flour. 27. Committee A grades 150 papers per hour; committee B grades 120 papers per hour. All the papers were graded in 2 ~ hours. 28. 2.6°C. 29. 230 L. 30. 36° , 56° , and 88°. 31. 100°. 32. 14. 14. and 24 cm. 33. 30.5 ft. 34. 24 in. 35. 14ft. 36. 10.6 cm. 37. 49°,49°,131°,131°. 38. 9.4 ft. 39. 73 floors; the angle is approximately 73.7°. 40. 29 ft. 41. 54. 42. The median weight is 184 Ib; the mean weight is 188.7 lb. 43. The mean is 27.8 seconds. the median is 28 seconds. and the mode is 3 I seconds. 44. 72. 45. 220. 153 46. 39/49. 47. 13. 48. 5/24 or 0.208333 .... 49. 0.42 SO. 12/306 or 2/51 or approximately 0.04. 154 Appendix A Brief Review of Solving Equations The information below will refresh your memory about some of the most important facts about solving equations. The two most common types of equations that you need to solve are linear equations and quadratic equations. For each there are several steps to follow that will ensure that you get the correct answer every time. Linear Equations An equation is linear if the highest power of the variable is 1 such as 4x - 5 = 11. In fact, you won't even see the power since we rarely write exponents of 1. The goal is to undo all the complications in the equations and work your way to having the variable appear on only one side, with a numerical term on the other. The following example illustrates the necessary steps involved: Example. Solve the equation 4(2x + 5) + 6x = 8x - 4 - 2x for the value of x. Step 1 Use the distributive property to "remove" any parentheses in the equation. Step 2 Combine like terms on each side of the equation separately. 8x+20+6x = 8x-4-2x 14x+20 = 6x-4 At this point the variable should appear at most once on each side, and a number should appear at most once on each side. 155 Step 3 Remove the variable with the smaller coefficient by adding its signed opposite to both sides of the equation, and combine the variable terms. Step 4 Remove the numerical term from the side of the equation with the variable by adding its signed opposite to both sides of the equation and combine the numerical terms. Step 5 Divide both sides of the equation by the coefficient of the variable and simplify. 14x + (-6x) + 20 = 6x + (-6x) - 4 8x+20 =-4 8x+20+(-20) = -4+(-20) 8x = -24 8x -.;..8 = -24-.;..8 x =-3 4(2x + 5) + 6x = 8x - 4 - 2x Step 6 Check your answer in the original equation by using it in place of the variable. 4(2 x (-3)+5)+6 x (-3) = 8 x (-3) -4-2 x (-3) 4(-6+5)+-18 -24-4+6 4 x (-1) + -18 - 22 -4 + -18 -22 ..( Example. Solve the equation 4(y - 7) - 2(8 - y) = 6(y + 3) - 3 Y + 1. Step 1 Use the distributive property to "remove" any parentheses in the equation. Step 2 Combine like terms on each side of the equation separately. Step 3 Remove the variable with the smaller coefficient by adding its signed opposite to both sides of the equation, and combine the variable terms. Step 4 Remove the numerical term from the side of the equation with the variable by adding its signed opposite to both sides of the equation, and combine the numerical terms. Step 5 Divide both sides of the equation by the coefficient of the variable and simplify. 156 4y - 28 - 16 + 2y = 6y + 18 - 3y + 1 6y-44=3y+19 6y + (-3y) - 44 = 3y + (-3y) + 19 3y - 44 = 19 3 y - 44 + 44 = 19 + 44 3y = 63 3y -.;.. 3 = 63 -.;.. 3 y = 21 Step 6 Check your answer in the (Try the check yourself. Each side has a original equation by using it in value of 82 when y = 21.) place of the variable. Key Fact: You can add, subtract, multiply, or divide only if you perform the same operation on each side of the equation. Simultaneous Linear Equations with Two Variables A pair of equations are said to be a simultaneous pair if you are told to look for the values of each variable that satisfy both of the equations. Only one pair of numbers will work. For example, the pair of numbers x = 7 and y = 2 is the only pair that will satisfy the equations x + y = 9 and x - y = 5 at the same time. We usually write the solution as an ordered pair, (x, y). In this 'case the solution is (7, 2). Simultaneous equations arise frequently in word problems that involve two unknowns. There are many methods for solving simultaneous pairs of equations. The two most common methods are by substitution and by combining. Substitution is usually used when one equation lets you easily write one variable in terms of the other. Example. Solve the following pair of simultaneous equations for x and y: 2x+ y= 6 Step 1 Solve one of the equations for one of the variables. Step 2 Use the expression for the solved variable in its place in the other equation. Step 3 Solve the "new" equation for its variable. Step 4 Use this variable to find the other variable from the equation you used in Step 1. y - 3x = -14 The second equation is easily solved for y by adding 3x to both sides, giving y=3x-14. Substituting 3x - 14 for y in the first equation gives 2x + 3x - 14 = 6. Combining and solving for x, we have Sx -14 = 6 Sx = 20 x=4 y-3x = -14 y-3(4) = -14 y-12=-14 y =-2 157 Step S Write your solution. Step 6 Check that your solution satisfies both equations. The solution is the pair x = 4 and y = - 2 or, as an ordered pair (x. y), (4, -2). In the first equation 2(4) + (-2)=6 8 + (-2) ,( 6 In the second equation -2 - 3(4) = -14 -2-12 -14 ,( Example. Solve the following pair of simultaneous equations for a and b: 3a - b = 20 Step 1 Solve one of the equations for one of the variables. Step 2 Use the expression for the solved variable in its place in the other equation. Step 3 Solve the "new" equation for its variable. Step 4 Use this variable to find the other variable from the equation you used in step 1. Step S Write your solution. Step 6 Check that your solution satisfies both equations. 2a+b = 15 The second equation is easily solved for b by adding -2a to both sides, giving b = -2a + 15. Substituting -2a + 15 for b in the first equation gives 3a - (-2a + 15) = 20. Combining and solving for x, we have Sa -15 = 20 Sa = 35 a = 7 2a +b = 15 14+b = 15 b = 1 The solution is the pair a = 7 and b = 1 or, as an ordered pair (a. b), (7, 1). In the first equation 3(7) - 1 = 20 21-1 20 ,( In the second equation 2(7) + 1 = 15 14 + 1 15 ,( Solving by combining is sometimes easier as it immediately eliminates one of the variables. Using the last example, we can add the given equations to eliminate b. 3a-b= 20 +2a+b = 15 Sa = 35 and we immediately see that a = 7. We can continue by using 158 this value in one of the equations to find the value of the other variable as in step 4 above. If the equations can't be added to eliminate one of the variables, we can "fix" the situation by doing a bit of multi- plying first. Example. Solve the following pair of simultaneous equations for x and y: 3x + 2y = 36 Step 1 Select one of the variables to eliminate and multiply each equation by the coefficient of this variable from the other equation. Step 2 Multiply both sides of one of these "new" equations by -1. Step 3 Add the equations to eliminate the variable, and solve the resulting equation for the other variable. Step 4 Use this value in either of the two original equations to find the value of the other. Step 5 Write your solution. Step 6 Check that your solution satisfies both equations. Quadratic Equations 4x - Sy = 2 If we choose to eliminate x, we will multiply the first equation by 4 and the second by 3, giving us the pair 12x + 8y = 144 and 12x - lSy = 6. Multiplying both sides of the second equation by -1 gives us the pair 12x + 8y = 144 and -12x + lSy = -6. Adding the two equations gives us 23 y = 138 and y = 138 -;. 23 = 6. Using y = 6 in the first equation, we have 3x + 12 = 36, 3x = 24, and x =8. The solution is the pair x = 8 and y = 6 or as an ordered pair (x, y), (8, 6). (Try the check yourself as practice.) In a quadratic equation, the highest power of the variable is 2. For example, the following are quadratic equations: x 2 = 25, y2 - Sy = 0, and n 2 - 6n + 8 = O. Key Fact: Every quadratic equation can be solved for two val- ues of the variable. Example. x 2 = 25 has the obvious solutions x = 5 and x = -5. Example. y2 - 5 Y = 0 has the solutions y = 0 and y = s. 159 Example. n 2 - 6n + 8 = 0 has the solutions n = 2 and n = 4. We sometimes find that these two solutions are equal. Example. x 2 - 8x + 16 = 0 has the solutions x = 4 and x = 4. The most common method of solving a quadratic equa- tion is by factoring using the steps described below. Example. Solve for all values of x that satisfy x 2 = 25. Step 1 Rewrite the equation so that one This is accomplished by adding -25 to side is O. both sides: x 2 + (-25) = 25 + (-25) or x 2 - 25 = O. Step 2 Factor the quadratic expression. This particular quadratic is a difference of two perfect squares, which factors as (x + 5)(x - 5) = o. The next step uses the fact that if the product of two expres- sions is 0, then one of the factors must be O. (How else could you multiply two numbers to get O?) Step 3 Set each factor equal to 0, and We have the two equations, x + 5 = 0 solve for the variable. and x - 5 = 0, which have the solutions x = -5 and x = +5. Step 4 Check each solution in the Clearly, x 2 = 25 when x is +5 or -5. ./ original equation. Example. Solve for all values of y that satisfy y2 - 5y = O. Step 1 Rewrite the equation so that This is already the case: y2 - 5y = O. one side is O. Step 2 Factor the quadratic expression. Step 3 Set each factor equal to 0, and solve for the variable. Step 4 Check each solution in the original equation. 160 Both terms of the particular quadratic have the common factor y, so it factors as y(y - 5) = O. We have the two equations, y = 0 and y - 5 = 0, which have the solutions y = 0 and y = +5. Using y = 0, we have 0 2 - 0(5) = 0 0-0 o ./ Using y = 5, we have 52 - 5(5) = 0 25-25 o ./ Example. Solve for all values of n that satisfy n 2 + 8 = 6n. Step 1 Rewrite the equation so that one side is O. Step 2 Factor the quadratic expression. Step 3 Set each factor equal to 0, and solve for the variable. Step 4 Check each solution in the original equation. This is done by adding -6n to both sides, giving us n 2 - 6n + 8 = O. This particular quadratic factors as (n - 2)(n - 4) = O. We have the two equations, n - 2 = 0 and n - 4 = 0, which have the solutions n = +2 and n = +4. Using n = 2, we have 22 + 8= 6(2) 4 + 8 12 12 .; Using n = 4, we have 4 2 + 8= 6(4) 16+ 8 24 24 .; Example. Solve for all values of x that satisfy x 2 = 8x - 16. Step 1 Rewrite the equation so that This is done by adding -8x and +16 to one side is O. both sides, giving us x 2 - 8x + 16 = O. Step 2 Factor the quadratic expression. This particular quadratic factors as (x - 4)(x - 4) = O. Step 3 Set each factor equal to 0, and We have the two equations, x - 4 = 0 solve for the variable. and x - 4 = 0, which have the equal solutions x = 4. Step 4 Check each solution in the Usingx = 4, we have4 2 = 8(4) - 16 original equation. 16 32 - 16 16 .; Key Fact: There are quadratic expressions that do not fac- tor. To solve these, you need to apply the following special quadratic formula. The quadratic expression ax 2 + bx + c = 0 has the two solutions: -b±Jb 2 - 4ac x = ------------- 2a (where a, b, and c represent the numbers that appear in the expression in those specific places; i.e., they are the coefficients of the terms). This is beyond the scope of this book, but the following example will show how it can be used. The equation x 2 - 9x- 22 = 0 has the coefficients a = 1, b = -9, and c = -22. These 161 values are substituted into the formula, and we have -(-9) ± )(-9)2 - 4(1)(-22) x = 2(1) 9±J81+88 9 ± J169 9 ± 13 = - - - - : ~ 222 = 9 + 13 = 22 = 11 2 2 9 -13 -4 x- ----2 - 2 - 2 - You should check these values to verify that they are solu- tions. Also, this quadratic, x 2 - 9x - 22 = 0, can be solved by factoring, and you should do so as practice. 162 Index The index below includes the most relevant pages to the items listed. Within the statements of problems or their solutions, there may be ideas related to these items not included here. Adjacent angles, 120 Age problems, 35-37, 49, 69, 82, 148 Algebraic solutions, 29-63 Altitude, 99, 120 Angle measurement, 91-94 Angle of elevation, angle of depression, 103-105, 110 Apothem, 117 Area definition, 4 Area formulas, 26 Area problems, 16, 79,96 Average (mean), 123-126 Average problems, 124, 134, 151 Bisector, 120 Capacity, 11, 148 Centigrade (Celsius), 13 Circles, 7, 17, 122, 147 Circumference, 7,26 Coin problems, 46, 50, 136, 149 Combinations, 129, 138 Complementary angles, 111, 120 Cone, 17 Congruent figures, 120 Consecutive integers, 33, 49, 148 Converting measurements, 2-6, 10-12,25 Cosine, 101 Cost, 14,44 Counting, 126-130 Counting problems, 135, 152 Counting principle, 127, 138 Counting principle, for probability, 133,140 Cubic units, 9 Diameter, 7 Direct variation, 77, 150 Discount, 14 Distance formula, 4, 26, 38 Distance measuring, 2 Distance problem, 17, 75, 147 Dry measure, 11 English system, 2 Equations with two variables, 32, 45, 51,90,157-159 Equilateral triangle, 121 Estimating, 3 Experiment, 131 Exterior angles, 121 Factorial, 128 Fahrenheit, 13 Formulas for measurement, 26-27 Fractional equations, 42 Geometric probability, 142 Geometry, 91-122 Geometry, table of relationships, 120-122 Interest, compound, 23 Interest, simple, 15 Investment problems, 15, 18,50,78,82, 148-149 163 Isosceles triangle, 109-110, 113, 121, ISO Likelihood, 139 Linear equations, 155-159 liquid measure, 10 Mass, 12 Mean (average), 123-126, 151 Measurement, 1-13 Measurement problems, with proportion, 73, 81 Median, 120-126, 151 Metric system, 2-13 Mixture problems, 43-46, 49, 71, 80-82, 149 Mode, 123-126, 151 Money-related formulas, 27 Motion problems, 37-41, 49, 82, Ill, 118,148 Number problems, 29-35, 48, 67-69, 81-82, 135, 148-149 Outcome, 131 Parallel lines, 93, 112, 120 Parallelogram, 111, 119, 122, ISO Pentagon, 110 Percentage, 65-90 Percentage, defined, 78 Perimeter formula, 2, 26 Perimeter problems, 16, 70, 82, 147 Permutations, 129, 138 Pi,7 Polygons, 94-97 Principal, IS Probability, 130-146 Probability problems, 135-136, 152 Proportion, 65-90 Proportion, defined, 72 Proportion problems, 95, 149-150 Pythagorean Theorem, 97-101, 114, 121 Pythagorean triples, 98 Quadratic equation, 34, 43, 51, 68, 112, ISS, 159-161 Quadratic formula, 161 164 Radius, 7 Rate, 38-42 Ratio, 65-90 Ratio problems, 91, 109, 113, 124, 141, 149-151 Regular polygon, 118, 120-122 Rhombus, 100, 122, 151 Right triangles, 97-109 Right triangle problems, 110, ISO Right triangles, special, 107-109 Rounding decimals, 1 Sales tax, 14 Scale model, 82 Shopping problems, 18 Similar figures, 74,95, 120, ISO Similar triangles, 121 Simultaneous equations, 157-159 Sine, 101 Speed, 4 Sphere, 11 Square units, 5 Statistics, 123-146 Supplementary angles, 94, 112, 120 Tangent, 101 Temperature problems, 13, 17 Transversal, 120 Trapezoid, 100, 109, 122, 150-151 Tree diagram, 127 Triangle problems, 16, 137 Triangle, sum of the angles, 91-92, 109-110,121, ISO Triangles, isosceles, 93 Trigonometric ratios, defined, 101 Trigonometry, 101-119, 121-122 Trigonometry, inverse, 107 Trigonometry problems, 110, 151 Units of measurement, 2 Vertical angles, 120 Volume, 9 Volume problems, 17, 148 Volume formulas, 27 Weight, 12 Weight problem, 18, 148 Work problems, 41-43, 49, 82, 149 How to Solve Word Problems in Mathematics David S. Wayne, Ph.D. McGraw-Hill New York San Francisco Washington, D.C. Auckland Bogota Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto Library of Congress Cataloging-in-Publication Data applied for. McGraw-Hill A ~ Division of The McGraw·RiU Companies Copyright © 2001 by The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 1 2 3 4 5 6 7 8 9 0 ISBN 0-07-136272-X DOC/DOC 0 9 8 7 6 5 4 3 2 1 0 The editing supervisor was Maureen B. Walker and the production supervisor was Tina Cameron. It was set in Stone Serif by PRD Group. Printed and bound by R. R. Donnelley & Sons Company. McGraw-Hill books are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs. For more information, please write to the Director of Special Sales, Professional Publishing, McGraw-Hill, Two Penn Plaza, New York, NY 10121-2298. Or contact your local bookstore. o This book is printed on recycled, acid-free paper containing a minimum of 50% recycled, de-inked fiber. Contents Preface Chapter 1-Measurement. and Probability Chapter 6-Miscellaneous Problem Drill Appendix-A Brief Review of Solving Equations v 1 29 65 91 123 147 155 iii . Proportion. Counting. and Percentage Chapter 4-Word Problems Involving Geometry and Trigonometry Chapter 5-Word Problems Involving Statistics. and Using Formulas Chapter 2-Using Algebraic Equations to Solve Problems Chapter 3-Word Problems Involving Ratio. Estimation. . This book is designed to enhance the student's ability to recall those facts and procedures that are necessary for success in solving word problems. and practical situations. Most state curriculum documents list as a key component mathematical reasoning and modeling. determining what skills and concepts need to be applied. formula application. students in grades 6 to 12 are undergoing statewide assessments in mathematics that differ significantly from tests that were taken prior to the late 1990s. relevant. data analysis. students are required to not only produce a correct answer but also to carefully explain their work. many of these assessment tests cover multiple grade levels and require students to recall facts and procedures that may not be part of their current year's course. The problems offer opportunities to review the important areas of measurement. These current tests are based on revised standards that were created to address a noticeable nationwide weakness in applying the principles and computational skills in mathematics to realistic. This book is not v . and how to present a solution that shows complete understanding. The solutions include strategies for decoding problems. and probability. Additionally. estimation. interpreting geometric situations. algebraic representation.Preface In nearly every part of the United States. and. These are reflected in the assessments in the form of word problems. in most cases. Chapter 5 presents examples of interpreting data and using the basic principles of probability and statistics to respond to questions about the data. Gayle. this book will provide students with opportunities to better prepare them for being successful in applying mathematics. Chapter 2 helps the students solve more complex problems using algebraic skills and modeling. and friends. Chapter 4 presents the student with word problems that involve geometric relationships and ways to model these situations. this endeavor would not have been possible. vi . Chapter 3 focuses on problems for which students need to recognize the use of ratios and proportions and apply them correctly.designed as a textbook that presents information to be learned for the first time. Each chapter in the book focuses on word problems that have a common element. there is sufficient explanation of key concepts to enhance the student's understanding. Carl Goodman and Bert Linder. my daughter. and these problems are solved by simple arithmetic and reasoning. I'd like to take this opportunity to thank my wife. and many other family members. my friends and mentors. There is also an appendix which contains a review of solving linear and quadratic equations. Without their support and encouragement. Phyllis. Chapter 6 is a collection of problems that are not categorized and provides additional practice for students in determining how these problems can best be solved. The information and concepts used in these word problems form the core of most problems that students face. Whether used in a classroom setting with guidance by a teacher or at home as a supplement to class. however. This chapter provides review and practice in solving linear and quadratic equations. Chapter 1 presents problems involving measurement and the application of simple formulas. colleagues. For example. or three decimal places depending on the accuracy you need in order to solve the problem. Your calculations should be performed using at least one decimal place to ensure that your answer is rounded correctly. Estimation. and Using Formulas You have probably found solving word problems to be the most difficult part of any math course. In general. Sometimes a problem will clearly tell you how many decimal places to use. you should use at least one more decimal place in your calculations than the answer requires.Chapter I Measurement. "Find the length to the nearest foot II requires you to give your answer rounded to the nearest whole number. The examples that follow will show you how to use some of the most common formulas and measurements that commonly appear in word problems. these involve applications of well-known formulas and an understanding of the relationships between quantities. When measurements and calculations result in numbers that are not whole and have long decimal parts. All the formulas and measurements are listed at the end of this chapter for reference. it makes sense to round answers to one. The key idea in solving these problems is to identify the way in which the quantities are being measured and to recognize which formula to apply. two. In most cases. If the problem I . It is also helpful to be able to estimate an answer so that you can determine whether your worked-out solution is reasonable. when a problem involves money. the units of measuring distance include inches. the units include millimeters. The perimeter is the sum of all the edges of the shape. In the case of a rectangle Perimeter = 2 x length + 2 x width ..S. you could use two decimal places since we rarely calculate with fractions of a penny. which is the distance around a region. another method will be shown.25 per foot. feet. Customary) system. International System of Units (SI)]. Therefore. For conversions that aren't as simple. meters. This is used in the example below.does not specify how you should state your answer. it is important to look for how an answer is to be given and to think about what would be a reasonable answer.. Example I A rectangular room measures 216 in by 150 in. Distance Measuring distance is the same as measuring length. and kilometers. For example. the number of feet (ft) is equal to the number of inches (in) divided by 12. and miles. For example. How much will it cost to buy the molding? Solution I The answer lies in finding the perimeter of the room first in inches and then in feet. In the metric system [Le. An important measurement of distance is perimeter. The interior designer wants to place a strip of molding around the room near the ceiling that costs $2. It is important to know how one unit of measurement can be converted to another. yards. British and U. An easy way to convert measurements is to divide the number of smaller units by the amount of those units that are in the larger unit. centimeters. In the English (Le. you should use a reasonable amount of decimal places so that your final answer will be meaningful and practical. or the number of meters (m) is equal to the number of millimeters (mm) divided by 1000. after "canceling" similar units that appear in the numerators and denominators. the resulting fraction should be the units you wish your answer to have. We convert this to feet by dividing by 12. we have Imi 1.61 km. 300 + 150 + 225 = 675. Jim wanted to travel from Rome to Milan to Venice and back to Rome. 150 km from Milan to Venice. Thus. Example 2 While planning a trip around Italy. and 225 km from Venice to Rome. This is a good device to help determine whether the items being multiplied are the correct ones and are in the correct order. The cost of the molding would be 61 ft x 2. The units can be multiplied as if they were fractions and. The . On a map he found that the distances were approximately 300 kilometers (km) from Rome to Milan.61 km = 0. for this problem. To the nearest mile. Jim was not comfortable with these measurements and wanted to know the distance in miles. we need to add the distances in kilometers. (feet/I) x (dollars/foot) = (dollars/I) = dollars.25 Note that the units in this product are written as fractions. the perimeter of our room is 2 x 216 + 2 x 150 = 732 in.25 d~llars oot = $137.Width Length Therefore. and we have 61 ft.62mi/km For our problem. and then multiply it by 0. what is the total distance of the trip? Solution 1 The conversion for miles to kilometers in the reference table states that 1 mile (mi) is 1. If we think of this as a fraction.62. for the purposes of this problem. Example 3 A car has been traveling at an average speed of 55 mph (miles per hour) for 7 hours. How many miles has the car traveled? Solution 3 Note the phrase "average speed." The car may have traveled at somewhat different speeds throughout the 7 hours. Distance traveled can also be measured indirectly by knowing the speed at which the travel was occurring and the time taken for the journey.answer is 675km x 0. or the space on which a house is built. we usually say that the car has traveled at a constant speed.) The question How many miles . The distance is computed by the simple formula distance = rate x time. you would know that your answer would have to be more than l/Z of 300 + 150 + 225 or l/Z of 675 = 337. and we are once again multiplying units to arrive at our answer in miles. (When making this assumption. you can use the formula directly: II Distance = 55 mph x 7 hours = 385 mi Notice that mph is really the ratio miles/hour.62 miles ki ometer I = 418. the space on a canvas of a picture... However.5mi = 419mi Remember that an estimate is always useful to have first to be sure that your answer makes sense. traveled?" is really asking "What is the total distance traveled?" Since you know the rate and the time. Area Area is the measurement of the two-dimensional space within a region such as the space on the floor of a room. 4 . Since the conversion factor is greater than 0.5.5 mi and you would feel comfortable with the answer of 419. you can assume that the car has been traveling at that average speed throughout the journey. 1 square foot (ft2 ) = 12 in x 12 in = 144 square inches (in 2 ) This means that within a square that measures 1 ft x 1 ft. Each plant requires 1 square yard (yd2 ) of space for planting. you can fit 10. How many square yards have been roped off? How many plants can be planted in this plot of land? Solution 4 Be sure to draw a rectangle and label the sides with the given measurements! Drawing diagrams is an important good habit. This leads to conversions different from those in measuring one-dimensional length. When measuring area. you can fit 144 squares that measure 1 in by 1 in. s . it is important to make sure that the two measurements that are being multiplied are in the same units.000 square centimeters (cm 2 ) This means that within a square that measures 1 m by 1 m. 1 ft 1 in D 1 ft 1 ft2 1 in 1 in 2 1 square meter (m2 ) = 100 centimeters (cm) x 100 cm = 10. This would require making conversions as we did when measuring length.000 squares that measure 1 cm by 1 cm.Note that the answer to an area problem uses square units. The measurements are 8 ft 4 in by 5 ft 8 in. Example 4 A rectangular plot of land is roped off for planting. 33 ft x 5.33 ft i- r----rPlant cs . If each plant truly reqUires a square shape in which to be planted. Thus. Since we don't plant fractions of a plant. Therefore. Since 1 square yard (yd 2) is 3 ft by 3 ft. The answer to the second question is a little trickier.25 yd 9 ft 2 in this plot of land. The measurements become 8.5ftBin 8ft4in In this problem it is important to convert all measurements to the same units. for example. then we have to look at a diagram to see that there is enough room for only two plants! 5.23 square feet or ft2. there are 9 ft 2 in 1 yd 2.66. 4 in is one-third of a foot (1/3 ft) or 0. We can also round our measurements to one or two decimal places since our answer must be in whole numbers. we should convert the inches parts of our numbers to their equivalent fractional or decimal parts of a foot. we will be looking for a whole-number answer.67 ft. There are approximately 1 yd 2 8. and 8 in is two-thirds of a foot (2/3 ft) or 0.67 ft Plant 8. The area is 47.33 ft by 5.67 ft x --2 = 5.33. we are not actually counting squares. but we approximate it with either 3. The red carpet costs $12 per square foot. The values of these measurements involve the number 7r (Greek lowercase pi). How much will the carpeting cost? Solution S Once again. The preceding diagram shows that there are only two actual squares in this plot of land that are 3 ft by 3 ft. The entire room is 12 ft by 12 ft.25 squares if it were possible. because the diameter is equal to twice the radius. The designer wants a red circle. the remainder of the carpet should be white (Le. (Technical work. (If this were a multiple-choice question. Example 5 An interior designer wants to cover the floor of a room with a carpet that has a two-color design. which creates an area of 144 ft2. and the space within the circle is its area. and area = 7r X radius 2. that leaves 2 ft to each of the four walls. which doesn't have an exact value. The room is square.) The diameter and radius of the circle are also involved in the calculation. it is useful to estimate. often requires better approximations.This points out that when we measure area. The rest of the space could create an additional 3. The distance around the circle is referred to as its circumference (rather than perimeter). there should be a 2-ft white-carpet border around the red-carpet circle in the center of the room). you might find that only one of the choices falls 7 . the cost should be between 144 ft2 x $10 / ft2 = $1440 and 144 ft2 x $12/ ft2 = $1728. such as engineering. The quantity that we call the area suggests that if we could rearrange the space.14. Circles are popular in word problems of this type.. Supercomputers have calculated 7r to billions of decimal places beyond 3. The formulas are circumference = 7r x diameter or 2 x 7r x radius. measuring 12 ft on each side. we could construct that number of squares. whose center is the center of the room. Using the two prices.14 or with the fraction 22/7. and the white carpet costs $10 per square foot. We need to convert this to acres. and find that the red circular area would be approximately 3. Example 6 A farmer has a farm liz mi x 11/4 mi.88.76 ft2 x$10/ft2 = $937. but it is implied.24 ft2.---+2=-1 12 ft Since we want to leave 2 ft to the wall. The remaining area in the room would be 144 ft2_ 50. area = 7r X radius 2.560 ft2). We can set up a multiplication of fractions using the correct units in the numerators and denominators to arrive at the 8 . This is referred to as an acre (1 acre = 43.14 x 42 = 50. We use the formula for the area of a circle. The conversion given above relates acres to square feet. its radius must be 4 ft.24 ft 2 = 93. and your work would be done!) A diagram is required to find the actual answer. There is a measurement of land area that doesn't directly state that there are square units...24 ft2 x $12/ ft2 = $602. That would have to be the correct choice.5 x 1.48.625 mi2.into this range. 1-=2+-----=-. if each crop requires l/Z acre? Solution 6 The farm has 0. the diameter of the circle must be 8 ft and.25 = 0. How many acres is this farm? How many different crops can the farmer plant.60. therefore. and the white carpet would cost 93. The cost of the red carpet would be 50.76 ft2.. The total cost would be $1540. 0." Note that the units of volume are cubic units since we are multiplying three measurements.560ft The farmer could plant 400 -. 1 ft 1 ft Similarly. width. 1 ft 3 = 12in x 12in x 12in = 1728in 3 This means that into a box that measures 1 ft by 1 ft by 1 ft. The most common objects for which we use formulas are rectangular boxes. Volume is a measure of the amount of material a threedimensional object can hold. 1 cubic meter (m3 ) = 100 cm x 100 cm x 100 cm = 1.625 miz 5280Z ftz 1 acre ml. . we can fit 1 million boxes that measure 1 cm by 1 cm by 1 cm.000 cubic centimeters (cm3 ).Volume l/Z = 800 crops. This means that into a box that measures 1 m by 1 m by 1 m. circular cones.number of acres on the farm.000. and height are all equal. we can fit 1728 boxes that measure 1 in by 1 in by 1 in. and spheres. z x z = 400 acres x I 43. A cube is a rectangular box whose length. The formulas are given in the reference table at the end of the chapter under the subheading "Measuring Volume. and the tank should be filled to a point that is 3 in from the top of the tank. we need to measure volume. we can empty 64 -:. How many smaller bins can be emptied into each cubical bin? Solution 7 Since we are interested in knowing what it takes to fill something. We often use different units to measure the volume of a liqUid. Example 8 A rectangular fishtank is 30 in long. We use the following conversion formula to understand how liters are related to cubic units. and 20 in high. 1 pt = 2 cups. These bins are emptied into larger cubical bins that are 4 ft on each side. The volume of the cubical bin is 4 ft x 4 ft x 4 ft = 64 ft 3 . in the metric system we use liters to measure volume. If we can determine the volume of each smaller bin and the volume of the cubical bin. 15 in wide. How much water can the tank hold to the nearest tenth of a gallon? 10 . Gravel is laid evenly at the bottom of the tank to a height of 2 in. The volume of each smaller bin is 24 in x 12 in x 12 in or 2 ft x 1 ft x 1 ft = 2 ft 3 .Example 7 Paper is shredded into rectangular bins that measure 24 in by 12 in by 12 in.2 = 32 smaller bins into the larger. For example. and 1 gallon (gal) = 231 in3 • Gallons are subdivided as follows: 1 gal = 4 quarts (qt). Therefore. 1 milliliter( mL) = 1 cubic centimeter (cm 3 or cc) This means that lliter (L) = 1000 cm3 • The English system of measurement uses the gallon as its standard. we can divide the volume of the cubical bin by the volume of each smaller bin to arrive at our answer. and 1 cup = 4 fluid ounces (oz). These don't seem to be cubic units. 1 qt = 2 pints (pt). but they really are. should be in liquid measurements. What is the maximum capacity of the balloon to the nearest liter? Solution 9 Capacity refers to the amount an object can hold. we calculate the volume to be 4/3 x 3. This is equivalent to 7234. 6750 in 3 x 1 Example 9 1 gal 3 = 29.23456 L.56 milliliters (mL) and.22 = 29. the tank can hold 30 in x 15 in x 15 in = 6750 in 3 of water. The reference table at the end of this chapter shows that the formula for the volume of a sphere is given in terms of its radius. Therefore. to 7. II . there- fore. Our answer would be 7 L.14 x 123 = 7234." 15 in After the gravel is laid. dividing by 1000. which is one-half of the diameter.14 for Jr. Our answer.(2 + 3) = 15 in. we need to measure its volume.2gal 23 lin A balloon when expanded is in the shape of a sphere with a diameter of 24 cm.Solution 8 20 in 2 in I ~_ _--.56cm 3 . the height available for water is 20 . which in this case would be 12 cm.. however. Using the formula for the volume of a sphere and 3. 6 or 10 weeks to reach the eligible weight. Example II The ingredients label on a box of cereal indicates that one serving of cereal contains 24 g of carbohydrates. Here. in how many weeks will he reach the required weight? Solution 10 In order to work through this problem. In order to understand this better. In the English system. weight is sometimes referred to as mass. 1 kg = 1000 g) and metric ton (1000 kg) for larger weights. Jim had to bring his weight to under 80 kg. how many ounces of carbohydrates will she be eating.2. Using the "fraction" scheme of conversion.Weight Weight is also measured in both the English and metric systems in different ways. The conversion between systems most often used is that 1 kg is equal to approximately 2. Rachel wants to convert this to ounces.2 lb. Example 10 In order to prepare for an international wrestling event.) In the metric system. 1 lb = 16 oz) and ton (2000 lb) for larger weights. 1 kg is the exact weight of 1 L of water. we use the gram (g) as the smallest standard unit and the kilogram (kg. we ought to convert the 80 kg to pounds.S lb per week. we use the ounce as the smallest standard unit and the pound (lb. To the nearest tenth.S lb/week = 9. Since the rate of losing weight is given in pounds per week. if she has two servings of cereal? 12 . we have 80kg x 2. (The ounce used for weight is usually referred to as a dry ounce and is different from the fluid ounce used when measuring volume.21b = 176lb 1kg 1 Jim would have to lose 200 .176 = 24 lb. If he currently weighs 200 lb and begins a program in which he loses 2. we need to convert the information to either pounds or kilograms. Also. This would take 24lb -:. Example 12 Find the freezing and boiling pOints of water in degrees Fahrenheit.6°F. The Celsius scale uses the freezing and boiling pOints of water to create a range from 0° to 100°e.8 C + 32.8 x 0 + 32 = 32°F = 1.21b 160z 2 servings 1 kg x .x 24 g 1 serving 2. The formula can still be used. what would be the Celsius reading on a thermometer. to the nearest tenth? Solution 13 The new Fahrenheit temperature would be 101.Solution II We can use the "fraction" scheme to answer this question with one calculation: ------''---. where F represents degrees Fahrenheit and C represents degrees Celsius.100°C ---+ F Example 13 = 1.. If a person developed a fever with a 3° temperature increase above normal.6°F. except that we would have to solve 13 . The conversion to Fahrenheit is given by the formula F = 1.8 x 100 + 32 = 212°F The human body maintains a normal temperature of 98.x ..6896 oz = 1.7 oz Temperature Other situations that arise in word problems involve temperature. Freezing . In the world today we measure temperature in two scales. Solution 12 This is a simple application of the formula.O°C ---+ F Boiling .x -------''1000 g 1 kg lIb 1 = 1. Fahrenheit and Celsius (or Centigrade). 6 C = IT = 38. °C Shopping Many word problems involve finding the amount one has to spend on a purchase when sales tax and/or discounts are applied.66 .99 = $48. Example 14 Jim wanted to buy several compact disks (CDs). The music store was having a sale that reduced the prices by 15 percent.6 -+ (add -32 to both sides of the equation) (divide both sides of the equation by 1. 101. the total cost of the four CDs needs to be determined.99. the tax should be in decimal form.8C = 69. A formula one can use for finding a discount price is Discounted price = original price . How much would this purchase cost if sales tax is 8 1/ 2 percent? Solution 14 Before figuring in the percentages.an algebraic equation to determine C.discount x original price The discount needs to be in decimal form in order to perform the multiplication correctly. He selected three CDs that cost $12. Using the first formula to find the new price 14 .6 = 1.8) (which is 38. The solution to most of these problems is found by finding the dollar amount of those percentages and adding or subtracting. This is (3 x $12..99) + $9.8C + 32 -+1.7°C to the nearest tenth) 69. Usually we are given an initial purchase price of an item and a percentage that indicates the discount and/or tax.96.99 and one that cost $9. A formula for finding the cost with sales tax included is Total cost = price + tax x price Again.. 15 x $48.5 percent of the price or multiplying $41. Interest Another common application of mathematics involves finances. People are interested in determining how much money will result when interest is applied.96 by 0.62 Note that this price could also have been calculated by finding 85 percent of the price or multiplying $48.16 Note that this could have been calculated by finding 108. Total cost = $41.62 x 0.with the discount. and t is the number of years that the investment is allowed to grow.085) = $45.96 . we can determine the total cost.85.62 + ($41.(0. What could she expect to happen to her original investment after one year? 15 .62 by 1.085. r is the interest rate as a fraction. but the simplest is found using what is called the simple interest formula: Interest earned = p x r x t where p is the original amount called the principal. decimal. Total amount = principal + interest earned Example IS Samantha has been working hard as a baby-sitter and has $540 collected. Using the second formula. She decides to invest this money in a fund that predicts an interest rate of 8 to 11 percent.96) = $41. we have New price = $48. There are many complicated ways to calculate interest. The total current amount is found by adding the principal investment and the interest earned. Interest is simply an amount of money that is added to the original amount and calculated by a percentage of that amount. or percent. Additional Problems I. Note that the percentage needed to be changed to a decimal in order to make the calculation.08 x 1 = $43. Samantha would receive 8 percent interest.40. At the most. This would result in 1= $540 x 0. what is the length of the shorter side of the framel The owner of a corner store wants to paint the upper portion of a side exterior wall that is 16 ft tall to make the store more noticeable. An actual solution would be as follows. 4. How much will she spend to fence her gardenl Gayle's triangular garden from Prob. she would have saved between $583. and take a look at the reference table at the end of this chapter to remind yourself of the formulas. The material she chose to use costs $4. At the least.25 per meter.20 and $599. This would result in 1= $540 x 0. If the longer side of the picture is I I in. If you didn't do this. At the end of one year. Be sure to draw diagrams to help you see the situation clearly. This would result in Samantha receiving $54 in interest for an estimate of having a total of $594 at the end of the year. Samantha would receive 11 percent interest.Solution IS You could first estimate an answer by supposing that the rate was 10 percent. using total amount = p + I. as the result would be 100 times larger! The problems that follow will give you plenty of opportunity to use the formulas and ideas that you just studied. the answer would be unreasonable.20. There is a 2-in border on all sides of the picture within the frame. The length of the wall is 25 ft.40. How much plastic should she usel The length around a wooden rectangular picture frame is 50 in. 16 . I needs a plastic ground cover in order to keep weeds from growing. She measures the distance from the 8-ft side to the opposite vertex of the triangle to be 6 ft. 3. He buys a gallon of yellow 2. Gayle would like to fence her garden.11 x 1 = $59. which is in the shape of a triangle with two sides measuring 10 1 ft and the other side /2 measuring 8 ft. Her front yard is rectangular and measures 12 m by 14 m. what is the largest area that can be used for the pool? Melissa is planning to have the front of her property landscaped. it was found that a compound boils at a temperature of 42°C. The speed limit / on the highway she drove is 65 mph. The heating apparatus can raise the temperature by 2°F per minute. In how long a period can the compound be brought to boil if the heating process begins at 68°F? While planning a winter vacation for his family. The average speed of the plane was 320 mph. She plans to put down sod over the entire yard. how much will Melissa need to spend to cover the entire yard if sod sells for $6 per square foot? How high would the sides of a rectangular water tank have to be if the tank base has an area of 2 m2 and the tank must hold 5520 L of water? An ice-cream parlor owner charges $1. If the yard is 30 ft by 26 ft and the pool is to be centered in the center of the yard. What portion of the wall will be yellow? A circular swimming pool is being installed in a neighbor's yard. 17 . At what speed. He wants to travel to a place where the temperature would be between 75 and 85°F. how many miles has the plane flown? On the first day of a cross-country automobile trip with her family.95 for an ice-cream cone. The county requires that the pool be 6 ft from the end of the property. 13. paint to do the job and is told that each gallon can cover 400 ft2. 1 I. The top of the cone has a diameter of 2 in. The information he finds on the Internet is given in degrees Celsius. 8. 7. 12. The scoop also has a diameter of 2 in. To the nearest dollar. 6. Frank investigates temperatures in different parts of the world. How much profit is the owner making. Jennifer covered a distance of 680 mi in 9 12 hours. except for a brick circular design in the middle that has a radius of lOft. 10. 9. How much longer or shorter time would it have taken Jennifer if she drove at the speed limit? The current school record for the 100-m dash is 9 seconds. and the cone is 6 in tall. and then a scoop of ice-cream sits like a ball on top. The cone is first packed in. would a runner have to run in order to beat the speed record for the 100-m dash? In a laboratory.5. If it took 2 hours to reach Phoenix and another 2 hours and 45 minutes to reach Dallas. if the ice-cream costs her $30 per gallon? An airplane traveling from San Fransisco to Dallas makes a stopover in Phoenix. in miles per hour calculated to the nearest tenth. you can buy a similar item for half-price. if you buy one item at full price. The total perimeter of the garden is 10 1 ft + 10 1 ft + 8 ft = 29 /2 /2 feet. How many weeks should she wait before she can afford the outfit. using this credit card. 32. how much would it cost to buy two pairs of jeans. if sales tax is 8 3/4 percend Mr. how many rolls of insulation will be needed and how much weight. 20. Miller invested $15004 years ago in a fund and hasn't made any withdrawals. would like to purchase three pieces of furniture made in England that she saw in a catalog. 18. will this insulation add to the attic? Mrs. 15. In addition. A store is having a "half-price sale". Mike purchased a computer for $1300. which already includes sales tax. of luggage that the plane would be carrying if there are 200 passengers on board~ Solutions to Additional Problems I.What range of temperatures should he enter into the search engine~ 14. If the average annual rate of interest over the 4 years was 12 percent. Lindsey has $90 to spend on an outfit. The cost of shipping by air across the Atlantic Ocean is $365 per 100 lb. that is. Alexander. each costing $24. What is the maximum weight. how much money would it cost to import these pieces of furniture from London~ European Airways allows each passenger to check two suitcases that measure 50 cm by 30 cm by 72 cm. how long will it take Mike to payoff his debd What was his total cost for the computer~ Gary found that his attic needed more insulation to prevent loss of heat. there is a 3 percent international tariff. 19.99~ An owner of a clothing store attempting to clear out her old inventory marks down the outfits she wishes to sell by 10 percent each week. 16. He has to be careful about how much weight he adds to the attic. If the sales tax is 6 percent. 17. The airline assumes that luggage weighs 75 kg per cubic meter. and 5 I kg. who lives in New York. in pounds. Each roll will insulate 40 ft2. If his attic is rectangular and measures 15 ft by 18 ft. If he pays $200 every month. If the three pieces weigh 16. how much money is currently in the fund~ A certain credit card charges 18 percent annual interest on the unpaid monthly balance. She sees that it is currently marked at $135. Insulation comes in rolls that weigh 7 Ib 4 oz. to the nearest pound. In order to calculate the cost. Gayle needs to convert this to 18 . meters. Using the conversion scheme of creating fractions. we have 29ft 12in 2.54cm 1m -- x -- x x = 8.84 m I I ft I in 100 cm The total cost is 8.84 m x $4.25/m = $37.57 8ft The distance from the 8-ft side to the opposite vertex is the height of the triangle (draw this line in the diagram from the previous problem and label it 6 ft). The base is the I01/2-ft side. Using the formula for the area of a triangle from the reference table. A = 112 X base x height. we have A = 0.5 x 10.5 x 6 = 31.5 ft2. 3. The perimeter of the frame is given as 50 in. (Note how the perimeter is disguised as the amount of wood used to make the frame.) If the longer side of the picture is II in and there is a 2-in border on either side. then the longer side of the frame has to be 15 in. The two longer sides of the frame. therefore. use 30 in of the wood and the formula for the perimeter of a rectangle. P = 2 x length + 2 x width. tells us that each of the shorter sides has to be half of the remaining 20 inches. The shorter side of the frame is lOin. 10 in 2. 12 in 15 in 11 in 12 in 19 4. The gallon of paint would cover 400 ft'. Assuming that we would have a rectangular area painted yellow at the top of the wall, we could use the 25 feet to represent the base of the rectangle. The formula for the area of a rectangle, A = base x height, tells us that 400 = 25 x height or height = 400 -7 25 = 8 ft. This is exactly one-half of the wall. 25 ft 8 ft 16ft Yellow area --------------~ 5. 6. Since the pool has to be 6 ft from the property line, the diameter of the pool can be no more than 26 - 12 = 14 ft. Therefore, the radius of the largest possible circular area is 7 ft. Using the formula for the area of a circle, A = IT X radius', we have A = 3.14 X 7' = 153.86 ft'. To solve this problem, we need to calculate the area of the entire rectangular yard and subtract the area of the circle. Since the price of the sod is given in dollars per square foot, we need to convert the measurements of the yard to feet. These conversions are: --x I 12m 14m I 100cm 1m x I in 2.54cm x - 1ft 12in = 39.4ft = 39ft --x 1ft 100cm I in x x 1m 2.54 em 12in = 45.9ft = 46ft 7. Using the formula for the area of a rectangle, A = length x width, the area of the entire yard is 39 x 46 = 1794 ft'. Using the formula for the area of a circle, A = IT x radius', the circular design has an area of A = 3.14 X 10' = 314 ft'. Therefore, Melissa needs to buy 1794 - 314 = 1480 ft' of sod. This will cost her 1480 ft' x $6/ft' = $8880. We know the volume of the tank to be 5520 L. Since we are being asked to find a height, we have to be working in measurements of length. We therefore need to convert liters into cubic meters in order to work with the same units. From the reference table, we 20 8. 9. 10. I I. see that I m3 is equal to I kiloliter (kL) or 1000 L. Therefore. the tank will hold 5520 + 1000 = 5.52 m3 • We need to calculate the volume of the tank. The formula for the volume of a rectangular three-dimensional shape is V = area of the base x height. Using this. we have 5.52 m3 = 2 m2 x height (in meters). Therefore. the height is 5.52 + 2 = 2.76 m. Notice how. in this last equation. we have to have equal units on each side. m3 = m3 . This is a good way to be sure that you are applying formulas correctly. We need to know exactly how much ice cream is used in the ice-cream cone. Since gallons is mentioned in the problem. we know that we will be calculating volume. Using the formulas for the volumes of a cone and of a sphere from the formula reference table. we have to identify the radius and the height of the cone and the radius of the scoop. In both cases. the diameter is 2 in; therefore. the radius of each is I in. We apply the formulas Volume of cone = 1/3 x 3.14 X 12 x 6 = 6.28 in 3 and Volume of scoop = 4/3 x 3.14 X 13 = 4.18 in 3 • and the total volume is 10.46 in 3 • However. we need to convert this to gallons since the cost is given in dollars per gallon. The reference table tells us that there are 277.42 in 3 in I gal. Therefore. the number of gallons is 10.46 + 277.42 = 0.0377 or 0.04 gal. The ice-cream parlor owner's cost in making the cone is 0.04 gal x $30/gallon = $1.20. The owner is making a profit of $1 .95 - $1 .20 = $0.75 on the cone. The total time taken for the flight is 4 hours and 45 minutes. To use this in the formula for distance traveled. we have to convert this to a decimal number in hours. We can use (45 minutesll) x (I hourI 60 minutes) = 0.75 hour. Our time is. therefore. 4.75 hours. Applying the formula. we have Distance traveled = 320 mph x 4.75 hours = 1520 mi. Using the speed limit of 65 mph in the formula for distance traveled. we have 680 mi = 65 mph x the number of hours taken. Therefore. the number of hours is 680 + 65 = 10.46. or approximately 10 12 hours. It would have taken Jennifer I hour / more. If needed. you could also calculate her rate using the formula 680 = average speed x 9.5 or average speed = 680 +9.5 = 71. 57 or 72 mph. which means that she was traveling faster than the speed limit allowed. The formula for distance traveled is applied to determine the record speed in meters per second: I00 m = average speed x 9 21 9°C and a higher temperature of 5/9 x (85 . The price after the fifth week would be 22 . We can use the formula F = 9/5 X C + 32 to get F = 1. the temperature has to rise 108 .68 = 40°F.50.6°F or 108°F (where F = degrees Fahrenheit and C = degrees Celsius. Using the formula Discounted price = original price .58. 13.32) to 75°F and to 85°F. it would be $121.10 = $98. the price would be $109. 15.42. we need to convert the Celsius temperature to Fahrenheit. 10 = $88.10 = $109. In this problem it is important to remember that the $90 that Lindsey has to spend must include the 8 3/4 percent sales tax. it would take 20 minutes for the compound to boil. as in Example 12). a runner would have to run faster than 24.seconds or speed = 100 -.4°C.49 = $39.50 .58 + 0. 10 = $121 .42x O. By applying the formula C = 5/9 x (F . that is. After the second week.49 + . The total cost of the purchase is found from the formula Total cost = price + tax x price. one of the pairs of jeans will sell for the full price of $24.50.32) = 29.84 hmi our 12. After the fourth week.99 while the other will sell for one-half of that.06 x $37. Therefore. We can find the different costs of the outfit as the weeks pass by making successive computations and keeping a list. The total price would be $88. I I m 60 seconds 60 minutes I km I mi ---x x x x--I second I minute I hour 1000 m 1.74. After the third week.$121.9 mph to beat the record.61 km = 24.35. and we compute $37.$109.$135 x O.42 .35 x 0.49. This has to be converted to miles per hour. the price after the first week would be $135 . We have to compute the price with tax to see if Lindsey's $90 would be enough.33.discount x original price. we have a lower temperature of 5/9 x (75 . At 2° per minute. and we can use the "fraction" scheme for doing this: I I.32) = 23. To the nearest tenth.9 = I I .$98. $37.35 ..58 = $96.8 x 42 + 32 = I07. I I m/second. or $12. 14. Therefore. which is more than she could spend. $98. the price would be below $90.50 x 0. the 6 percent sales tax has to be applied to the sum of the prices. That is.0875 x $88. The discount has to be applied first. To find the time taken. 28 A table like this is a good tool for keeping track of calculations and gives you a way to look back on your work in an organized fashion. Since $1300 -.60 x 0.12 = $225. Each year there is a new amount that is earning interest.58 + $8 42 = $57000 $57000 .$200 = $936 29 0015 = $1404 $93629 + $14 04 = $950 33 $95033 .00 0015 = $5 55 $37000 + $5 55 = $375 55 $375. Before beginning the calculations.79 $2107.50 .55 .18 Pay $178 18 to complete debt 23 .12=$180 $1680 x 0.$200 = 6.00 $17555 x x x x x x x Total debt New balance 2 3 4 5 6 7 O.oJ5 = $19 50 $1300 + $19 50 = $1319 50 $1319.60 $2107. Lindsey would have to wait 5 weeks and hope that the outfit hasn't been sold.. $88.39 x 0.$200 = $75033 0015 = $1125 $75033+$1125 = $76158 $761 58 . we would expect that with interest adding to the amount. Therefore.$200 = $370. At end of year 1 2 3 4 Interest earned $1500xO.89 = $1680 $1881.$200 = $1119 50 0015 = $16 79 $111950 + $1679 = $1136 29 $1136 29 . (This is usually called compounding interest.72+0. it should take 7 or perhaps 8 months to payoff the debt. At end of month Interest added $1300 $111950 $93629 $750.39 + $252.12 = 1.$88.$200 = $561 58 0015 = $8 42 $561. A table is useful to follow the month-by-month situation.12 = $252.12 = $201.58 x .89 New amount $1500+ $180 = $1680+ $201. This problem is similar to one asking for a total amount earned with interest.60 $1881.70.0875 x $79.$200 = $175 55 0015 = $263 $17555 + $2 63 = $178. a new amount is owed each month after adding on interest on the unpaid balance and $200 deduction. we need to use 18 percent -.16. 17.72. Miller has at the end of each year.60 = $1881.5.60+ $225.58 ..33 $56158 $370. That is. we can determine how much Mr.10 = $79.72 = $86. Since the interest is added monthly.5 percent to calculate the added interest each month.) Using the formulas Interest earned = principal x rate of interest x time invested and Current amount = principal + interest earned. an easy estimate should be made. The total price with tax would be $79.39 $2360.79 = $2107. Step 2 Step 3 Step 4 Step 5 24 .21b/kg = 217.3 m x 0. Step I The dimensions of the suitcase should be converted to meters in order to use the weight rate.03 = $23. First we need to determine how many rolls of insulation are needed. Therefore. 4 oz is 1/4 or 0.72 m. That is. 19. For the second part of the question.25 lb.5 m x 0. The suitcase has a volume of 0.9375 or 49 lb.97 + $23.18.2 would give us the number of pounds.65/1b = $794. We see that our estimate was a good one. The answer is found by breaking the problem down into smaller parts. Therefore.72 m = O.Slb x $3. The total cost of the computer is found by adding the interest amounts to the original price of $1300.3.75 or 7 rolls.25 Ib/roll = 4S. that is.IOS m3 • The weight of one suitcase would be . Dividing each measurement by 100 gives us 0. Step 2 Find the cost per pound.S2. $3. IS.97 x 0. we need to convert 7 Ib 4 oz into a decimal number in pounds. and 0.IOS m3 x 75 kg/m 3 = S. we need 270 -:. $794. The shipping rate is given with respect to 100 lb.97.2lb/kg = 712Slb. The weight is therefore 6. The area of the attic is 15 ft x IS ft = 270 ft2. Multiplying by 2. Converting to pounds. 0. IS = $137S. I kg.5.25 lb. Dividing by 100 gives the cost per pound.40 = 6. There could be up to 200 x 2 = 400 suitcases on the plane for a total weight of 400 x S. 3240 kg x2.75 since we do not have to use the remaining portion of the roll. Step 5 The total charge is $794.75 rolls x7. Step 3 Find the shipping charge. First. we have 99 kg x 2.Slb.S5 = $8IS. I = 3240 kg. Since there are 16 oz in one pound. Step I Find the total weight in kilograms and convert it to pounds. 20. we can use the exact number 6. Step 4 Compute the tax to add.65 per pound.85. each roll weighs 7. The formula from the reference table Total dollar amount of item = quantity of item x dollar amount of one item is applied to get 217. $1300 + $7S. The total weight is 16 + 32 + 51 = 99 kg. 06 quarts 1 liter Basic unit 1 millilter (mL) 1 cm 3 or cc 1000 mL or 1 dm 3 1 liter (L) 40z" 2 cups = 8 oz 2 pt = 32 oz 1 kiloliter (kL) 1000 L or 1 m 3 4 qt or 277.= 1 ..47 acres Volume or Capacity in Cubic Units Space within a three-dimensional cube: @ 1 unit 1 cubic unit 1 unit English system 1 ounce (oz)* 1 cup 1 pint (pt) 1 quart (qt) 1 gallon (gal) Metric system Conversions between systems 1. tThese are referred to as dry ounces. ~IlUon I' 231 cubic mche.61 km Area or Space in Square Units S"".8 ft2 2.54 cm 1m 1.OOO m 2 Conversions between systems to. ocmp'" by • ".21b 1 kg "These are referred to as fluid ounces. D 1 unit 1 unit 1 "l""'" . ~IlUon Weight or Mass Metric system 1 gram (g) 1 kilogram (kg)t 1 metric ton (t) Basic unit 1000 g 1000 kg English system 1 ounce (oz)t 1 pound (lb) 1 ton Basic unit 160zt 2000lb Conversions between systems 2. lOne kilogram is equal to the weight of one liter of water. 25 .""'m.42 in 3 bnperud (UK) / The US hqmd.' 1 m2 1 ha English system 1 square inch (in 2 ) 1 square foot (ft2) 1 square yard (yd2 ) 1 acre Basic unit 144 in 2 9 ft 2 43..Table of Common Measurements and Conversions Length or Distance in Linear Units Straight-line distance between 2 points: 1 Unit English system 1 inch (in) 1 foot (ft) 1 yard (yd) 1 mile (mi) Basic unit 12 in 36 in 5280 ft Metric system 1 millimeter (mm) 1 centimeter (cm) 1 decimeter (dm) 1 meter (m) 1 kilometer (km) 1/1000 m 1/100 m l/to m Basic unit 1000 m Conversions between systems 1 in 39 in 1 mi 2.""=" 'q..560 ft 2 Metric system 1 square meter (m 2 ) Basic unit 1 hectare (ha) to.. " trl. D.U.14) Distance traveled = average rate x time traveled Measuring Area Area of a rectangle = length x width or base x height "<e./3/ 4 x .log<.ull".Ido'.14): Continue 26 . of "qu~= . p". of..m = 00.Id" .ngio=.n ". x lrelgh" JH..Table of Common Formulas Seen in Word Problems Measuring Length Perimeter of a rectangle = 2 x length + 2 x width or 2 x base + 2 x height: Ii Length or base ~height Width or Circumference of a circle = 2 x 7r x radius (7r is approximately 3. D Side Area of a trapezoid = average of the bases x height or l/Z x (lower base + upper base) x height: /l \ / l Height ~ Lower base Upper base Area of a circle=7r x radius2 (7r is approximately 3.~ht Base Base / Area of a triangle = liz x base x height: ~ Area of . do A<e. discount x original price Interest earned = principal x rate of interest x time invested Current amount = principal + interest earned 27 .Surface area of a three-dimensional sphere = 4 x Jr x radius2 : Radius Measuring Volume Volume of a rectangular solid = length x width x height or area of the base x height: Height Base Length Width Volume of a cube = side 3 : Side Volume of a sphere = 4/3 x radius 3 Radius Volume of a circular cylinder = Jr X radius2 x height: Height Volume of a circular cone = 113 x Jr x radius2 x height: Money-Related Formulas Total dollar amount of item = quantity of item x dollar amount of one item Total cost = price + tax rate x price Discounted price = original price . . The key will be to identify all the relationships between the numbers that are mentioned in the problem.) It is important to carefully identify the unknown variable in the problem and determine the relationship between it and the other information in the problem. or making an arithmetic mistake. this will require modeling the situation with an equation in which what you wish to find will be expressed by a letter representing an unknown quantity. The equations will come from applying basic formulas and well-known concepts. It is also important that your answer really does solve the problem! There are many places to make mistakes in these problems. solving the equation incorrectly. We call this the variable of the equation. you will learn how to apply algebra to solve problems. such as creating the wrong equation. . you studied problems that were solved by making basic arithmetic computations. (You can review basic tools of algebra by visiting the Appendix in the back of the book. In most cases. In this chapter.Chapter 2 Using Algebraic Equations to Solve Problems In the previous chapter. Most of these problems will tell you how two or more numbers are related. Problems Involving Numbers The simplest of all word problems are those that involve finding numbers. dividing. Larger number = 3 x smaller number + 14 We can now substitute our variable expressions for the smaller and larger parts. Let the smaller number = n We are now ready to start translating the first sentence into an algebraic equation. and equality when it is easily recognized. A good idea is to first rewrite the problem leaving most of the words and using mathematical symbols for adding. Since the problem specifically tells us how the larger and smaller numbers are related. The letter n is always a good choice for a number. We must realize that to have more than something means to add to that something. It makes sense. It is also a good idea to select a letter for the variable that reminds us of what we're looking for. Larger number = 3n + 14 30 . we can compute the larger one. The first sentence of this problem tells us that if we know the smaller number. Remember to drop the multiplication symbol when multiplying by a variable. therefore. Find the smaller number if their difference is 38. multiplying. This would give us Larger number = 14 more than 3 x smaller number Now we have to take a look at the phrase on the right. This is often called the "Let" statement.Example I The larger of two numbers is equal to 14 more than 3 times the smaller number. subtracting. We should also start our work with a statement of how we are assigning variables. we can use one variable. to use the variable to represent the smaller one. Solution I The first step of any word problem is to find a way to use a variable to represent the unknowns in the problem. Clearly. we are taking the smaller from the larger. "their difference is 38. 70 .n = larger part 31 . and the larger number is 3 x 12 + 14 = 36 + 14 = SO. The "Let" statement is Let n = smaller part 70 .12 = 38"( Sometimes one of the relationships regarding the numbers is given indirectly. the smaller number is 12. we get 2n+ 14 = 38 2n= 24 n= 12 Therefore. Example 2 Separate 70 into two parts so that 5 times the smaller part is equal to 14 more than twice the larger part. To check.n.n = 38 Simplifying and solving for n. We can assign a variable. it might be hidden in a key phrase. n." Again we should first translate the operation symbols and write the remaining words. Solution :1 The problem is really asking for two different numbers whose sum is 70.smaller number = 38 Substituting our variables gives us 3n+ 14 . and if the difference is positive. SO . We get Larger number . We should realize that finding a difference means to subtract. The larger part must be what is left when n is taken away from 70.Both numbers in the problem are now represented by variables. This will come from the other relationship given in the problem. to the smaller part. and we are ready to form the equation that we will use to solve the problem. we need only verify that their difference is 38. that is. 2n + 14 Add 2n to both sides to get 7n = 154 n=22 The smaller part is 22. When we use two 32 . and the larger part is 70 .n) + 14 Solving in the usual way. "more than" and "twice.Translating only the operations gives us 5 x smaller part = 14 more than twice the larger part Now we have to take a look at the phrases on the right: specifically. The check is easy. Remember to drop the multiplication symbol and to use parentheses when we need to multiply by an expression with more than one term. 14 more than twice the larger is 2 x 48 + 14 = 96 + 14 = 110"'! Alternative Solution 2 Another way to solve the problem is to use two different variables to represent both the parts of 70. 5 times the smaller is 5 x 22 = 110. 5 x smaller part = 2 x larger part + 14 Now we are ready to substitute our variable expressions for the smaller and larger parts. 5n = 2(70 . Simply follow the numbers and words of the problem. we get 5n = 140 . 5 x smaller part = twice the larger part + 14 Now we use the fact that to have twice something means to multiply that something by 2.22 = 48." We must realize that to have more than something means to add to that something. The next example shows you another way to express one of the relationships indirectly. Example 3 In this example. Sx = 2y+ 14 We work with this pair by solving the first equation for y y= 70-x and substituting in the second equation Sx = 2(70 .x) + 14 Note that we have the exact same equation to solve (even though the variable here is x instead of n) and we should arrive at the solution. The translation should still be performed as outlined above. Find all three integers. To understand this. x = 22 and y = 48. x+ Y= 70 The second equation comes from the way these parts are related. we need two equations. The numbers 33 .) Let x = the smaller number Let y = the larger number The first equation comes from the fact that the sum of the numbers is 70.variables. but now we use the variables x and y to substitute for the smaller part and for the larger part. (See the Appendix for a review of how to work with equations with two variables. The key phrase in this problem is consecutive positive integers. let's look at each word. Solution :I It is clear that we are dealing with three different numbers.S less than 6 times the largest of three consecutive positive integers is equal to the square of the smaller minus 2 times the middle integer. 2 times the middle Less than something means to take away from that something 6 x largest . 6(n + 2) . This is the set of negative and positive whole numbers and zero {.5 = smallest2 - 2 x middle Now we can substitute our variables and remove the multiplication symbols. Since the numbers are positive.. we won't accept any answers that are fractions.2 x middle Square of means to raise the number to the second power 6 x largest . decimals.3.5 = n2 - 2(n + 1) This is a quadratic equation since we have an equation with a power of 2.1. Let n = the smallest integer n + 1 = the middle integer n + 2 = the largest integer Now we can use our translation scheme. }.2 34 . -1. -2.. ° 6n + 12 . This means that the three numbers are all one apart. . The most important word in the phrase is consecutive.. We can then represent our numbers with the same variable. -3.5 = n2 - 2n .. First only the symbols 5 less than 6 x largest = square of the smallest . (See the Appendix for a review of solving quadratic equations.. }. 3 .) It must be simplified and have on one side in order to solve it by factoring. 2.5 = square of the smallest .. that is. n.0.2.have to be integers. we are looking only for numbers from the set {I. . or roots. 9 = 0 We can now factor and solve for n. add 2n. (n . Since n was the smallest.) Since we need to have two relationships. In 10 years.5 = 66 ..1 (Not all consecutive integer problems will give rise to quadratic equations as you will see in the supplementary exercises.9)(n + 1) = 0 Therefore n=9 or n= -1 Remember. Example 4 Sandy is 4 times as old as Robby is now. we subtract n2 . the three integers are 9.. we simply use the numbers and follow the words of the problem: 5 less than 6 times the largest is 6 x 11 . so we can reject -1 as an answer. Sandy will be 6 years older than Robby. and add 2. these problems may cause you to consider ages at the present time and at another time in the past or future.) Problems Involving Age Problems that involve the ages of two or more people are very similar to number problems.2 x 10 = 81 . an age is a number. How old are they now? 35 .To both sides of the equation.5 = 6I. (After all.20 = 61. The square of the smallest minus 2 times the middle is 92 . 10. and II. we are looking for positive integers. To check. to arrive at We can now multiply every term in the equation by -1 to get n2 - 8n. . Robby's current age is 2 and Sandy's current age is 8./ In 10 years. we have Let Robby's age =x Sandy's age = 4 x Sandy's age = 4x Robby's age "In 10 years" means that we will add 10 to each of their ages. . Sandy will be 6 years older than Robby.Solution 4 Since we are told how to find Sandy's age if we know Robby's age. Sandy will be 18 and Robby will be 12. How many years ago was Jessica twice as old as Melissa? 36 . We ought to state what their ages will be. Robby's age will be x + 10 and Sandy's age will be 4x + 10 We should now translate the second sentence using these future ages and substitute the future variables Sandy's age 4x + 10 = Robby's age +6 = x + 10 + 6 Simplifying and solving for x. Following the procedure outlined in the previous problems. we get 4x + 10 = x + 16 3x =6 x=2 Therefore. To check. we should simply follow the words of the problem. ./ Example 5 Jessica is 27 years old and Melissa is 21 years old.. Sandy's current age is 4 times Robby's age or 8 = 4 x 2 . In 10 years. we should use a variable for Robby's age and determine the symbolic representation of Sandy's age with the same variable. Clearly. x years ago. or moving in the 37 .15 = 12 years old and Melissa was 21 . we get 27 . jessica was 27 . The two objects can be moving toward each other. the variable is the number of years in the past. jessica was twice as old as Melissa"'! In Chap.2x x = 15 years ago To check. we will see that the key to the problem is finding out which of the quantities mentioned are meant to be equal or which quantities are added to get a total amount.x years old. Let x = the number of years in the past Therefore. The relationship x years ago was jessica's age = 2 x Melissa's age Substituting the past ages. we have to compute the girls' ages 15 years ago. substitute known values. jessica was 27 . Those problems merely required that we identify the appropriate formula. 27 .x = 42 . and solve for the missing value. Clearly.x) Simplifying and solving for x. Now we will examine situations where those formulas are used as the basis for the equations needed to solve more complex problems. Problems Involving Motion In problems that involve motion. and we have to subtract this amount from their current ages.x = 2(21 .15 = 6 years old. we get the equation for the problem.Solution S In this problem. In each problem. 1 we solved problems that make use of many of the basic formulas that arise in mathematics. moving away from each other.x years old and Melissa was 21 . there are usually one or two objects moving along a straight line at a constant rate (speed) for a period of time. and the unknown is the time. while the other is traveling at SO mph. As before. we have to decide what is given and what we have to find by examining each of the parts. to the nearest minute. The car traveling at 60 mph will cover more of the distance than the car traveling at SO mph. there is a total distance traveled by the objects. Therefore. will have passed at the moment when they pass each other? Solution 6 In order to use the formula. Drawing a picture can help. The governing relationship for these problems is Distance = rate x time Example 6 Two cars are traveling on the same highway toward each other from towns that are 150 mi apart. If they started at the same time. One car is traveling at 60 mph. Let t = time traveled by both cars 38 . the time they've both traveled is the same. t. In any situation. we can use one variable for this time. Since the cars started at the same time and pass each other at the same time. -------------------------------x---------------~-----------------150mi------------------~ Fast car Slow car The picture shows that the total distance traveled is a known amount and that it must be equal to the distances traveled by each car. how much time. it is a good idea to state the use of the variable before any work is begun.same direction as in a race. Total distance = distance traveled by faster car + distance traveled by slower car We know the rates of each car. This is the basis of the model for this problem. the answer is 1 hour and 22 minutes. we will use the actual fraction 15/11 for the time. Remember that this answer must be in hours since the rates were given in miles per hour. Allison ran 10 minutes more. which will . Note that the units of t must be hours. The faster car traveled 15/11 x 60 = 900/11 . our model becomes 150 = 60t + SOt 150 = 110t 150 t= 110 t = 1.Using the relationship Distance = rate x time for each car..37 x 60 = 22. Allison had a 10minute head start and was running at 3 km per hour. we will represent it by t. The problem asks for the answer to be given to the nearest minute. we need to compute the distances traveled by each car. To check the answer.363636 . The decimal part of the hour can be converted to minutes by multiplying . The model for the problem is Distance traveled by Allison = distance traveled by Rachel Since we are asked to find Rachel's time. when rounded. and the slower car traveled 15/11 x SO = 750/11 • By adding these fractions. Rachel ran at 5 km per hour. we find that together they traveled 1650/11 = 150 mLy" Example 7 Allison and Rachel were competing in a race. the key factor is that at the moment that Rachel catches up to Allison. To be accurate. since the rates are given in hours. they will have traveled the same distance..2 minutes and. In how many minutes did Rachel catch up with Allison? Solution 7 In this problem. . This would allow t to represent minutes. Let t = Rachel's time t + 1/6 = Allison's time Using Distance = rate x time... and Allison traveled 3 x 5/12 = 1.time 40 .25 km. (:0) (t+ 10) = (:0) t 3t 60 30 5t = 60 + 60 3t+ 30 = 5t 30 = 2t 15 = t The relationship used in these problems can be rewritten to express the rate as a quotient: distance traveled Rate = .be represented by t + 1/6 since 10 minutes is one-sixth of an .. and we could use 10 minutes in the equation instead of 1/6 hour. Note that we could have converted the rates to kilometers per minute by dividing each rate by 60. hour. we compute the distance traveled by each girl using t = 1/4 for Rachel and t = 1/4 + 1/6 = 5/12 for Allison. we have 3 (t+ 1/6) = 5t 3t+ 1/2 = 5t 1/2 = 2t or t = 15 minutes To check.. Rachel's distance was 5 x 1/4 = 1.25 km as well. in that they both work for the same amount of time and together they complete the given total amount. This problem is actually similar to Example 6.This form of the relationship may be necessary to use in problems when the rates aren't explicitly given. equivalently Rate Example 8 = rate of work x time number of jobs completed = ---------'--time If Joe can paint five rooms in 8 hours and Samantha can paint three rooms in 6 hours. For example. Problems Involving Work There are other situations which follow the same principle using a rate. we can use Number of jobs completed or. how long will it take them to paint seven rooms working together? Solution 8 Using the rate formula. Let t The model is Total number of rooms painted rooms painted by Joe + rooms painted by Samantha 7 = 5/st + 1M = 5/St + 4/st 7 = 9/st 56 = 9t = the time to complete the job together = 41 . we see that Joe's rate of work is 5/Srooms per hour and Samantha's rate of work is I/Z room per hour. when someone works at a job. how long would it take Suzanne to do it alone? Solution V Since Laurie takes less time to complete the job. we have 1 _ _1_ x 15 .4 t+ 10 +t ~) -=--+15/4 1 1 1 t + 10 t (Note that this line tells us that the combined rate is equal to the sum of the individual rates. where the total number of jobs completed is the sum of the amount of work done by each girl in the 2 hours. Therefore.) Simplify the fraction on the left to get the fractional equation 411 15 = t+ 10 +t Multiplying both sides of the equation by the least common 42 . we will let her time be represented by t. If it takes 3 3/ 4 hours for them to do the job together. If we can find this time. the amount done by Suzanne is l/t+ 10 x 15/4 .Example 9 It takes Suzanne 10 hours longer than Laurie to take their store's inventory.t+ 10 4 +t ~ x 15 _ 15 (_1_ 4 . and the amount done by Laurie is Ift x 15/4 . Substituting into the preceding model. The model for the problem is similar to Example 8. then the answer to the problem will be t + 10. you can use it as a model when the problem involves only completing 1 job. Laurie's rate is l/t and Suzanne's rate is l/(t + 10). If you remember this fact. 1 job = amount done by Suzanne + amount done by Allison Using 15/4 for 3 3/ 4. The number of jobs considered in this problem is 1. he creates type C. If he mixes 2 quarts (qt) of type A. What percent of type C is chocolate? Solution 10 In this situation. Problems Involving Mixing Quantities A situation similar to those in the problems above is finding out what occurs when we mix different types of objects. Suzanne can complete the job in 15 hours working alone. Therefore.5ort=5 Since the problem requires physical time. we have the relationship Amount of chocolate in the type = percent of chocolate x amount of the type 43 .150 = 0 both sides gives us Dividing all terms by 2 simplifies the equation to We can now factor and solve (2t + 15)(t . with 1 qt of type B. which contains 10 percent chocolate. but it is not a rate involving time. Example 10 At the chocolate factory. we reject the negative answer and we see that t = 5 is Laurie's time. For each type.5 =0 t=-7. we get 4t(t + 10) = 15t + 15(t + 10) Simplifying. we get a quadratic 4t2 + 40t = 30t + 150 equation Subtracting 30t and 150 from 4t2 + lOt .denominator. the rate is the percentage of chocolate. In these problems there is also a rate involved.15t(t + 10). John's job is to experiment with mixing different amounts of milk and chocolate to find different flavors.5) = 0 2t+ 15 =Oort . as we will see. which contains 15 percent chocolate. 1333 . Another type of mixture problem involves the price of a mixture of two items where each sells for different individual prices.P = number of pounds of honey-roasted peanuts The model for the problem is that the total cost for the mixture should be the same as the sum of the costs for each individual kind of nut: Cost of mixture = cost of cashews + cost of honey-roasted peanuts 44 .1 qt of chocolate Together. The underlying formula is also a kind of rate relationship since price is usually given as dollars per pound. We can solve the problem using only one variable since we know that the total amount must be 8 lb.4/3 = 0. Let p = number of pounds of cashews 8 . How many pounds of each type of nut should Nancy use to make 8 lb of a mixture that will sell for $3/lb? Solution II The problem asks us to find the amount of both kinds of nuts. in the mixture type C. we have 0. we have 0.1 + 0. amount =pnce Example II or cost = price x amount Nancy's Nut Stand sells cashews for $5/lb and honey-roasted peanuts for $2/lb pound.where the percentage is taking the form of the rate as we saw in previous problems... For instance Cost . the answer is easily found to be 0.10 x 1 = 0.15 x 2 = 0.3 qt of chocolate In type B. To do this. we have 0. we rewrite the relationship (amount of chocolate)/(amount of type) = percentage of chocolate.4 qt. Since the mixture has exactly 3 qt of liquid.4 qt of chocolate. or 13.3%.3 = 0. Since there are 0. we calculate the percentage that is chocolate. cents per pound. In type A. or dollars per kilogram. and using the relationship Cost = price x amount, we have (3)(8) = Sp + 2(8 - p) 24 = 5 P + 16 - 2 P 8= 3p P = 8/3 or 22/3 5 The mixture should have 22/3 lb of cashews and 8 - 22/3 = lb of honey-roasted peanuts. Alternative Solution 11 1/ 3 Another way to solve the problem is to use two variables and get two equations to work with. (See the Appendix for a review of how to work with equations with two variables.) Letx = amount of cashews in mixture Let y = amount of honey-roasted nuts in mixture The first equation is the total amount of the mixture x+y=8 The second equation comes from the cost relationship Cost = price x amount, and the model Cost of cashews + cost of honey-roasted peanuts = cost of mixture. Sx + 2y = (3)(8) = 24 The two equations together form a simultaneous system of equations and can be solved easily by the following process. Multiply the first equation by 2 and subtract it from the first Sx + 2y = 2x + 2y = 3x 24 16 =8 x = 8/3 = 22/3 45 Use this value of x in the first equation and solve for y: 22/3 + y = 8 Y = 51/3 It's okay if you are not familiar with working with two equations. As we saw, all of the problems can be solved with one variable and one equation. However, you will be a better problem solver by having several ways to solve a problem in your arsenal. Problems Involving Money One other kind of problem that is easily solved by an algebraic equation involves combinations of money of different denominations. These are problems similar to the previous ones because the amount of money that a coin is worth is a sort of rate. For example, a quarter is worth 25 cents per coin and a dime is worth 10 cents per coin. The amount of money we have is found by the relationship Amount of money = value of coin x number of coins Here, the value of the coin is taking the form of the rate. Example 12 Joey has $2.35 in nickels, dimes, and quarters. If he has two less quarters than dimes and one more nickel than dimes, how much of each coin does he have? Solution 12 The key to the problem is the way in which the numbers of quarters and nickels are related to the number of dimes. We can find the amounts of quarters and nickels if we know the amount of dimes. Therefore, we will assign the variable to this amount. Let x = number of dimes 46 Since the number of quarters is two less than this amount, we have x - 2 = number of quarters Since the number of nickels is one more than the number of dimes, we have x + 1 = number of nickels The governing relationship is Total amount of money = amount in quarters + amount in dimes + amount in nickels Using the Amount = rate of coin x number of coins, we have 2.35 = .25(x - 2) + O.lOx + 0.05(x + 1) Since we are given the total amount in dollars, we need to use decimals to express the rate of each coin as a decimal fraction of a dollar. If we multiply every term by 100, we can clear the decimals 235 = 25(x - 2) + lOx + 5(x + 1) Simplifying and solving for x, we have 235 = 25x - 50 + lOx + 5x + 5 235 = 40x - 45 280 = 40x 7=x Therefore, there are seven dimes, five quarters, and eight nickels. This is easily checked: 70¢ + $1.25 + 40¢ is indeed $2.35.1' 47 48 . Step 4 Substitute the representations of the unknowns into the mathematical statement you created in step 2. Step 2 Identify the formula or key concept that is needed in the problem and rewrite the problem. keeping most of the words but using basic arithmetic symbols for operations and equality. Step 5 Use algebra to solve the equation(s) and determine the value of the unknowns you represented by a variable. What are the two numbersl Separate the number 17 into two parts so that the sum of their squares is equal to the square of one more than the larger part. Step 6 Check your answer to see if it really does solve the problem and satisfy all the necessary conditions. Step 3 Assign a variable to one of the unknowns with a "Let" statement and determine the symbolic representations of all the other knowns with this variable using the relationships in the problem. the mathematical model of the problem is sometimes more easily seen when using two variables and creating a system of two equatiOns to solve.Summarizing what we have done with all our word problems. there are at least two relationships among the unknowns. we saw that we need to follow some basic steps: Step 1 Read the problem carefully and identify the unknowns. If only two unknown quantities are asked for. The larger number is as much greater than 34 as the smaller number is greater than 10. 2. It is worth remembering that in many problems. Additional Problems I. The sum of two numbers is 62. Begly's office has an old copying machine and a new one.3. If 2 is added to 7 times the smaller. The difference between two numbers is 12. 4. 5. The team in the east moved forward at 2 mi per day over mostly flat land. Melissa has two kinds of coffee she'd like to mix to make a blend. and how many miles of track did each team layl On an international flight across the Atlantic Ocean. How many years ago was Sarah's aunt 3 times as old as her moml A cyclist traveling 20 mph leaves 'h hour ahead of a car traveling at 50 mph. At which speeds do these planes fly in miles per hourl Mr.00/lb that isn't selling well. One sells for $1. How long did it take for the teams to meet and the last spike to be nailed in. How old are they nowl Sarah's mom is 42 years old. In 15 years he will be only twice as old as his son. How many pounds of the expensive candy should he mix with the 100 Ib in order to produce a mixture worth $1. The older machine takes 6 hours to make all copies of the financial report.75/lbl At her specialty coffee store. The manager notices a different candy worth $2. and her aunt is 48 years old. 7. 8. a certain type of plane travels at a speed that is 220 km/hour more than twice the speed of another. while the new machine takes only 4 hours. The faster plane can fly 7000 km in the same time that the slower plane can fly 2700 km. Find the numbers. 9. Find four consecutive even integers such that the sum of the squares of the two largest numbers is 34 more than the sum of the other two. What are theyl Find three consecutive odd integers whose sum is 75. Two teams at different ends of the country started at the same time. The sum of three consecutive integers is I I greater than twice the largest of the three integers. the result is the same as when 2 is subtracted from 3 times the larger. 14. how long will it take for the car to pass the cyclisd In the early days of building railroads in the United States. If both machines are used.60/lb and the other.20/lb. and the team from the west moved at 1. how long will it take for the job to be completedl The candystand at the multiplex has 100 Ib of a popular candy selling at $1. 6. the first transcontinental railroad was started from two points at the ends of the 2000-mi distance over which tracks had to be laid. 13.5 mi per day over mountainous terrain. A father is now 5 times as old as his son. 1 I. He decides to form a mixture of both types of candy to help clear his inventory of the more expensive type. If they go in the same direction. 12.10/lb. 10. $2. How many pounds of each kind must she use to make 49 . Let x = the smaller number. 75 Ib of mixed coffee that she can sell for $1. 18. 19.10 (62 . how many of each kind are there? The triplets john. How many days did he work? Solutions to Additional Problems I.05 in 34 coins on his dresser. The larger number . how much does she invest at each rate? Three people each received part of $139 so that the first received twice what the second received and the third's portion exceeded the sum of the other two by one dollar. If the coins are quarters and dimes. jen had only dimes and gave 4 times as many coins as john gave. he received $2260.x is the larger number. 17. She invested part of $5000 in one fund that had a return of 9% interest and the rest into a different fund that earned I I%. we have the following. john had only quarters to give. I. how much did each child contribute? Mrs.10 28 .x) . At the end of the 60 days. the expression "as much greater than" indicates that we want the difference between the numbers. and the larger number is 62 .19 = 43.x = x . joey had only nickels and gave seven more than twice the number of coins that jen gave.15.10 38 = 2x x = 19 The smaller number is 19. For each day that he worked he would receive $45. but for each day that he did not work he would pay $10 for his room and board. 16.90/Ib and make the same profit? joey has $5. If the total amount of the contribution was $22.34 = x . In Prob.40. 50 . therefore.34 = the smaller number . Stone likes to diversify her investments. 62 . jen. Using one variable. If her total annual income from these investments was $487. and joey gave their father all their coins toward a present for their mother. How much money did each person receive? A farmhand was hired for a 60-day period and would live on the ranch. Substituting into the first equation. Let x = smaller number and let y = larger number.34 = x ./ An alternative solution using two variables is as follows. Recognizing the sum of their squares to mean larger part2 + smaller part2 .10 = 9 .Check: 43 .12 = 0 x = 12 or or x . we have the quadratic equation Squaring the binomials./ 51 . Adding the first equation to this new form of the second equation gives us 2y = 86 and y = 43.24) = 0 x .. We should use the single variable to represent the larger part.24 = 0 x = 24 Since the larger part of 17 must be less than 17.12)(x .x = smaller part and x + I = one more than the larger part. The second equation could be rewritten as -x + y = 24. we have x 2 + 289 .36x + 288 = 0 We can factor this and solve (x . The larger part is 12 and the smaller part is 5. 2. Let x = the larger part and 17 .34 = 9 and 19 . The equations would be x + y = 62 and y . (12 + If 122+5 2 = 144+25= 169 and = 13 2 = 169.34x + x 2 = x 2 + 2x + I Combining terms gives us 2X2 .10. we reject the answer of 24.. since another piece of the puzzle relates to it.34x + 289 = x 2 + 2x + I Subtracting terms on the right from both sides of the equation produces x 2 .. Both "parts" of the number 17 can be represented using one variable. . Check: 12+5= 17. we have x + 43 = 62 and x = 19. y = 12 and y = 8. Therefore Let x = first (smallest) number x The model is + I = second (middle) number x + 2 = third (largest) number First number + second number + third number =2 x third number + II (Be sure not to confuse the phrase "is I I greater than" with" II is greater than.2.smaller number = 12 7 x smaller + 2 = 3 x larger . The problem is best understood and modeled when two variables are used for larger and smaller numbers. Consecutive integers imply that the three numbers are one apart.y = 12 and 7y + 2 = 3x .2 = 60 . (At this point. and we will be able to eliminate one of the variables by adding equations if we multiply every term in the first equation by 7: 7x . We can rewrite the second equation as -3x + 7y = -4. we have 20 . The other relationships are 7 x 8 + 2 = 56 + 2 = 58 3 x 20 .2 = 58\""" 4.) 3x + 3 = 2x + 4 + I I 3x + 3 = 2x + 15 x = 12 The numbers are 12. The two equations are x . see if you can identify what steps are being taken. The model is Larger number . Adding the equations gives us 4x = 80 and x = 20." which translates into an inequality" I I > "!) Then x + (x + I) + (x + 2) = 2(x + 2) + I I.7y = 84. This linear equation can be solved in the usual way. S2 . Using the original first equation to find y.8 = 12. Check: The difference of the numbers is 20 .2 Let x = larger number and let y = smaller number. and 14.3. The numbers are 20 and 8. 13. By definition. Therefore Let x = first (smallest) odd number x x +2= +4 = second (middle) odd number third (largest) odd number The model is simply First odd number + second odd number + third odd number = The equation is therefore 75 x + (x + 2) + (x + 4) = 3x + 6 = x 75 75 3x = 69 = 23 The three consecutive odd numbers are 23. an odd number is an integer (fractions are not considered to be odd or even). and 14 are consecutive integers. To be consecutive odd numbers the numbers must differ by 2. 12 + 13 39 and 2 x 14 + II = 28 + I I = 39. Let x = first (smallest) even integer +2= x +4 = x +6 = x second even integer third even integer fourth (largest) even integer The model for the problem is (Third integer)2 + (fourth integer)2 = first integer + second integer + 36 53 . Consecutive even integers must be 2 apart. Check: 23 + 25 + 27 = 75. Therefore.1' 6. and 27. 25.1' + 14 = 5.Check: 12. 13. The equation is therefore (x + 4)2 + (x + 6)2 = X + (x + 2) + 34 x 2 + ax + 16 + x 2 + 12x + 36 = 2x + 36 2x2 + 20x + 52 = 2x + 36 + 16 = 0 x 2 + 9x + a = 0 (x + I)(x + a) = 0 x+I= 0 or x + a= 0 x = -lor x =-a Since we want even integers. 40 = 2 x 20 . One relationship can serve as the basis for assigning a variable. -6.( S4 . -4. we reject x = . Check: In 15 years the son will be 20 and the father will be 40. 5x = father's age now The second relationship provides us with the model Father's age now + 15 = 2 x (son's age now + 15) 5x + 15 5x = 2(x + 15) 2x + 15 = + 30 3x = 15 x=5 The son is now 5 and the father is 5 x 5 = 25. The four even integers are -a.. and -2.I and our smallest even integer is -a. Since the father's age is given in terms of the son's age Let x = son's age now 7. There is no reason to assume that all number problems involve only positive integers! The two relationships are clearly stated in the problem. Check: (_4)2 + (_2)2 = 16 + 4 = 20 and 2x2 + lax -a + -6+ 34= 20 ..( Don't be disturbed by having negative numbers as answers. we make use of the formula Distance = rate x time.8.5) = 20t + 10 = 10 t SOt SOt = 30t = 1/3 hour or 20 minutes 55 . 9 = 3 x 3.5 = time traveled by cyclist The equation we have from the model is 20 x (t + 0.39 = 3. The "hidden" concept here is that at the moment when the car passes the cyclist.x) 48 . Since time is the unknown and the cyclist travels 0. The unknown in this problem is the number of years in the past from now. The model for the problem will be Distance traveled by car = distance traveled by cyclist For each distance. The model for the problem is Sarah's aunt's age in the past = 3 x Sarah's mom's age in the past Let x = the number of years from then until now.x = 3(42 . they have traveled the same distance. Sarah's aunt's age was 48 . Check: 39 years ago. Sarah's aunt was 48 . 39 years ago. Dividing both sides by 2 gives us x = 39. we have Let t = time traveled by car t + 0."f 9. Sarah's aunt was 3 times as old as Sarah's mom. Therefore. The equation is therefore 48 .3x To both sides we add 3x and subtract 48 to get the simple equation 2x = 78. Sarah's mom's age was 42 .x.5 hour longer than the car.x.39 = 9 and Sarah's mom was 42 .x = 126 . 43/365 = 1.43 = 857. There are clearly two relationships given in the problem.43 days or approximately 571. The first is straightforward and allows us to use one variable 56 . then at the point when they meet. Notice that the problem specifies that we give our answer in miles per hour. I I.5 = 571. Then./' 10. The car travels for 113 hour and covers a distance of 50 x 113 = sOh mi. This adds only one more step in the solution as we can find the rates in kilometers per hour and convert to miles per hour at the end. The model for the problem is based on the fact that the total distance is equal to the sum of the distances covered by both teams or Distance covered by team from east + distance covered by team from west = 2000 The "hidden" fact here is that when two objects are moving along the same path toward each other. The team from the east laid approximately 2 x 571. using Distance = rate x time for each team's distance.Check: The cyclist travels for 113 + Ih = 5/6 hours and covers a distance of 20 x 5/6 = 100/6 or 5Of6 mi.5t = 3.8 or I 143 mi of track. This is really a "motion" problem when thinking of the teams as very slow moving vehicles.5 x 571. we let t = the number of days spent by both teams. they have spent the same amount of time traveling! Since the rate is given in miles per day. Working in the metric system does not change the way in which we solve the problem. we have 2 mi/day x t + 1.5t 2000 2000 = 2000 t = 2000 -:.1 or 857 miles of track.56 or Ilh years.43 = I 142.5 mi/day x t = 2t + 1. the only necessary check is that 1143 mi + 857 mi = 2000 mi.3. The team from the west laid approximately 1. Check: Since the problem called for the amount of track laid. we have that the slower plane is traveling at approximately 230 mph and the faster plane can travel at 597 mph. Let r = rate of slower plane 2r + 220 = rate of faster plane The problem gives us the distances for the same amount of time traveled by both planes. MUltiplying each of these rates by the conversion factor of 0.000 r=371. The slow plane travels 2700 km in 2700 -. hours.5 x 7. and we can derive the model Distance of fast plane Rate of fast plane and the equation distance of slow plane rate of slow plane 7000 2r + 220 Cross-multiplying. From Distance = rate x time.1' 12.371. The model here states that if the machines are working together for the same amount of time. we will use the metric answers.62 mi/km.2727 .25 = 7..25 The slower plane is traveling at 371.. we have 2700 r 7000r = 2700(2r Simplifying and solving.2727 . = 5400. = 594. Check: For the check..50 km/hour.to represent both rates. 57 . we obtain + 220) 7000.. = 7000 km.000 1600. + 594. the fast plane travels 962. we have the alternative form of Time = distance/rate.25 km/hour and the faster plane is traveling at 962.. In this time. each will do a fraction of the job. The equation from the model is 1.6 = I job.:-. Using the Number of jobs completed = rate of work x time for each machine.4/4 = 0.00x. thus 0.4 + 0.---=--time The rate of the older machine is '/6 job/hour. we have I = t = '/6t + '/4t I = s/'2t '2/5 or 2.500 = 12.4 hours the older machine has done 2."! 13.000 + 200x 5500 = 25x x = 220lb 58 .75 (x + 100) = I .4/6 = 0.75 per pound should equal the total of having sold each of the two candies at their normal price or Dollar amount from mixture = dollar amount from cheaper candy + dollar amount from expensive candy We use the formula Dollar amount = price x amount sold.4 of the job and the newer machine has done 2.Therefore I job = fraction done by old machine + fraction done by new machine The rate of each machine can be found from number of jobs completed Rate = ----. Let x = the number of pounds of expensive candy and we have that x + 100 be the amount of the mixture.. The underlying concept here is that the amount of money earned from selling the mixture at $1.4 hours Check: In 2. and the rate of the new machine is '/4 job/hour.20 x 100 + 2. Let t = the number of hours the machines work together.6 of the job. Simplifying the equation and mUltiplying every term by I 00 gives us 175x + 17. Therefore.x = number of pounds of expensive coffee in blend Each amount is found by the formula Dollar amount = price x amount used.20 x 100 = $120. our equation is 1.50.1' 14.60 x 30 = $48.50 = + 2. and 220 Ib of the expensive candy would bring in $2. Check: The amount from the blend is $142.1' = $94. specifically. The two relationships in this problem are Total number of coins = 34 Value of the coins = $5.IOx -15 = -0.00 and $2. specifically.50. 100 Ib of the cheaper candy would bring in $1 .5 . The phrase "make the same profit" indicates that Melissa is seeking to receive the same amount of money from the blend as she would if she sold the same amounts that are in the mixture separately.x) 1.5x x = 30 Melissa needs to use 30 Ib of the cheaper coffee and 45 Ib of the more expensive coffee.75 x 320 = $560.50 = 142.60x + 157. $1.00 + 15. Let x = number of pounds of cheaper coffee in blend 75 .2. the mixture would bring in $1.00 x 220 = $440.10(75 .90 x 75 = 1. The model is therefore Dollar amount received from blend = dollar amount from cheaper coffee + dollar amount from the expensive coffee The two amounts of coffee can be represented by the same variable easily.60x 142.10 x 45 $94.50.05 59 . $120 + $440 = $560. $48.Check: When all candy is sold. 40 Each amount is calculated by Amount of money = value of coin x number of coins.. so 4x = number of coins (dimes) Jen gave 60 . We should try to express all three numbers of coins with the same variable instead of using three different variables. Amount of money = value of coin x number of coins Total amount of money = amount in quarters + amount in dimes The second equation is therefore 0.) Amount from John's quarters + amount from Jen's dimes 16.25 x + I I quarters = 34 coins and $. Therefore. The model for the problem is given by + amount from Joey's nickels = $22.10 x 23 II = $2. jen gave 4 times as many coins as john.d for the number of quarters. we multipy the first equation by 25 and the second equation by 100 to get 25d + 25q 10d + 25q = 850 = 505 and we subtract to get 15d = 345.( (Try solving the problem. using d for the number of dimes and 34 . d = 23.05. The first relationship gives us the equation d + q = 34.Let d = the number of dimes and let q = the number of quarters.30 + $2. The second relationship uses the following concepts. Let x = number of coins (quarters) john gave. and from the original first equation 23 Check: 23 dimes +q = 34 or q = II + $. we need to study the given relationships carefully.75 = $5. To get both equations into a form in which we can add or subtract them to eliminate one of the variables.1 Od + 0. To do this.25q = 5.05. The unknown in the problem is the amount invested in each 61 ./ Also $0.40 + $8. we have 25x + 40x + 40x + 35 = 105x + 35 = 105x 2240 2240 = 2205 x =21 Using this in our "Let" statements. and Joey gave 8 x 21 + 7 = 175 nickels.25 + $8. .05 x 175 = $5. so 8x +7 = number of coins (nickels) Joey gave Using these representations in the model.25x + 0.Joey gave 2 times the number of coins Jen gave + 7.75 = $22./ and 175 is 2 x 84 + 7 . Check: 84 is 4 x 21.40. we deduce that John gave 21 quarters.10 x 84 + $0./ 17. Jen gave 4 x 21 = 84 dimes. we have 0.10 x 4x + 0. This problem is similar to those we've looked at in that the underlying model is Total interest earned = interest earned from 9% fund + interest earned from We use the formula I I % fund Interest earned = principal x rate of interest x time invested and note that the time over which this interest was earned was I year.05(8x + 7) = 22.40 Simplifying and multiplying every term by 100.25 x 21 + $0. The situation is clearest when we use the second person's amount as the single variable since the first is easily determined when we know this amount.00.50 +$203. The model gives us the equation 487 = 0. therefore Third person's portion = first person's portion + second person's portion + $1 139 .700 = 9x + 55. then $283.000 . Since the total amount is $5000.11 (5000 .50.(x + 2x).700 = 55. Stone invested $3150 in the 9% fund and the balance. This forces us to try to use only one variable to model the problem.2x 2x = $6300 x = $3150 Mrs.3x = 3x + I 138 = 6x x = 23 62 .09x + 0.50 = $487.3x = amount received by third person Note that the phrase "exceeds by some amount" can be translated as "is that amount more than.11 x $1850 = $203.Ilx 48.x) 48.1' 18. we let x = the amount invested at 9% and 5000 .50.$3150 = $1850 in the I I percent fund. and the interest from the II percent fund was 0.x = the amount invested at I I percent. 139 .3x = 2x + x + 1 139 .000 ." The model is. Let x = amount received by second person. Clearly.fund. 2x = amount received by first person.09 x $3150 = $283. The third person must have received the remaining portion of the $139 or 139 . It would become very complicated to try to use three variables to represent the different amounts that each person received. $5000 . Check: The interest from the 9 percent fund was 0. 45x .) Let x = number of days that the farmhand worked Let y = number of days that the farmhand did not work We know that the total number of days is 60.The second person received $23. Check: $23 + $46 + $70 = $139. then $23 + $46 + $1 = $70.1' = $2340 - 63 . then 52 x $45 -8 x $10 $80 = $2260. The farmhand worked for 52 days and did not work for 8 days.1' 19. This problem doesn't belong to any category that was mentioned in this chapter. and we find x = 52. the first person received $46. (See the Appendix for a review of how to work with equations having two variables.lOy = 2260 Adding the two equations gives us 55x = 2860. therefore. Having two variables makes the modeling easy. but it is still basically a "rate x amount" problem and is easily solved with some algebra. The rates are $45 earned per day and $10 paid per day. Therefore.lOy = 2260. Check: 52 + 8 = 60.amount paid for room and board = amount received The second equation is. and the third person received $139 . Multiplying every term in the first equation by 10 and adding this to the second equation gives lOx + lOy = 600 45x .$69 = $70. the first equation is x+y=60 The basic concept is that Amount earned from working . . Writing groups of BBBGGGGG and keeping track of how many we have in a cumulative way gives us: 65 . but the fraction that indicates how much of the larger group is the smaller or how two groups are related to the whole. Sometimes when ratios are given. how many boys and how many girls are there in the class? Solution I One way to understand this problem is to restate the ratio in the following way. These fractions are referred to as ratios. there are 32 students. and Percentage Many situations presented in word problems involve comparison of the sizes of groups and objects. This is equivalent to the fraction a/b. If the ratio of boys to girls is 3:5. Proportion. The answer sought may not be the number of objects. Example I In a math class. With this we can create a visual representation of the situation using B for a boy and G for a girl. they are given by the statement a:b. The fraction allows us to use arithmetic in the usual way to solve problems.Chapter 3 Word Problems Involving Ratio. For every three boys there are five girls. but it would be tedious if the group were much larger than 32. we have the equation 3x + Sx = 32 8x =32 x=4 66 .) If we think of the number of groups as the unknown. The model for the situation is Class size = number of boys + number of girls where each number is found by Rate of boys or girls per group x the number of groups (In Chap.6 = 3/s . 2. The check for this answer is to create the fraction of boys/girls and see if it is equal to 3/s. we can solve the problem algebraically by assigning a variable to this and modeling the situation with an equation..Groups BBBGGGGG BBBGGGGG BBBGGGGG BBBGGGGG Running count 8 16 24 32 Adding the Bs and Gs tells us that there are 12 boys and 20 girls. The problem really involves finding the number of groups and applying a rate. Let x = the number of groups Number of boys = 3x Number of girls = Sx Using these quantities in the model./ Alternative Solution 1 The method described above was quick. Check: 12/z0 = 0. The ratio 3 boys:S girls can be translated into two rates:3 boys/group and 5 girls/group. there are many problems using similar models. therefore'. Rather than listing groups. Therefore. Ratios can appear in many of the types of problems we saw in the previous chapter. we easily realize that there has to be 180/5 = 36 groups and.There are four groups. each consisting of 3 boys and 5 girls. there are 3 x 4 = 12 boys and 5 x 4 = 20 girls. How many dentists would they expect to not recommend this brand. there would be 36 Ds if the claim were accurate. The more structured algebraic approach would be Let x = the number of groups Number of those who recommend = 4x Number of those who do not recommend = x 4x +x = 180 5x = 180 x =36 There would be 4 x 36 = 144 dentists who recommend the brand and 1 x 36 = 36 who do not. A consumer group wanted to check the accuracy of this claim and surveyed 180 dentists. assuming that the square of the smaller is 15 less than 10 times the larger. Number Problems Example 3 Two positive numbers are in the ratio of 5:8. we would have to list and count groups of RRRRD. if the advertisement's claim is accurate? Solution 1. The phrase "4 out of 5" indicates that the ratio of those who recommend to those who do not recommend is 4: 1. Example 2 An advertisement for toothpaste claims that 4 out of 5 dentists recommend a certain brand. Find the numbers. Using the picture scheme R for recommend and D for doesn't recommend. 67 . Multiple Smaller Larger Smaller2 10 x larger-IS 1 25 5 8 80 -15 = 65 2 10 160-15 = 145 100 16 240 . listing them in this fashion would be tedious. we can list pairs of multiples and check the conditions for each pair.15 25x 2 .15 = 225'( 15 24 225 3 As before.16x +3 = 0 (5x .15 25x 2 = 80x .1 = 0 or x . specifically. this is an easy solution because the actual numbers came up early in the list.80x + 15 = 0 Dividing every term in the equation by 5 gives us the simpler quadratic 5x 2 .3 = 0 x = 1/5 or x=3 68 . If the actual numbers were larger.15 and we would have the quadratic equation: (5X)2 = 10(8x) .3) = 0 5x ." we can use a solution similar to that for Probs.Solution :I If we think of multiples of 5 and 8 as "groups. 1 and 2. The algebraic approach would be Let x = "multiple" needed 5x = 8x = smaller number larger number The equation would come from the model Smaller2 = 10 x larger .l)(x . liz of Jason's age -31/z +3 1 /z +4 "f 15 15 20 15 23 32 16 Alternative Solution 4 Using algebra Let x = the multiple Chris' current age = 5x Jason's current age = 9x The model is Chris' age + 5 = lIz Oason's age + 5) + 4 Substitution gives 5x + 5 = 1/z(9x + 5) + 4 5x + 5 = 4. Therefore.5 + 4 69 .15 = 240 . Check: 15 2 = 225. we reject the answer of lIs. How old are the boys now? Solution 4 The ratio indicates that the current ages of the boys are multiples of 5 and 9.5x + 2."f Age Problems Example 4 Jason is Chris' older cousin. Now Chris Jason 5 10 9 18 27 In 5 years Chris Jason 11 lIz Jason's age 71/z 11 1 1z Chris' age . then 10 x 24 . In 5 years. Chris' age will be 4 more than half of Jason's age. it must be a whole number.15 = 225. Our numbers would be the third multiples of 5 and 8 or 15 and 24. The ratio of their ages is 9:5. Making a table as follows will lead to the answer quickly.Since x is a multiple. ) 2~ 10x Let x be the multiple. The sides of the triangular plot are 2x. A surveyor is making a map of the land and notes that the ratio of the sides is 2:9:10.5 0. Check: In 5 years their ages would be 20 and 32. and Jason's age is 9 = 18.5x = 1. An algebraic approach gets to the heart of the problem quickly.v' Perimeter Example 5 The perimeter of a triangular plot of land is 273 m. The model is Sum of all sides We have the equation = 273 2x+9x+10x=273 21x = 273 x = 13 70 . 9x. = liz x 32 + 4 = 16 + 4 = 20. (Make sure to draw a diagram that reflects the ratios to help visualize the problem accurately. the use of a table would be a lengthy task.5 x3 x=3 Chris' age is 5 x 3 20 = 15.5x + 6. Specifically. and lOx. What is the actual length of each side of the plot? Solution S In solving this problem.5x + 5 = 4. we might ask.95 = $928.20 71 . and 26 + 117 + 130 = 273. which would be the same as if she sold each candy separately.85 -:. How many pounds of candy does she have in supply. In other words. The price of the mixture would be $17. 9 x 13 = 117 m. she would have revenue of $928.23/lb (rounded). but it helped us model the problem. we should try to use a simple number and work with the information in the problem to help get a better understanding of what is occurring. Sometimes when a problem seems complicated.The sides are therefore 2 10 x 13 = 130 m.1' Mixture Example 6 Carole's candystand sells a popular mixture of two of her candies which sell for $2.95/lb.85. The model is Number of pounds of expensive candy x $2.70 + 5 x $1. let's assume that 8lb was sold.95 = $17. The key to the problem is that the revenue would be the same as if she sold each type of candy separately. We easily see that 8 lb is not the answer we are looking for. Check: x 13 = 26 m. For example.70 + number of pounds of cheaper candy x $1. the answer to this question is 3 x $2. and at what price does she sell it for? Solution 0 This seems like a more complicated problem than the previous ones in that more information is given. "What would be the revenue if 3 lb of the expensive candy and 5 lb of the cheaper candy were sold?" Clearly. If Carole sold her entire supply of the mixed candy during the coming month.8 lb. since the ratio of the candies in the mixture is 3:5. The ratio of the expensive candy to the cheaper candy in the mixture is 3:5.20.70/lb and $1. giving us $2. When dealing with such situations. which would have a revenue of 156 x $2. a numerically similar situation could be working with a recipe that lists quantities required to make four servings needs to be scaled to make more servings when more than four are required. This would be the price for any amount sold.23/lb.20.00. (Note that this is the same price as when 8 lb is sold.20 17.. This often requires a scaling of the quantities in a proportional manner. 72 . since we are maintaining the same ratio in all quantities of the mixture. we create a mathematical statement indicating that the ratios are the same.416 = $2. A geometrically similar situation could be working with two rectangles that are of different sizes but of the same length:width ratio.20 + $507.70 = $421...20. which would have a revenue of 260 x $1. and 5 x 52 = 260 lb of the cheaper candy. 8 x 52 = 416lb ofthe mixture is in supply and sells for $928.00 = $928.85x = 928. For example.1' Proportion Word problems sometimes involve comparing two situations that are numerically or geometrically similar.20 the multiple 3x 5x 8x x = 52 Therefore.) Check: The mixture contains 3 x 52 = 156 lb of the expensive candy. The total revenue is $421.20.Using algebra: Let x = Number of pounds of expensive candy = Number of pounds of cheaper candy = Total number of pounds sold = The model gives us the equation 3x x 2.95 = 928.70 + 5x x 1.95 = $507. This is called a proportion and would appear in ratio form as a:b = c:d or in fraction form as alb = cld. how much of each type of sugar will be needed? Solution 7 The underlying assumption in the problem is that the recipe needs to be scaled proportionally when seeking to make more servings.From the ratio form. Let x be the amount of regular sugar for 12 people. Using (a). we see that if we multiply both sides of the equation by bd. This also helps us see that we will need to solve two different proportions: one for regular (refined white) sugar and the other for brown sugar. therefore (a) The amount of regular sugar for 8 people is proportional to the amount of regular sugar for 12 people. The model for the problem is. The first step is to decide which two of the three quantities given in the original recipe are necessary for the ratios that will appear in the proportion: 2 liz. It is helpful in solving problems of this kind to rephrase the question so that the problem is better understood: "What happens when the number of people changes to 121" This indicates that our ratios and proportion must involve 8 and 12. From the fraction form. 1/4 . This will be the governing principle in solving the following problems. The model 73 . or 8. If there will be 12 people to serve. The more mathematical phrase is: In a proportion. we call a and d the extremes while band c are referred to as the means. The directions indicate that this will serve 8 people. we can assign a variable and proceed algebraically. the product of the means is equal to the product of the extremes. (b) The amount of brown sugar for 8 people is proportional to the amount of brown sugar for 12 people. This is sometimes referred to as crossmultiplying in order to solve a proportion. Measurement of Quantities Example 7 Among the ingredients in a recipe for a sweet dessert are 21/2 cups of regular sugar and 1/4 cup of brown sugar. we arrive at ad = bc. we have 8x = 3 and y = 3/8 cup of brown sugar.1/4 For brown sugar. Check: One way to check the problem is to see if the ratios of the original amounts to the new amounts is the same as 12:8 or 3/z. ./ Measurement of Similar Figures Example 8 On a blueprint for a new office building. Once we find the actual width. we can solve in a similar fashion. Therefore. that is. we actually 74 . . Note that in computation of w.5w = 180. we have 4. For regular sugar. Using (b). we need to compute the area of the floor of the room. I. Let y be the amount of brown sugar for 12 people 1/4 y 12 8 Cross-multiplying. 33/ 4 -:./ 3/8 -:.2 1/z = 15/4 -:. The shorter wall of the actual room measures 15 ft. w = 180/4.5/2 = 15/4 X 2/5 = 30/z0 = 3/2 . = 3/8 X 4/1 = 12/8 = 3/z .5 = 40 ft.5 in by 12 in. we were concerned about how to keep track of units of measure. the sides of the room on the blueprint are proportional to the actual sizes or the ratios of the corresponding sides are equal. How much carpeting will be needed to cover the floor of the actual room? Solution 8 The underlying assumption is that a blueprint is a scaled drawing of the room.5/15 = 12/w. In Chap. Cross-multiplying gives us 4. The model for the problem is (length on blueprint)/(actuallength) = (width on blueprint)/(actual width). The area is 15 x 40 = 600 ft2. Letting w be the actual width.gives us the equation 21/2 8 x 12 Cross-multiplying. the rectangular conference room measures 4. we have 8x = 30 and x = 30/8 = 33/ 4 cups of regular sugar. distance Tl m e = . the ratio of the areas is the square of the ratios of the lengths. This fact is useful to remember. we can use the ratio of distance/rate for each situation and create a proportion to 75 . and the ratio of the lengths is 180 in/4. including units: w = (12 in x 15 ft)/4..400 in 2/54 in 2 = 1600.have the following calculation. The length of the actual room is 15 feet = 15 ft x 12 in/ft = 180 in. we can set up a proportion between the distances and rates. In general. The area of the room on the blueprint is 4. one of which was 20 mph greater than the other. .400 in 2. We see that something interesting happened when we compare the areas of the blueprint drawing to the actual drawing.5 in = 40 ft. Travel From the formula for motion Distance = rate x time we have the following equivalent formula. and what was the test time to the nearest second? Solution g Since the test times were the same. What we notice is that 1600 = 402. The ratios of the two areas is 86.5 in x 12 in = 54 in2. Example 9 Test drives of a new automobile indicated that while traveling at two speeds. and the area of the actual room is 600 ft2 = 600 ft 2 x 144 in 2/1 ft 2 = 86.rate In a problem where the time taken is the same for similar modes of transportation. which shows that we have made the correct calculation as it gives us our answer in feet when we cancel units. What were the tested speeds.5 in = 40. the higher speed covered a test distance of 1000 ft while the lower speed covered a distance of 800 ft in the same time. while the faster tested speed was 95 mph.19r.002 hour 60 minutes 60 seconds 7 2 7 d -------. Calculating the time tested actually gives us the check on the problem since both distances and speeds should produce the same time.189 = 0. Tlme of slower test = distance rate .19 mi = 0.---=-=- 0.1 . The model gives us the equation 0. Multiplying by 100. 1000 ft 1 mi 1 x 5280 ft 800 ft 1 mi .002 hour = -95 = 0. 15r + 300 = 19r 300 = 4r r = 75 mph The slower tested speed was 75 mph. distance Tlme offaster test = rate .152 = 0.002 hour 0.model the problem: ----~-----------=----~--------- Distance of slower test Slower rate distance of faster test faster rate Before beginning.15 mi Let r be the slower rate and r + 20 be the faster rate. we have 0.15 To convert to seconds. = secon s 1 1 hour 1 minute 76 .19 r +20 Cross-multiplying. we use 0. = -75 = 0.15(r + 20) = 0.x x = . and we can solve this equation in the usual way. we have 15(r + 20) = 19r.x 5280 ft = 0. however.15 r . we need to convert the distances into miles since the rates are given in miles per hour.19 0. 83 and 20 -. 12 -. that is. Example 10 A construction firm knows that the number of new houses that it can build in a month varies directly with the cost of labor due to overtime charges. v' (Note that the calculator shows differences to the fourth decimal place. we have 12c = 290 and c = 290/12 = $24.. The model gives us 20 12 = c 14.24.17 per hour.14. that is. how much force.83 is referred to as the constant of variation.50 Cost of labor Let c be the cost of labor we are seeking. The check would be that the ratios would be the same.50 Cross-multiplying.. If a force of 15 dynes (the metric measure of force) stretches a spring 2. if it needs to build 20 houses? Solution 10 Since we are told that the quantities vary directly.3 cm. we can immediately set up the model proportion.50 = 0. how much should the company expect to pay its workers.83. when one aspect of a situation changes. This is due to the rounding we did in the problem. Check: Example II The force applied to stretch an elastic spring varies directly with the amount that it is stretched. would be required to stretch the spring 6 cm? 77 . to the nearest dyne.17 = 0. another aspect changes in a proportional way or at the same rate.Direct Variation Another way in which problems make use of proportions is to refer to a direct variation among changing quantities. 12 Number of houses = 4.50 per hour.) The ratio 0. If 12 houses can be built when it pays an average of $14. 3 = 39. you can think of it as asking "How much would we have if we scaled our base amount to be 100?" For example. When asked for a percentage of a number.3 and f = 90/2. and we have f 6 15 2.Solution II Since we are told that the quantities vary directly. when asked "What is 35% of 80?" we can translate this to mean "If we have 35 objects out of a set of 100.3 f be the required force. what would be the annual rate of return for this fund? Solution 12 In order to determine the annual rate. we can immediately set up the model proportion: Amount of force Stretch Let 15 2. Check: The constant of variation must be the same: 15 -:. Percentages appear in many of the problem situations we have seen before.6 = 6.5 and 39 -:.2.5.3 = 6."! Percentage Problems that involve percentages are really problems involving proportions. Assuming the same rate of return.13 = 39 dynes. he had $1240 at the end of 3 months. we have to make use of the given assumption and conclude that over the year 78 . This gives us 100x = 35 x 80 = 2800 and x = 28. how much would we have in a similar group of 80?" The problem is then easily modeled by the proportion 35/100 = x/80. Investment Example 12 When Gary invested $1200 in the Lion Fund. The model gives us the proportion 3376 x = 3690 100 79 . Using the percentage model.33 .133333 x . 4 x $40 = $160. then multiply by 0. In other words. The percentage model is then used to give us Area of carpeted space Area of entire room x 100 The area of the entire room is 45 x 82 = 3690 ft2. which is the amount of carpeted space. Therefore.000 and x = 16.314 = 3376 ft2. which is approximately $40.14 for rr. and their difference.000/ 1200 = 13 113%. (The difference arose because we did not have to use the entire repeating decimal. we have 160 1200 = x 100 Cross-multiplying gives us 1200x = 16.14 x 102 = 314 ft2. the carpeted area is 3690 .( Area Example 13 The floor of a social hall that is 45 ft by 82 ft is carpeted except for a circular dance floor in the middle that has a diameter of 20 ft.25 = $39. The radius of the dance floor is 10 ft and using 3. the area of the circle.he would earn 4 times the interest received. the area of the circular dance floor is 3.. To the nearest tenth of a percent.. $1200 x 0. since 3 months is one-fourth of a year.25 to determine what the quarterly interest is.9999.) . what percent of the floor is carpeted? Solution 13 The solution requires finding the area of the entire floor. that is. Check: We can calculate the interest for one year at 13. percent and. 900 .. Regarding 30% as 30/100. we have 30(950 .x is the amoun t of remaining water and the total weight. is 900 . we have 28. How much water. The model gives us the equation: SO 30 .500 . the model for the problem is the proportion ~--------------------~~= Amount of acid Remaining water + amount of acid 30 100 Let x be the amount of water that has evaporated.-.Solving for x. Fortunately.x) = 5000. Check: The check is simple: 3690 x 0.v' More complicated problems involving percentages can be found in mixture problems involving liquids that contain dissolved ingredients such as chemicals.49 = 91. has to evaporate for this to be accomplished? Solution 14 First. the solution is made with SO kg of dry acid and 900 L of water. in the metric system. we have x = 3376 x 100/3690 = 91.30x = 5000 23. 1 L of water weighs exactly 1 kg.915 = 3376 (rounded). Example 14 A certain chemical solution used in manufacturing batteries contains water and acid.500/30 = 783 113 Therefore. 80 . in kilograms.5%. Therefore. Initially. Simplifying and solving..x + SO = 950 . to the nearest liter.x.x Cross-multiplying.500 = 30x x = 23. we need to make sure that our units of measurement are the same. 783 L of water has to evaporate. The solution is mixed and left to sit so that the water evaporates until it becomes a 30% acid solution.= 100 950 . how long is the wingspan for the actual plane to the nearest foot? 3. 8% of the girls and 5% of the boys went on a field trip. Additional Problems Separate 133 into two parts so that the ratio of the larger to the smaller is 4:3. In the sixth grade. in the eleventh grade. If the wingspan of the model is 9 in. On a box for a model-airplane kit.299 = 29. 50/167 = . I. Shapes are similar only when corresponding sides are in proportion. Solve for the mUltiple and compute the actual amounts. 2. that is. assign a variable to be the multiple of each number in the ratio and use the specific multiples given by the ratio as the amounts.50 and the profit from a classical CD is $1. Computing a percentage or using a given percentage will usually be solved by creating a proportion in which the ratio of the actual amounts is equal to x / I00. of the school's population went on this trip? I." or "is proportional to" appear in the problem. and the total weight is 50 kg of acid + 117 kg of water = 167 kg. it is 3:2.9%. What percent. Model the problem and substitute these variable amounts to derive an equation to solve. to the nearest tenth.80. Of this group. when they are in eleventh grade. 3. how much profit was made if 2345 rock and classical CDs were sold during November? 4. CD Emporium finds that CDs of rock artists outsold CDs of classical music by 5:2 during November. the ratio of students who found math easy to those who didn't was 3: I. 2.Check: The remaining water is 900 . This will be the case when phrases such as "in the same ratio as:' "varies directly as. which is okay since we used a rounded amount. now.783 = 117. labels indicated that the scale used is I:72. Identify when a proportion is implied in the problem. How many fewer students find math easy now than before? 5. 81 . A survey to determine attitudes of students of mathematics was given to 420 sixth-grade students 5 years ago and to the same group now. Glenn High School has 1620 students in which the girl:boy ratio is 5:4."! Summary When given a ratio of two or more objects. If the profit from a rock CD is $2. 4. where x is the percentage. Regular chocolate ice cream has 240 calories (cal) per serving and contains 64 g of sugar. If the digits are reversed. How much will it cost to fill the region on the actual patio~ The sum of the digits of a two-digit number is 12. During the same week. Given that the longer side of the second rectangle is I I in more than the longer side of the first and the shorter side is 18 in more than the shorter side of the first rectangle. and each bag can be used to cover 8 ft2. Carl's crew completed 12 more jobs than Ben's crew. 9. 8. the respective returns were 8. Find both numbers. At what speed does each vehicle travel~ 82 . The amount of calories in a serving of ice cream varies directly with the amount of sugar the ice cream contains. A plane can make the same IOOO-mi trip as an express train in 6 hours' less time. Over the course of a year. The ratio of the ages of a girl. Carl's crew can complete 7 jobs in the same time that Ben's crew can complete 5 jobs. Six years ago the ratio of the girl's age to her mother's age was 4: 13. The ratio of another rectangle's longer side to its shorter side is 4:3.000 in three different stocks in the ratio of 3:4:7. A company makes a regular chocolate ice cream and a "Lite" version. Simmons tries to be a shrewd investor and invested her $28. Mrs. and a side of the triangular region is 1.6. How many grams of sugar does the Lite chocolate have. What was her overall percentage return for the year~ 14. find the ratio of the perimeters of the first rectangle to the second. 12.25 per bag. and the girl's grandmother is 2:5:8. The ratio of their speeds is 5:2. 1 I.5 feet. there is an equilateral triangle in its center that is to be filled with a specially colored concrete mix costing $15. a larger number is formed and the ratio of the larger number to the smaller number is 7:4. 7. her mother. On a scale model of an outdoor patio at the home design center. and 10%. How many jobs were completed by both crews~ Fifty liters of an acid solution contains 18 L of pure acid. The ratio of the longer side of a rectangle to the shorter side is 7:2. The model is scaled at a ratio of 1:8 compared to the actual size of the patio. How many liters of acid must be added to make a 4% acid solution~ 12. How old are the three women now~ 10. if it has only 180 cal per serving~ 13. 0139.Solutions to Additional Problems I. = = 'h. and the smaller part is 133 . Using the calculated inches. Cross-multiplying.80.648 = . The problem gives us the exact proportion x : 133 . we have 4(133 . This is a simple problem that will use the following proportion as its model.1' 3. 648 in x ~ I 12 in = 54 ft Check: We need to make sure that the ratio of the actual span to the model span is 1:72 or 0. we have x = 9 x 72 = 648 in.x) = 3x 532 . we have 9 in/x = lin.f 2. we have 9 -. Therefore.. Therefore.345. Step 2 Total profit = the number of rock CDs x $2. Using the product of the means is equal to the product of the extremes. The total profit is the sum of the profits from each type.76 = 57. We need to first find the amount of each type of CD sold. Wingspan of model Actual wingspan = 72 We can find the actual length in inches and then convert to feet..x = 4 : 3. 83 . Check: 76 + 57 = 133 and 76/57 = 1.333 . Let x be the larger part of 133. 133 . Let x be the number of inches in the actual wingspan.0139.4x = 3x 532 = 7x x = 532/7 = 76 11/3 The larger part is 76.. Step 1 Number of rock CDs + the number of classical CDs = 2.x is the smaller part. The model for the problem consists of two steps..50 + the number of classical CDs x $1. Let y = multiple 3y = number of students who find math easy 2y = number of students who do not 84 .Let x = multiple of 2 and 5 2x = number of rock CDs sold in November 5x = number of classical CDs sold in November From the model for step I 2x + 5x = x 2345 7x = 2345 = 2345 -:. specifically.7 = 335 Therefore. For the sixth grade.00 = $5393.5.50 + 670 x $1. there were 3 x 105 = 3 15 students who found math easy (and 105 who did not). We have to compute the actual numbers of students who find math easy and those who do not at each grade level on the basis of the given ratios. 3x be the number of students who found math easy.50 + $1206.5. 3x +x = 420 4x = 420 x = 105 Therefore. 4.50 Check: The check is concerned mostly with the amounts of each type of CD sold. and Ix be the number of students who did not. At the eleventh grade (we should use a different variable so to avoid confusion between our two answers). From the model for step 2 Total profit = 1675 x $2. 2 x 335 = 670 classical CDs were sold and 5 x 335 = 1675 rock CDs were sold. let x be the multiple.. at the sixth grade.670 = 2. 1675 -:.flt is also necessary to recheck the arithmetic for step 2. We need to make sure that the ratio is indeed 5:2 or 2.80 = $4187. .. and 5% of the boys would mean 40 boys on the trip.1620 = 0.. 315 . we see that simply adding 8% + 5% = 13% is most likely not the correct answer. The problem must be solved by finding the number of girls and boys who went on the trip. The sixth-grade ratio should be 3 : I or 3.3y + 2y = 420 5y = 420 y = 84 Therefore. there would be 810 girls and 810 boys. The difference 3 15 . The ratio 5:4 indicates that slightly more than half of the students are girls. An estimate of the total number of students on the trip would be 120 and 120 -:. (This is the most common error in a problem like this.074. that is.168 = 1... Let x = multiple 5x = number of girls 1620 4x = number of boys 5x + 4x = 9x = 1620 x = 180 85 . The algebraic scheme for finding the ratios will give us the numbers of girls and boys in the school. If the groups were equal.fThe eleventh-grade ratio should be 3:2 or 1. which is approximately 7 percent..252 = 63 is our answer...51' 5. Using 10% as an estimate for 8%. Check: The check involves making sure that the numbers we determined are in the ratio given in the problem. 252.. about 80 girls would be on the trip.5.105 = 3.) It would be helpful to have an estimate before beginning. using the percentage model Number of girls on trip + number of boys on trip x 100 Total number of students Realizing this. that is. at the eleventh grade there are 3 x 84 = 252 students who find math easy (and 2 x 84 = 168 who do not). Since the only information about the first rectangle is the ratio of the sides. we will need to use the information about the second rectangle to help find the sides. we could have set up our answer in variable form immediately.05 = 36. and 2x be the shorter side. The model gives us the proportion: IOS/ 1620 = x/I 00 and x = 10. The answer is reasonable on the basis of our initial estimate. Specifically.SOO -:-1620 = 6.. the perimeter of the first rectangle is 2(7x + 2x) = ISx and the perimeter ofthe second rectangle is 2(7x + II + 2x + IS) = ISx + 58. We can set up the following proportion for the second rectangle.There are 5 x 180 = 900 girls and 4 x ISO = 720 boys. Alternative Solution After having assigned the variable and obtained representations for the sides. we have the equation 3(7x Simplifying and solving. Its perimeter is 2 x (21 + 6) = 54 in. Let x be the multiple needed for the first rectangle.66 . 7x 2x + II + 18 - 4 3 Cross-multiplying.. 7x be the longer side of the first rectangle. which reduces to 27/56 or 27:56. The sides of the second rectangle are 21 + I I = 32 in and 6 + IS = 24 in. we obtain 21x + II) = 4(2x + IS). Its perimeter is 2 x (32 + 24) = I 12 in. Therefore. The ratio we want is 54: I 12 or 54/1 12. Then 7x + I I is the longer side of the second rectangle and 5x + IS is the shorter side of the second rectangle.7% rounded to the nearest tenth. 72 + 36 = 108 students went on the trip. Thus S% of the girls is 900 x O. 6. Check: The best check for a problem with these many steps is to recheck the arithmetic at each step and to feel sure that the answer is reasonable.OS = 72 and 5% of the boys is 720 x 0. = 6. The ratio of 86 . the dimensions of the first rectangle are 7 x 3 = 21 in and 2 x 3 = 6 in. + 33 = 8x + 72 13x = 39 x=3 Therefore. u and when substituted into the equation. the ratio is quickly seen as 54/1 12. the numbers are 48 and 84. We know that the original number is the smaller of the numbers.4 x 144 = 62. The ratios to consider are 93/39. Once we found that x = 3. Using the formula for the area of an equilateral triangle from the reference table in Chap.785 bags. we find that the area is . lOt + u is the original number and IOu + t is the number when the digits are reversed.75. The cost is $15. The ratio 7:4 is to equivalent to 1. Simplifying gives us 40u + 4t = 70t + 7u.the perimeters would be 18x / (18x + 58).75 is 84/48. The only one of these equal to 1. Check: We need to ensure that the ratio of the numbers is 7/4 = 1. and 57.63u 99u = 792 u = 792/99 = 8 (and t must be 12 . we have an equation in one variable to solve 40u + 4( 12 . The only two-digit numbers whose sum of digits is 12 and which are smaller than the number whose digits are reversed are 39.73 -:.75. and 75157. 7. we have 4( IOu + t) = 7( IOt + u).28 -:.8 = 4) Therefore. An algebraic solution to problems involving the digits of a number are best solved by using two variables. one for the tens digit and one for the units digit.( 87 . I.u) + 7 u 40u + 48 . from the proportion. Therefore. 8. we will need 62. A simple way to solve the problem is to try all possibilities. The ratio plays only a small part in this problem. Using (b) and cross multiplying.00. the actual side of the triangle will be 8 x 1.5 = 12 ft. From (0) we see that t = 12 .70u + 7u 36u + 48 = 840 .4u = 840 . Since each bag covers 8 ft2. the original number is 48 and the larger number is 84. Let u be the units digit and t be the tens digit of the original number. The problem gives us two pieces of information to use: (0) t + u = 12 and (b) (IOu + t)/ (lOt + u) = 7/4.u) = 70( 12 . which means that we have to buy 8 full bags.48.25/bag x 8 bags = $122.J3/4 x 122 or approximately 1. Therefore. 84/48.8 = 7. Therefore..75.28 ft2. A scale of 1:8 simply means that the actual patio will be 8 times larger than the model. or 84/48 = 1. We find that 42/30 does reduce to 715. 7x be the number of jobs completed by Carl's crew. The quotient ratio 12/39 is reducible to 4113.6) = 4(5x .( 10.6).. we have the equation 7x = 5x + 12 and. we have 26x . Six years ago. the mother is 5 x 9 = 45 years old.9. Simplifying and solving.6 4 ---= 5x . Let x be the mUltiple for the current ages.6 13 Cross mUltiplying gives us the equation 13 (2x .78 = 20x .6 and the mother's age was 5x . respectively. With these. The total number of jobs was 72.6. The second sentence gives us the simple model Number of jobs completed by Carl's crew = number of jobs completed by Ben's crew + 12 Inserting the variables. Check: We need to ensure that 42:30 is equivalent to 7:5. we have the proportion 2x . the girl and her mother were 12 and 39 years old. Check: Six years ago. the girl is 2 x 9 = 18 years old. The first sentence of the problem gives us the ratio Jobs completed by Carl's crew 7 = Jobs completed by Ben's crew 5 Let x be the multiple. the girl's age was 2x .24 6x = 54 x=9 Currently. x = 6. and 5x be the number of jobs completed by Ben's crew during the week. Therefore. 5x be the mother's age. 2x be the girl's age.. clearly. and the grandmother is 8 x 9 = 72 years old. and 8x be the grandmother's age. Carl's crew completed 7 x 6 = 42 jobs and Ben's crew completed 5 x 6 = 30 jobs. An algebraic solution is the fastest way to the answer.( 88 . Cross-multiplying gives us the equation 100(IS + x) = 40(50 + x).000. we have to solve this problem in several steps. 89 . letting g be the number of grams of sugar in the Lite version. and the largest investment earned 10%. and so on..II.OOO and x = 2000. the middle amount earned 12%. and $14. Check: 240/64 = 3.75 and ISO/4S = 3.520 and g = 11.--=-x13 13 50 + 3 / 53 / 3 3 3 160 64 = = 0. 3x + 4x + 7x = 14x = 2S. The three amounts invested are 3x. that is. (0) Find the amount of each investment. it is helpful to realize that the original solution is 36 percent acid. 4x. This enables us to create the proportion (IS + x) / (50 + x) = 40/ 100. Cross mUltiplying gives us 240g = ISO x 64 = 11.75j 13. To better understand the problem. it means that the smallest investment earned S%.520 -. Let x be the multiple. This is usually meant as a direction to make a correspondence between the first items in each list. Therefore.4 = 40%1' 160 12. we obtain ISOO + 100x = 2000 + 40x .240 = 4S g of sugar. we have the proportion 240/64 = ISO/g. and 7x. Therefore. Therefore. Note the word respective. the second items in each list. The three amounts are actually $6000. we let x be the amount of acid added. Amount of acid = = 36 = 36% Amount of acid + water 50 100 Since we are adding more acid.!! 60x = 200 Check: / 211/3 64 160 64 3 IS + 3 13 ---=--=--. Simplifying and solving for x. In this problem. $SOOO. The phrase "varies directly" indicates that the ratios of calories to amount of sugar are proportional. Each percentage return must be calculated for the money in that investment. Simplifying gives us 2xt = 5xt . For the $8000 investment. Check: We need to check the ratios of the speeds and the time taken by each vehicle. that is. which is 6 hours less.. 12 = $960.08 = $480.28. Checking for a ratio involving three or more terms can be done pairwise: $8000/$6000 = 4/3.14%.000/$8000 = 7/4.30x or 30x = 3xt.( 90 . $14. Let t be the time of the train and. we have the equation 2xt = 5x(t .(b) Find the new amounts on the basis of the percentage returns.000 x.. Let x be the multiple for the speeds. For the $14.100 mph = 10 hours. For the $6000 investment.rate.000 investment. (c) Find the overall percentage using 2840/28. Therefore. and the time taken by the plane is 1000 mi -.6). Since the plane is obviously the faster vehicle.( Using Time = distance -. We can use two variables to set up this problem quickly. therefore. x = 2840 x 100 -:.000 = x/I 00. Using Distance = rate x time. t . and realizing that the distances traveled by each are equal. 5x is the speed of the plane and 2x is the speed of the train.. Using xt = 500 from above.. the train travels at 2 x 50 = 100 mph and the plane travels at 5 x 50 = 250 mph. The rest of the check would be to recheck the arithmetic in finding the earnings and overall percentage.6 is the time of the plane. she earned 6000 x .1 0 = $1400. she earned 14.000 = 10. We can use the formula A = p+ prt. where t = I.250 mph = 4 hours. the time taken by the train is 1000 mi -. The total return was $480 + $960 + $1400 = $2840. we have 30x = 1500 and x = 50. she earned 8000 x . Check: The most important aspect of the problem to check is the amounts of each investment represented in the ratio of 3:4:7.( 14.. The quotient ratio 250/ I00 can be reduced to 5/2. Both are true. we have 2xt = 1000 or xt = 500.. is a major component of mathematics as it enables us to understand the physical world around us. and you should refer to Chap.Chapter 4 Word Problems Involving Geometry and Trigonometry the study of the relationships among shapes. therefore. This requires careful reading and following the description. Geometry. Word problems that involve geometry and trigonometry mostly require an understanding of a few basic relationships between lines. Find the measure of the largest angle. the shapes are simple to draw. others are mentioned in the solutions to the problems as they arise. Diagrams are essential in problems involving shapes. The key step is to identify the relationship that applies. the two areas are closely related. Problems Involving Angles Example I The ratios of the measures of the angles of a triangle are 2:7:9. At other times the problems describe the geometric figure in detail in order to test your ability to visualize the situation. 91 . Many of the basic relationships are listed at the end of the chapter in the Table of Fundamental Geometric and Trigonometric Relationships. 2 and to the Appendix for review and practice if you have difficulty following the steps in their solutions. Trigonometry is the study of the relationships within triangles and. In most cases. and polygons. Algebraic equations will arise. angles. and 90° solve the problem. we make use of the method seen in Chap. 7x. therefore. If the cables form a 40° angle with each other. what are the measures of the angles formed by the cables and the rooftop? Solution :1 A diagram is necessary to visualize the situation and identify the shape involved. Each is attached to the top of the antenna and to a spot on the rooftop. A diagram is also helpful in understanding the problem. The model gives us the equation 2x + 7x + 9x = 180 18x = 180 x = 10 The angles are indeed 2 x 10 = 20°. It is nearly obvious that the angle measures 20°. and 9 x 10 = 90°. and 9x. Example 2 Two cables of equal length are supporting a vertical antenna on a roof. 7 x 10 = 70°. 92 . To be sure. 2x Let x be the multiple. 3 to solve ratio problems. 70°. 2x.Solution I The model for the problem comes from the fact that the sum of the measures of the angles of a triangle is 180°. The angles are. Let x be the measure of each of these congruent angles. the triangle has equal sides and is.Since the cables are of equal length. The relationships that arise with parallel lines concern the angles formed when a third line is introduced that intersects both parallel lines. In an isosceles triangle the angles opposite the congruent sides are congruent. isosceles. Using this along with the fact that the sum of the measures of the angles of a triangle is 180 we can derive an equation to solve. The next example demonstrates such a situation. 0 Solution 3 All the facts about the angles that are formed when a transversal crosses two parallel lines indicate that only two different angle measures appear. This line is called a transversal. 0 . Find the measures of these angles. x+x +40 = 180 2x + 40 = 180 2x = 140 and Many word problems involve a pair of parallel lines. Example 3 The largest of the angles formed by a transversal that crosses two parallel lines is 4S less than twice the smallest of the angles. therefore. 93 . Check: The angles must be supplementary: 75° + 105° = 180° . The model uses this fact and the following relationship in the problem: Measure of larger angle + measure of smaller angle = 180° Measure of larger angle = 2 x measure of smaller angle = 45° Since the problem relates the larger angle to the smaller angle.45 = 105°.Therefore. These are clearly adjacent angles that lie on the same straight line and are. supplementary.45 = larger angle x + 2x . the "largest" angle is the larger of these two measures and the" smallest" is the smaller of these measures. Let x = smaller angle 2x .. quadrilaterals. we will represent the smaller angle by the variable. and other polygons depending on the 94 . therefore.45 = 180 3x = 225 x = 75 The smaller angle is 75 and the larger angle is 2 x 75 .( Problems Involving Line Segments of Polygons Many relationships exist among the line segments found in triangles. angle bisectors. To the nearest foot. diagonals. Example 4 At some time in the afternoon a 6-ft person casts a shadow that is 10 ft long while a nearby flagpole casts a shadow that is 22 ft long. and others. how tall is the flagpole? Solution 4 In order to identify the shapes we are working with. 95 . The table at the end of the chapter lists many of these relationships. and we have 6/10 = h122. The most common word problems involve similar figures that are found in the diagram that models the problem. altitudes. Some basic assumptions must be made: (1) the person and the flagpole are standing perpendicular to level ground and (2) the sun's rays form the same angles with the head of the person and the top of the flagpole. Person Flagpole h 6ft 10 ft J Person's shadow . Cross-multiplying gives us 10h = 132 and h = 13. we have Height of person Length of person's shadow height of flagpole length of flagpole's shadow Let h be the height of the flagpole. J Flagpole's shadow 22 ft Note that two similar right triangles are formed on the basis of the assumptions. These assumptions are standard for this kind of problem and lead to the following diagram. we need to draw a diagram. medians. These line segments could be sides of the figure.type of figure in question. The height of the flagpole to the nearest foot is 13 ft. Since corresponding sides of similar triangles are in proportion.2. . This segment cuts the side it is drawn through into two segments of 15 and 24 units. The problem tells us that one side is 24 + 15 = 39 units.. which cuts side DC at F. This requires careful reading and following the description. The vertices of the starting rectangle ought to be labeled.--_ _ _ _ _---. and we draw segment AE.The next example demonstrates the need to carefully read the description of the shape involved in the problem. The directions given in the problem must be followed with care in order to arrive at the correct diagram. The problem tells us that CE = 10 units.8 DL----~------iC 24 10 E Suppose that we extend side BC through C to a new endpoint E. The vertex opposite C is A. We need to study a diagram of the problem in order to find the means to find the width. A . The problem tells us that DF is 24 and FC is 15. What is the area of the rectangle? Solution 5 Finding the area of the rectangle requires finding the length and width. Part of the goal of such a problem is to test your ability to visualize the situation. The diagram makes it clear that there are two 96 . and a line segment is drawn from its new endpoint through the rectangle to the vertex opposite the vertex extended through. and as the directions are followed new labels should mark pOints of intersection. Example 5 One of the shorter sides of a rectangle is extended 10 units beyond a vertex. 97 . you will have to compute square roots. The area of the rectangle is 39 x 16 = 624 square units. we have AD/ EC = DF /CF or AD/I0 = 24/15. It relates the two legs and the hypotenuse of a right triangle in the following way: The square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. Since corresponding sides of similar triangles are in a proportion. a b Remember that the hypotenuse is always the side opposite the right angle and is the longest side of the right triangle. Using the Pythagorean Theorem The Pythagorean Theorem is used when we know two sides of the right triangle and need to find the third. Problems Involving Right Triangles The most frequent real-world applications of geometry include situations that involve right triangles. This is due to finding right triangles in abundance in the physical structures of the world around us and the importance of finding perpendicular distances. Thus AD = 240 -:.ECF. we have the familiar statement a2+~=c2.ADF and t'}. The two most important tools in right-triangle problems are the Pythagorean Theorem and the trigonometric ratios: sine. Therefore. and tangent. t'}. be sure to give the answers to the degree of accuracy asked for in the problem.15 = 16 units.similar right triangles. Using variables a and b to represent the lengths of the legs and c to represent the length of the hypotenuse. In solving problems with the Pythagorean Theorem. cosine. These are seldom going to be whole numbers. Building h 10 More specifically. h2 + 100 = 900 = 800 h = )800 = 28. are useful to remember. how tall is the house to the nearest foot? Solution 15 A simple diagram of the situation clearly reveals that we are working with a right triangle in which we know two sides and have to determine the third.5 5. Several sets of numbers.Example 6 A 30-ft ladder leans against the side of a building and reaches the flat roof. we know one leg and the hypotenuse.28 The height of the building to the nearest foot is 28 ft.24. The four most common triples are 3. Each triple is a set of three whole numbers that satisfy the Pythagorean Theorem.12.17 7. referred to as Pythagorean Triples.25 32 + 4 2 = 52 + 122 = 8 2 + 15 2 = 72 + 242 = 9 + 16 = 25 = 52 25 + 144 = 169 = l3 2 64 + 225 = 289 = l7 2 49 + 576 = 625 = 25 2 98 .4. we have the equation h 2 + 102 = 302 . l3 8.15. If the ladder stands at a point 10 ft from the house. If we let h be the height of the building. There is no other possibility for the leg but 12. Be sure that the smaller numbers represent the legs and the largest one represents the hypotenuse! Actually.If you notice two of the three numbers in a problem. Example 7 The sides of a triangle are IS. The Pythagorean Theorem shows up in other figures where right triangles are formed. Therefore. IS. We can also solve the problem algebraically to be convinced of this. you can save computation time. . Then h = J144 = 12. it bisects the side. and 18. What is the length of the altitude to the noncongruent side? Solution 7 The altitude creates two congruent right triangles and. For example 6. Let h be the length of the altitude. A multiple of the 3:4:5 triple is 9:12:15. hence. h 2 +81 = 225andh2 = 144. Look for this when "dropping" altitudes or working with perpendicular lines. The following diagram indicates that we are looking for a leg of a right triangle whose other leg is 9 and whose hypotenuse is 15.8.10 62 + 8 2 = 36 + 64 = 100 = 102 Look for these scaled triplets as well. any triplet that is in the same ratio as any of the above will also satisfy the Pythagorean Theorem. h2 + 9 2 = 15 2 . in which one leg is 5 units and the other leg is the altitude of the trapezoid. Therefore. the longer base is divid~d into segments of 5. the diagonals create four congruent right triangles in the interior. Each nonparallel side of the trapezoid is a hypotenuse of the right triangle. The manufacturer makes these tiles according to the lengths of the diagonals. how long are the nonparallel sides? Solution 8 The answer is found by dropping altitudes from the shorter of the parallel sides to the longer. and 5. How long is each side of the tile. Since a rhombus is a parallelogram. The hypotenuse must be 13 from the familiar 5. h = 156/13 = 12.13 Pythagorean Triple. 8 111\ 5 8 5 We can calculate the length of the altitude from the formula for the area of a trapezoid. The shape formed in the center is a rectangle. We now have a right triangle whose legs are 5 and 12. 100 . Example 9 A floor pattern requires tiles that are in the shape of a rhombus. and the right triangles formed on its sides are congruent.12. Area = liz x (base 1 + base 2) x h We have 156 = liz x 26 x h = 13h. 8. Therefore. if the diagonals are 10 and 16 in? Solution 9 The solution to this problem lies in the fact that the diagonals of a rhombus are perpendicular. If the area of the trapezoid is 156 square units. Therefore.Example 8 The parallel sides of an isosceles trapezoid are 8 and 18 units. the diagonals bisect each other. The other side is.43. The square root .J89. This is reasonable since it is larger than either of the other two sides of the right triangle. The problem doesn't ask for a specific degree of accuracy. of course. to leave the answer in radical form.where each side of the rhombus is a hypotenuse of the right triangle..-----'----length of hypotenuse (cosine is usually abbreviated as cos) 7'. the hypotenuse. Letting s represent the side of the rhombus.J89 is approximately 9.----.. The ratios are referred to with respect to the angle in question and are called the sine of the angle. It is best.::. The legs of the right triangle are identified as being either the side opposite the angle or the side adjacent to the angle. and the tangent of the angle...---::------:-=-=--=----length of hypotenuse (sine is usually abbreviated as sin) · length of adjacent side Cosme 0 f an angIe = ---"------. f I length of opposite side langent 0 an ang e = . the cosine of the angle. Using the Trigonometric Ratios Trigonometric ratios are ratios of two sides of the right triangle and are used when we have problems that involve two sides and an acute angle of a right triangle.. the Pythagorean Theorem gives us the computation S2 = S2 + 82 = 2S + 64 = 89 and s = . it is a good idea to get its value to see if the answer is reasonable.. therefore. However. The ratios are specifically · length of opposite side Sme 0 f an angIe = --. length of adJacent sIde (tangent is usually abbreviated as tan) 101 . Finding Sides of a Right Triangle Example 10 The warning on the side of a 2S-ft ladder says that the angle it makes with the ground should be no smaller than 40° before it becomes unsafe. and it also can tell you the angle if you know the ratio. 102 .Letting the legs be represented by a and b and the hypotenuse by c as in the diagram below. what is the height along a wall that this ladder makes at such an angle? Solution 10 The diagram for the problem presents us with a right triangle. or tangent of any acute angle. b c a cosB =- c tanB = - a Your calculator or a table of trigonometric values can tell you the approximate value of the sine. To the nearest foot. B c a A b C we have . a sm A =- c cosA =- b c tanA= b b a while smB =- . cosine. We should label the vertices of the triangle for reference. A tram from that point carries people to the top.The side of the building is the side opposite the angle we are using.6428 (four decimal places is standard).". we often find the phrases "angle of elevation" and "angle of depression.::-::-length of ladder or using h for the height..6428 = 16. The angle of depression is the measure of the downward tilt off the horizontal to view an object below. Therefore." In Example 10. and how long is the tram's cable? 103 . the angle of elevation to the top of the hill is 28°.07. To the nearest foot. It is important to note that the angle of depression lies outside the triangle and is the complement of its adjacent angle... the angle of elevation is the measure of the upward tilt off the horizontal required to view an overhead object. Therefore. we have h = 25 x 0. we need to use the sine ratio.::.. sin 40° = h12S. Object Angle of elevation Horizontal Angle of depression -------~Object Horizontal In both cases a right triangle is formed when drawing the height of the viewer or the object. how high is the hill. Example II From a point 150 ft from the center of the base of a hill. In word problems involving trigonometry. the 40° angle that the ladder makes with the ground would be the angle of elevation of the ladder to the point where it reaches on the wall. and the ladder is the hypotenuse. In other words. it is also equal to the other acute angle of the right triangle. Therefore.. The verbal model is height along wall · Slne 0 f 400 = ---=---::------:-. The calculator tells us that sin 40° = 0. and the trigonometric ratios can be used. our ladder will reach a height of 16 ft. To the nearest foot. .8829 = 227 feet. The height of the hill is measured as the perpendicular axis from its top to the center of its base. you don't have to be an artist.C 200 We know the length of the side adjacent to the angle AC = 200. Letting h be the height. and the length of the cable is the hypotenuse AB.. Letting c be the length of the cable. In order to find the height. since we were given AC and we found Be.---~. we have c x cos 28° = 200._ _ _. 106.34 = 106 ft. one using trigonometry and the other using the Pythagorean Theorem . we easily determine that the appropriate trigonometric ratio is tangent since it is the one of the three that involves the opposite and adjacent sides. we should use the value of BC to more decimal places. The following diagram models the problem. • We can use the Pythagorean Theorem to find the hypotenuse of the triangle.. We can find the length of the cable in two different ways.' _ L -_ _ --'~----' ~. and work with several decimal places until we round off at the end of 104 . to the nearest foot.. To be accurate. (Remember. and we label the vertices for reference. The height is the side opposite the angle BC.34. we have tan 28. we have cos 28° = AB/ AC = 200/c. h = 200 x tan 28° = 200 x 0. 0 • We can use the cosine ratio since it is the one that involves the adjacent side and the hypotenuse. Note that we are using a value that we had to determine and we had rounded that answer. When we cross multiply here. to the nearest foot. = BC / AC = h/200.5317 = 106. Therefore.Solution II The angle of elevation is the angle that the tram cable makes with the ground. use simple lines to make the right triangle easy to identify!) B Cable A L. Therefore c = 2007 cos 28° = 200 7 0. (If we would have used 106 ft for Be. we have c2 = 200 2 + 106.20 = 51.20 = 226.7 = 101 ft. It also indicates that the height includes the cliff and the tower. Letting d be the distance. The height of the cliff is the adjacent side of the angle. to the nearest foot. Boats must be warned if they come too close to the cliff. we should use the tangent ratio. but not correct. c = ". Therefore.308. Letting c be the cable length. 80 + 40 = 120 ft.) Example 12 An observer from the top of a lighthouse that is 40 ft tall and stands on a cliff 80 ft above sea level notices a boat off shore.308. our answer would have been 226 ft. and the distance between the boat and the cliff is its opposite side.8391 = 100. how far from the cliff is the boat? Solution 11 The diagram clearly shows the right triangle that must be used.342 = 40. which is close. 40 80 Boat- d The diagram also indicates that the angle within the triangle we need to use measures 40°.20 Therefore. lOS .51.the calculations.000 + 11. because of the rocks below the waterline. that is. If the observer notes that the angle of depression is 50'.308.51 ft = 227 ft. to the nearest foot. we have tan 40° = d/120 and d = 120 x tan 40 = 120 x 0. We can safely say that is 5 ft. Since we are concerned with the angle of elevation of her head.C.5 = 550 ft. which is 555 feet tall. 550 ft Ground The monument is the side opposite the angle. They are equally as useful when trying to find the acute angles of a right triangle. She is standing 50 yd away. is awed by the size of the Washington Monument. you have to know the lengths of two sides. Example 13 A young tourist visiting Washington. we must consider the horizontal axis to begin at the height of her head. At what angle. The tourist. The second aspect is more subtle. is she elevating her head when she looks at the top of the monument? Solution 11 Two aspects of this problem must be accounted for before we can accurately identify the right triangle. Therefore. the vertical height used is 555 . which is equal to 150 ft. is standing 50 yards (yd) from the monument. D. and the distance is the side adjacent to the angle. The important step is to identify the position of the given sides relative to the angle you seek to find. to the nearest degree. who is 5 ft tall. We can make use of the tangent ratio to solve the problem. To accomplish this. One is the obvious change of units in the distance that the tourist is from the monument. 106 .Finding Angles of a Right Triangle The previous examples demonstrated how the trigonometric ratios are used to find the lengths of the sides of a right triangle. Let x be the measure of the angle. We have tan Xo = 550/150 = 3.6667. Your calculator should have a key press that will find the inverse tangent (sometimes written as tan- 1 or arc tan). This will tell you what angle has the calculated number as its tangent. We could write x = the inverse tangent of 3.6667 or x = tan- 1 3.6667. Enter 3.6667, and press the key to get x = 74.7° or, to the nearest degree, 75°. Special Right Triangles Just as there are some easy triplets of numbers to help with using the Pythagorean Theorem, certain special right triangles are well known and used frequently in problems involving trigonometry. The most easily remembered trigonometric ratio is sin 30° = 0.5 or l/Z. This means that in a right triangle that has a 30° angle, the hypotenuse is twice the size of the side opposite the 30° angle. If these sides of the triangle were 1 and 2, the Pythagorean Theorem would be used to find that the other leg is v'3. 60" 2 We refer to this as the special 30-60-90 right triangle. From the diagram, we have six trigonometric ratios that are easy to remember and have exact radical values. sm30° = sin 60° = . 2 1 cos 30° = 2 v'3 1 tan 30° = tan 60° = v'3 v'3 1 ~ cos 60° = 2 and we have the following ratio: leg opposite 30° : leg adjacent 30° : hypotenuse = 1 : v'3 : 2. This makes problems like the following easy to solve. 107 Example 14 The diagonal of a rectangle is B in long and makes a 30° with the longer side. What is the perimeter of the rectangle? Solution 14 The diagram with the vertices labeled shows that we have a right triangle, tiBAD, with hypotenuse BD. Since the angle is 30°, it is a 30-60-90 right triangle and we know that AD:AB:BD= 1: J3: 2. 8 o c Therefore, since BD = B in we have AD = 4 in and AB = 4J3 in. The perimeter of the rectangle is 4 + 4 + 4J3 + 4J3 = B + BJ3. This is an exact answer and does not require a decimal approximation unless one is asked for. (Note that J3 is approximately 1.73 if it is needed.) The other special right triangle is the isosceles right triangle. Since the acute angles must be equal, they are both 45°. Since the legs are equal, their ratio is 1. Finally, since the legs are the sides opposite and adjacent to a 45° angle, we have that tan 45° = I! The Pythagorean Theorem gives us the hypotenuse exactly as ../2. The diagram below depicts the special 45-45-90 right triangle. We have the three trigonometric ratios sin 45° = 1/../2, cos 45° = 1/../2, and tan 45° = I, and the sides are in the ratio leg:leg:hypotenuse = 1 : 1 : ../2. 108 Example IS What is the exact area of a square whose diagonal is 8 in? Solution IS x The diagonal creates two isosceles right triangles within the square. Using the ratio leg:hypotenuse = l/vIz, if we let x be the side of the square, we have the proportion x/8 = l/vIz. Crossmultiplying gives us xvlz = 8 and x = 8/vIz inches as the side of the square. The area of the square is, therefore, (8/vIz)2 = 64/2 = 32 in 2 . Summary Make sure to draw simple but accurate diagrams. Look for right triangles. isosceles triangles. parallel lines. and other shapes that have known relationships. 3. Assign a variable to the unknown length or angle. and create a model equation that uses the identified relationship. 4. Solve the equation for the variable. and be sure to use it to determine all the lengths or angles asked for in the problem. S. Check your arithmetic and algebra. and make sure that the answer is reasonable for the problem. I. 2. Additional Problems In a triangle. two of the angles are in a ratio of 5:3. The third angle is 40 less than the smaller of the other two. Find the measure of the largest angle. 2. The acute angles of a right triangle are such that the square of the smaller is 25 less than 10 times the larger. Find both angles. 3. Two angles formed on the same leg of a trapezoid are such that one is 30 less than 5 times the other. Find the measures of the angles. 0 0 0 I. 109 to the nearest square centimeter. assuming that the length across the patio extends 12 ft from the house and that it is 3. 1 I.4. 8. This player takes the shot from that point. a ventilating shaft is constructed vertically upward and breaks through at a point 80 m along the hillside. If the smaller of these angles is 20° less than the base angle of the isosceles triangle. At a point 45 m along the entrance. When pointed at the top of the antenna on the building. In an isosceles triangle. on different sides of the pole. to the nearest meter? A basketball player is standing directly in front of the basket at the foul line. what is the distance between the points at which the cables are anchored to the ground? What is the angle. 5. How tall is the ventilating shaft.7 in higher at the side adjoining the house than on the other side? A 25-ft vertical pole is to be supported by two cable wires. which is 15 ft from the basket. of a regular pentagon whose perimeter is 100 cm. a segment drawn from the vertex angle to the base cuts the vertex angle into two angles whose ratio is 5:3. to the nearest foot. 12. If the angle of depression from that row to the stage is 28° and the vertical height above the stage is approximately 50 ft. to the nearest degree. each 30 ft long. The player passes the ball to another player 8 ft directly to the right. will this shot count for three points if it is made? A kite flying on a 150-ft string is anchored to the ground. how far. To the nearest foot. 1O. At one point the string is making a 70° angle with the ground. the angle of elevation given by the transit is approximately 80° from a point 260 ft away from the building. 9. what is the angle at which the patio inclines. does the sound have to travel to reach someone sitting there? An outdoor patio along the back of a house has to be constructed on an incline to allow rainwater to drain away from the house. 110 . 6. what are the measures of the angles formed by the line segment? An entrance to a mine is dug horizontally into a hill. To the nearest foot. 7. how tall is the Empire State Building including the antenna? In testing the acoustics for a theater. If the three-point line is 23 ft 9 inches from the basket. To the nearest tenth of a degree. How high above the ground is the kite? A surveyor uses a transit to determine the height of the Empire State BUilding in New York City. the distance from the stage to the last row in the balcony has to be determined. that these wires make with the ground? Find the area. That is./ The key point is that the acute angles of a right triangle are complementary. how many seconds. On a baseball diamond. and 100°: 60° is indeed 5:3. the distances between the bases is exactly 90 ft. + 20° = 180° . are 5x. to the nearest tenth. Let x be the multiple to use with the ratio. If the catcher throws the ball at 80 mph. The fact that the sum of the measures of the angles of a triangle is 180° will give us the equation to solve. the angles.40° = 20° and 100° + 60° 2. Therefore. 5x + 3x + 3x - 40 = 180 Ilx = 220 x = 20 The largest angle is 5 x 20 = I00°. The catcher has to be able to make an accurate throw from homeplate to second base. 14. Solutions to Additional Problems I. will it take the ball to reach second base? Find the exact area of a parallelogram whose sides measure lOin and 20 in and the acute angle formed by the sides is 60°.13. and 3x . Check: The middle angle is 3 x 20 = 60°.. I I I . 3x.40. in descending size order. their sum is 90°. The remaining angle is 60° . and the larger measures 65°.x be the measure of the larger angle. A bit of trial and (x . and all terms must be brought to one side) x 2 = 900 . The leg of the trapezoid can be regarded as a transversal that. .The relationship given provides the model (Measure of smaller angle)2 = lOx measure of larger angle .25 Let x be the measure of the smaller angle and 90 . Using these in the model. The smaller angle measures 25°. 112 . hence.25 (which needs to be simplified. The model is.25)(x + 35) = 0 -35 x = 25 or We can reject -35 since angles must have a positive measure.x) . Check: 25 2 = 625 and lOx 65 .lOx .25 = 650 . Therefore + 10. if extended. therefore. we have the quadratic equation x 2 = 10(90 . I The two angles are.875 = 0 We seek factors of -875 whose difference is error will yield -25 and +35.25 x 2 + lOx . and they are supplementary.25 = 625.1' 3. interior angles on the same side of the transversal. would cross the parallel sides. To avoid confusion. we need a second equation involving both of them. giving us 8x + 2(3x + 20) = 180. Let y be the measure of the base angle. Simplifying and 0 0 113 . Therefore. This comes from the other piece of information. We can use the sum of the angles of a triangle = 180 to model the problem: 8x + y + y = 180 or 8x + 2y = 180. To solve a problem with two variables. 3x is the smaller angle formed by the segment and 5x is the larger one.the sum of the two angles = 180 0 • Let x = one angle 5x .30 = 145 0 • Check: We need to make sure that the angles are supplementary: 35 0 4. since the altitude to the base would also cut the vertex angle exactly in half. + 145 0 = 1800 . namely. the base usually refers to the noncongruent side and the angles formed on this side are called the base angles. we will assign a variable for the multiple as follows.20 or y = 3x + 20. ( In an isosceles triangle. Therefore. we can represent the base angle by a different variable. Let x be the multiple. we have the equation 3x = y . the vertex angle is 8x.30 = the other angle x + 5x - 30 = 180 6x = 210 x = 35 The angles are 35 and 5 x 35 .30 0 = 175 . Therefore. 5y\\X ill Since we are dealing with ratios. The third angle is called the vertex angle of the isosceles triangle. the smaller angle formed by the segment is 20 less than the vertex angle. . We can substitute for y in the first equation. The diagram for this picture requires an isosceles triangle with a line segment that is clearly not an altitude. we have the following diagram. The phrase "vertically upward" indicates that the shaft is perpendicular to the horizontal (axis) and a right triangle is formed whose legs are the entrance and shaft and whose hypotenuse is the hillside. Thus x' = 4375 and x = J4375 = 66.75 ft (since 9 in is three-fourths of a foot). We have x'+ 45' = 80' or x' + 2025 = 6400. 50 0 • The sum ofthe angles ofthe isosceles triangle is 50° + 50 0 + 80 0 = 1800 . we obtain 8x + 6x + 40 = 180 14x +40 = = 180 10 0 14x = 140 x Therefore. the angles formed by the segment are 3 x 10 = 30 and 5 x 10 = SOc. the shaft is 66 m. ( 5.45 m----l 6. Let x be the length of the vertical shaft. To the nearest meter. hence. the distance to the basket has to be greater than or equal to 23 ft 9 in or 23.solving for x. that is. f--. In order for the shot to be worth three points. The use of the word directly enables us to assume that we are dealing with perpendicular distances and. ~~L L Entra(l~ ~ ~. Check: The base angles would each have to be 20° more than the smaller angle. . A f-r-_---'1"'S"'ft _ _-7 c Bft d B 114 . Therefore.14. we can apply the Pythagorean Theorem. (The actual height is 1472 feet. Let x be the height of the building.9397.1 lIS 260 It . The diagram shows that we have a right triangle where the height of the kite is the side opposite the 70° angle and the string is the hypotenuse.5 ft. To the nearest foot. 17 Pythagorean Triple. Let d be the distance. The first player is at A. and the hypotenuse is 17.) 11 i 80° . h = 150 x 0.96. The difference is due to rounding the angle given by the transit. From a calculator or table. Be is the distance we are trying to determine. we make use of the sine ratio. The diagram indicates that we have a right triangle.6713 = 1474. we need to use the tangent ratio. We recognize these to be the legs of the 8. I II ! . and we have d 2 = 82 + 15 2 = 64+ 225 = 289 and d = J289 = 17 ft. which is too short for a three-point shot. Therefore. the second player is at B. the height is 1475 ft. Then sin 70 c = h /150. i . Therefore.9397 = 140. IS. we find sin 70 = 0. We see that 260 ft is the length of the side adjacent to the given acute angle and the height of the Empire State Building is the side opposite this angle. you can use the Pythagorean Theorem. To the nearest foot.7. the kite is 141 ft above the ground. If you don't recognize the triple. and the basket is at C. Let h be the height of the kite. and we have tan 80° = x /260 or x = 260 x tan 80° = 260 x 5. 0 h 150 It 8. and it is the hypotenuse of a right triangle whose legs are AC and AB. Therefore. 0.. (We must work in the same units when taking ratios of sides. to the nearest tenth of a degree. 1..l ~_(_ _ 12tt=144.sin 28" = 50. We can make use of the sine ratio. 12 ft or 144 in away.. The diagram indicates that we have the sides opposite and adjacent to the angle of incline. we will have to find a right triangle in which we know two of the sides.0257.7/144 = 0. since this is measured perpendicularly to the ground. Therefore. The altitude of a right triangle forms two congruent right triangles. Let x be the measure of the angle of incline.4695 = 106. we can use the Pythagorean 116 . we have a right triangle that extends to the other side of the patio. and.9.7 in. we have a right triangle where the hypotenuse is the distance we wish to find and the vertical height is the side opposite the known acute angle.5". Let d be the distance.. and.47° or. Stage 10. since we know two of its sides. The sound will have to travel approximately 107 ft. The model for this situation is easily seen to be an isosceles triangle where the cables are its congruent sides and the pole is its altitude. which indicates that we need to use the tangent ratio. we have x = I . and we have tan XO = 3. To find the angle of incline. At the house side of the patio we have a vertical height of 3.) ':. and we have sin 28° = SOld or d = 50.49. The angle of depression from the last row in the balcony to the stage is the same as the angle of elevation from the stage to that row. Using the inverse tangent. The problem asks us to find two pieces of information.n I I. 117 . The inverse sine gives us y = 56. Let y be the angle.58 The distance between the cables is. 33. 33 ft. Therefore. in case we computed incorrectly. 12. to the nearest foot.4° or. therefore. we should use the given information rather than the length we computed. 56°. The Pythagorean Theorem gives us x 2 + 25 2 = 302 x 2 + 625 = 900 x 2 = 275 x = J275 = 16. to the nearest degree. The diagram shows us that the interior of the pentagon can be divided into 10 congruent right triangles by drawing from the center all the radii and perpendicular segments to each side.8333. However. we could create a ratio with any two of the sides. Such a perpendicular segment is called an apothem.Theorem to find the third.16 ft or. This side of the right triangle is one-half the distance between the cables on the ground. we should use the sine ratio. The area of the pentagon will be equal to lOx the area of each right triangle. and we have sin yO = 25/30 = 0. The given sides represent the side opposite the angle and the hypotenuse. To find the angle. Cable Let x reprelient the third side of the right triangle. The radius bisects this angle and. We can use trigonometry. Second base 90 ft Third First base base Home plate 118 . to the nearest square centimeter. the acute angle in the right triangle is 54°.5 = 20 cm.Since all sides are congruent. the apothem bisects each side..764. The area of the pentagon is therefore. A basic fact about a regular polygon with n sides is that an exterior angle formed by extending a side is equal to 360/n°.. The problem will therefore require several steps. Since the radii form isosceles triangles. The apothem is the side opposite this angle and the "half side" of the pentagon is the side adjacent to this angle. The speed of the throw will come from the distance formula: Distance = rate x time.. specifically..rate. and we have tan 54° = a/IO or a = 10 x tan 54° = 10 x 1..82 = 688.764 = 68. The area of the right triangle is 'h x lOx 13. we can use the tangent ratio. Therefore. this is 360. and we will have to convert units. For a pentagon. we know that each leg of the right triangle is 10 cm. The rate is 80 mph.. Time = distance .2 cm 2 or. lOx 68. 180° . therefore. 13. 688 cm 2 . each side has a measure of 100 cm .72° = 108°. We need to find the length of the apothem in order to find the area of the right triangle.... Therefore. if we can find the acute angle formed by a radius and a side. Let a be the length of the apothem.82 cm 2 . The interior angle is its supplement.5 = 72°..3764 = 13.. In either case.. . A ~t1 E 0 /0 c The diagram shows that we have a 30-60-90 right triangle and we know that the sides are in the ratio leg opposite the 30° angle: leg opposite the 60 angle: hypotenuse = I :. (b) Convert the rate to feet per second. therefore.. 14.J3 in = 100.4142. we should be looking for a special right triangle. the area of the parallelogram is 20 in x 5 .J3 in...J3 in 2• 0 0 • 119 . Therefore. 1. first base. and our distance is approximately 127. Since this is a special 45-45-90 right triangle. I) 85 mi 5280 ft I hour 1 minute ---x x x---1 hour 1 mi 60 minutes 60 seconds feet = 124. Home plate.(a) Find the distance from home plate to second base.. is 5.fi x 90.0 second. Since we will be making conversions later.278 -.02 seconds or.667-second (c) Calculate the time.J3 : 2.fi is approximately 1.rate = 127. and the side opposite the 60 our altitude. The area of a parallelogram is the product of one side (the base) and the altitude drawn to that side (the height).667 = 1. Time = distance -. we should keep several decimal places along the way.124..278 ft. The word exact is an indicator that we will be able to express our answer either as a number that we did not have to round off or as a number involving radicals. we know that the hypotenuse is . 5 in. to the nearest tenth. For this we can use the dimension scheme (see Chap. and second base form an isosceles right triangle with the distance we seek as the hypotenuse. We know that the hypotenuse AB is lOin. The side opposite the 30 angle is. Relationships between Lines and Angles 1. Two angles whose sum is 90°. A line segment from a vertex of a Altitude triangle perpendicular to the opposite side. therefore. Its length is the height of the triangle when the opposite side is considered to be its base. Vertical angles formed by intersecting lines are congruent. Continued 120 . Proportional in measure. Median A line segment from a vertex of a triangle drawn to the midpoint of the opposite side. A line or line segment that divides a Bisector side or angle into two congruent sides or angles. Adjacent angles forming a right angle are complementary. Parallel lines Two lines that are always the same distance apart and. Regular polygon A closed figure in which all sides and interior angles are congruent. Two nonadjacent angles formed by Vertical angles intersecting lines. Adjacent angles forming a straight line are supplementary. 2. Two angles whose sum is 180°. Two angles that share a common side and vertex. 3. therefore.Table of Fundamental Geometric and Trigonometric Relationships Congruent Similar Supplementary angles Complementary angles Adjacent angles Definitions Equal in measure and form. will never intersect. but identical in form. bisecting the opposite side. The ratios of corresponding sides of similar triangles are equal. all sides are congruent and each angle measures 60 0 • 0 • Relationships Involving Right Triangles 15. Congruent angles interior to both parallel lines and on alternate sides of the transversal c. 13.4. the altitude drawn to the noncongruent side is alsoan angle bisector and a median. length of adjacent side length of hypotenuse c. length of hypotenuse b Cosine of an angle = . s. two sides are congruent and their opposite angles are congruent. A median divides a triangle into two triangles of equal area. 17. Perpendicular lines form right angles. 11. 7. 10. The square of the length of an altitude drawn to the hypotenuse of a right triangle is equal to the product of the segments it creates on the hypotenuse. The trigonometic ratios are a Sine of an angle = length of opposite side . The sum of the angles of a triangle is 180 An exterior angle is equal to the sum of the two nonadjacent interior angles. Tangent of an angle = ::~~!~~~~~fa~~i~~:~~: Continued 121 . The Pythagorean Theorem states that the square of the hypotenuse is equal to the sum of the squares of the legs. Congruent angles in corresponding positions on each parallel line b. 12. In an equilateral triangle. 16. Supplementary angles interior to both parallel lines and on the same side of the transversal 6. 8. Two triangles are similar if they have the same three angles. In an isosceles triangle. A transversal that intersects a pair of parallel lines forms a. 14. 9. Relationships Involving Triangles The sum of any two sides of a triangle must be greater than the remaining side. In an isosceles triangle. A rectangle. A square is both a rectangle and a rhombus and has relationships 21 and 22. 0 • 28. A rectangle has four right angles. 31. and its diagonals are perpendicular. A rhombus has four congruent sides. 24. The segments are drawn from the center of an n-sided regular polygon to the vertices from n congruent isosceles triangles. The diagonals of a parallelogram bisect each other. Relationships Involving Circles 32. A trapezoid has only one pair of parallel sides.2). 30.Relationships Involving Quadrilaterals 18. The diameter is the longest line segment with endpOints on the circle. Consecutive angles of a parallelogram are supplementary. All radii of the same circle or congruent circles are congruent. Opposite sides of a parallelogram are parallel and congruent. 29. 22. 35. 27. a rhombus.n. The ratio of the areas of similar polygons is equal to the square of the ratio of corresponding sides or of the perimeters. 34. 21. and the interior angle is its supplement. The measure of each exterior angle of an n-sided regular polygon is 360 -7. Opposite angles of a quadrilateral inscribed in a circle are supplementary. 20. Opposite angles of a parallelogram are congruent. 23. 33. and its diagonals are congruent. Relationships Involving Regular Polygons The sum of the interior angles of an n-sided regular polygon is 180 x (n . The sum of the angles of any quadrilateral is 360 19. 122 . and a square are all parallelograms and have relationships 16 through 19. 26. 25. A radius meets a tangent to a circle at a right angle. you will see the most common ways of doing this and the variety of situations that appear most frequently when working with statistics. median.Chapter 5 Word Problems Involving Statistics. if you were in a car traveling at an average speed of SS mph. You have probably computed your test average sometime in your life and know that for a set of scores A sum of scores verage score = number of scores The average score is descriptive of the data in the following way: If all the data were the same. In the problems that follow. and probability. Median. Mean. counting. A necessary ingredient in all of these is the ability to count objects accurately. 123 . they would all be equal to the mean or average score. Counting. and Mode The words mean and average are synonyms. In reality. and mode and computing probabilities. These calculations include familiar statistical ideas such as mean. For example. it would be as if you traveled exactly SS mi every hour of your journey. you may have traveled different distances during each hour. and Probability Many word problems will ask you to analyze data and to make calculations that describe the data. However. The largest of the three numbers is 7 x 12 = 84. her new average will be 748 -7. and 7x. 3). We need to apply the algebra of ratios (see Chap. What is the largest of these numbers if their average is 52? Solution 1 The relationship tells us that the sum of the three scores is 52 x 3 = 156. what will her new average be if she earned a 92 on her next test? Solution I Using the relationship described above. To the nearest whole number. Example I Dana's average in math is 82 after taking eight tests. If the new score is a 92.3 = 156 -7. If the list has an even number of terms. When the data list is large. the sum will be 656 + 92 = 748. 2x + 4x + 7x = 13x = 156 and x = 12. 48.) Example 2 Three numbers are in the ratio of 2:4:7. 4x.3 = 52.9 = 83. The average of the three numbers is (24 + 48 + 84) -7. In the last example. your computation would be 55. 48 is the median of the three scores 24. Check: The other two numbers are 2 x 12 = 24 and 4 x 12 = 48. It is useful to remember that the sum of the scores = the average score x the number of scores. the sum of her scores is equal to 82 x 8 = 656.if you divided the total distance by the number of hours of your trip.1' The median of a set of data is the score that occupies the middle position when the scores are arranged from smallest to largest. Therefore. the median is the average of the two in the middle of the list.1 or 83. (The best way to check this answer is to calculate the average of eight scores of 82 and one score of 92. it may be presented in the form of a table which indicates the score and the number of 124 . and we are given the numbers represented by 2x. Let x be the multiple. and 84. Therefore. the list has 28 scores. -4. that is.5 The average of these two scores is the median.4.4.3. Find the mean. median. For these data. 3. -4. The sum of all the scores is the sum of the products of each score and its frequency. Example 3 The Fahrenheit temperatures at 8 A.3.3. -4. its frequency would be 8.3. Since this is an even number. -2.M. The mean can also be computed from the table. which include the two middle positions. The -4°s and -2°s occupy the first 10 positions. If no 125 . an easier way to find the median is to count through the frequencies in the table. The data are given in the table below.4.3.3. each score has a frequency of 1. The mode of the data would be the score with the highest frequency. -2. and mode temperatures.5. were recorded for each day during February in a Canadian city.5. -2.5. if all the test scores were 82. -4. Rather than listing all the scores. Temperature _4° _2° 3° 4° 10° Solution 3 Frequency 4 6 9 4 5 The mode is easily seen to be 3° since it has the highest frequency. By adding the frequencies.3.4. -2. we would average the scores in the 14th and 15th positions in the list. -2. the sum is -4 x 4 + -2 x 6 + 3 x 9 + 4 x 4 + 10 x 5 = -16 + -12 + 27 + 16 + 50 = 65. the score that appears in the list most often. In Example 1.3.5. In Example 2. The 3°s occupy positions 11 to 19. In this case. The median would require listing all the scores in ascending order. we know the total number of scores in the list.times it appears in the list. 3. -2. This number is called the frequency of the score. we saw in the previous examples that counting doesn't necessarily have to be done this way if we know how many times a specific object appears in the collection. the sum is equal to 76 x 7 = 532. we get x + 352 = 532 and. it is customary to compute the mean to one more decimal place than that of the original scores. to the nearest tenth..28 = 2. The average of the three lowest scores is 180 -.-1 Counting Objects Word problems that ask for a count of objects would be very tedious if we had to list all the possibilities to know how many we have. Check: We need to check that the sum of the seven scores we now know is 532: 3 x 60 + 82 + 3 x 90 = 180 + 82 + 270 = 532. In order to find the average of the three lowest scores. after subtracting 352 from both sides. Of James' seven tests this quarter in biology.3 = 60. what is the average of his lowest three scores? Solution 4 Since we know the mean. either of two corridors to the main lobby. If the average of his highest three tests is 90. Therefore. Example 5 To get from class to outside the main entrance of the school.specific directions are given. The following problems ask for a count of the possibilities of making selections from a set of objects. the median score was 82 and the mean score was 76. the mean is 65 -. How many different paths can you choose from? 126 . we can let x be their sum. we can compute the sum of all seven scores. 76. we see that x = 180. In fact. you can use any of three stairways to the first floor. Therefore. that is. x + 82 + 3 x 90 = 532. Example 4 This example demonstrates how the median and the mean can be very different for a set of data. and either of two doors. Multiplying and combining terms.3.. /' \ '" '" C2 '" C1. One way to create a visual for counting problems such as these is to think of filling slots./' . and one beverage? Solution «5 The Counting Principle leads us to believe that the number of different meals is equal to the product 8 x 4 x 5 = 160./' '" '" C2 '" At the local fast-food restaurant you can choose from eight different sandwiches. How many different meals can you have consisting of one sandwich. Note that the number of branches of the tree is based on the product 3 x 2 x 2 = 12. For the problem described above. Stairway 1 C1 Stairway 2 C1 Stairway 3 / \ ./' ./' ./' . There are clearly 12 possible ways from the classroom to the outside of the main entrance. The Counting Principle states that The total number of different selections consisting of picking one item from each of several different types is equal to the product of the possibilities for each item selected. D2 represent corridors 1 and 2 and doors 1 and 2. we would draw x x Sandwich side order beverage 127 . Example 6 '" C2 ./' . and five different beverages./' 02 D1 D2 01 D2 01 02 01 02 D1 D2 01 (where C1. respectively). It's that simple! Drawing a tree diagram would take much longer.Solution S This situation can be visualized by a tree diagram or flowchart that diagrams all the possibilities. which is a popular way to solve such problems. C2 and 01. four different side dishes. one side order. This is the basis of the Counting Principle. Example 8 How many ways can four out of seven students be seated in a row of four chairs? How many ways can all seven be seated in a row of seven chairs? Solution 8 Filling slots for the first situation.1 = 840.230. All the rest are filled with a number from a set of nine each since we can repeat digits. 8.x -9- Letter letter number number number number number number or 6 X 5 X 96 = 15.and fill the slots with the number of different choices for each item.x -9.x -9. namely. 128 . it is important that you are sure of the way you reasoned through the problem. 4. This will help with more complicated problems. such large numbers will turn up.x -9.6 . The remaining six characters are digits from 1 to 9 which can be repeated. However. while the second is filled with a letter from a set of five since we cannot repeat the letter we already used.x -9. The first two characters are different letters chosen from the letters A through F.x -5. How many different 10 numbers are possible to create? Solution 7 The situation is the same as filling eight slots. (A tree diagram would have been impossible to draw!) Sometimes answers such as the previous one will seem incorrect. when considering all possibilities. This product is symbolized by 7! and is called "7 factorial": 7! = 5040. and they cannot be checked by simple means. 7 x 6 x 5 x 4 x 3 x 2 x 1.943. For the second situation we continue the product until we reach I. The first slot is filled with a letter from a set of six. Example 7 Student ID numbers all have eight characters. The model is . and 5. we have Z X Q x ~ x . Therefore.x -9. Fredricks wants to have a mock election in his social studies class. Example 9 Mr. cab. Combinations The previous problems were solved by considering the order in which selections were made. Since it does not. we should try to reduce before multiplying and eliminate all the factors in the denominator. You would enter the number and press the key for the computation. nCr. the threeletter arrangements abc. bca. This makes sense since there have to be less to count if we don't care in what order they appear. which is a synonym for arrangement. but the same combination. we would be counting permutations. 129 . We use 9C4 = 9 P4 = 9 x 8 x 4! 7x 6 4x 3x 2x 1 Since our answer is going to be a whole number. For example. The formula is nCr =nPr/r!. For these problems. The fraction reduces nicely to 126. If nine students were nominated. The answer to the first question would be symbolized by 7 P4 and the second answer by 7P7. acb. Another symbol sometimes used is n Pr . Students have to elect four people to serve as class officers. bac. and cba are different permutations. our answer. we will find the number of combinations. where C stands for combinations. how many possible different groups of four could the class elect? Solution 9 If the problem asked for specific officer positions. where n is the number of objects in the set and r is the number of objects chosen to be arranged. Some situations involve counting groups and are not concerned with a specific order. we often make use of another symbol. which is 7!. n!.Your calculator may have a key for computing factorials. The P stands for the term permutation. BF. EH. you can label the vertices A through H and start listing the diagonals systematically: AC. CF. Several words are used 130 . BG. If you think of a diagonal as being a connection between two vertices. Therefore. eight of these pairs are not diagonals. You will often be able to find a way to restate the problem that makes it obvious. CE. DG. (CA is already in the list). That is sCz = 2! = 2 x 1 = 28 sPz 8x 7 However. CH. DH. the answer is 28 . but sides of the octagon. CG. AG. Example 10 How many different diagonals can be drawn in a regular octagon (an eight-sided figure with equal sides)? Solution 10 F E G o H A c B If you don't recognize the use of combinations. Probability Most questions about probability are given as word problems with descriptions of practical situations.Sometimes you won't be told explicitly that you are looking for combinations. BD. AD. the problem is really one of finding all the different pairs that can be formed from the eight vertices. DF.8 = 20. BH. AF. EG. (DA and DB are already in the list). We count 20 diagonals. FH (the rest with F are already in the list). BE. (the rest with E are already in the list). AE. The situation usually describes making selections and is referred to as the experiment. To find the number of successful outcomes. an outcome. we can use the Counting Principle. the formulas for permutations and combinations. we should expect to make use of the Counting Principle and. Example II A bag of candy contains candies of different fruit flavors.to help clarify the situations so that a model for probability problems can be formed. we find ourselves concerned with counting. Once again. A successful outcome: The flavors are the same. f number of successful outcomes P b bT total number of outcomes ro a 1 lty 0 an event = This will always be a number ranging from 0 to 1. Experiment: Picking two candies. and three are cherryflavored. An outcome: The two flavors picked. We now need to count the total number of outcomes. There are 4 x 3 = 12 ways to pick strawberry candies. we have to add the number of ways we can pick two candies of each flavor. we have to identify the experiment. IJI . eight are lemon-. there are 15 x 14 = 210 different ways to pick 2 of them. Therefore. Four candies are strawberry-. Since we are not replacing the first candy and there are 15 candies in the bag. there may be several outcomes that are considered to be successful in achieving the desired event. If two candies are picked at random without replacement. if necessary. since the numerator can never be larger than the denominator. The probability is a simple ratio. When considering all the ways in which the situation can unfold. what is the probability that they both will have the same flavor? Solution II To solve this problem. and what constitutes a successful outcome. For this. The probability of picking two candies of the same flavor is 74/210 or. queen. the sum of the two top sides will be any number from 2 to 12. A fair die is one in which there is an equal chance of landing on either side. and king are referred to as picture cards.) • A fair coin is one with which there is an equal probability of landing on either side." although it could have any kind of markings on either side. a pair of dice. to the nearest percent. Some points to remember are • A regular deck of 52 playing cards consists of four suits. The top side is considered to be the result of the toss of the die.8 x 7 = 56 ways to pick lemon candies. (Sometimes you may be asked to express the answer as a decimal or percent. These regions may not necessarily be of equal area. Of these 13 labels. 12 + 56 + 6 = 74. the answer is 35%. but there are different probabilities for each of these sums. To the nearest thousandth. the numbers 2 through 10. and a king. The probability of getting either heads or tails when tossing a fair coin is liz. The number of successful outcomes is. Sometimes we are told that the coin is not fair and there is a bias for one side. The sides are usually distinguished by "heads" and "tails. (The ratio of areas is used for probabilities in any geometric setting. or a two-sided coin. a queen. 37/105. a circular spinner divided into colored regions. and 3 x 2 = 6 ways to pick cherry candies. The probability of landing in any region is the ratio of the area of the region to the area of the entire circular region and is directly proportional to the angle formed by the radii that border the region. This may require rounding. When a pair of dice is rolled. when reduced. the jack. • A circular spinner is divided into regions by radii drawn from the center. therefore.352. each having and ace. a jack. • A single die is a cube with each face representing a number from 1 to 6. 132 . the answer is 0.) Many probability problems involve familiar objects such as a deck of 52 playing cards. Jokers are not part of the 52 cards. Therefore. What is the probability that both are picture cards? Solution 12 Identifying the components of the problem. An outcome: The types of cards picked.Example 12 A card is drawn from an ordinary deck of 52 playing cards and replaced in the deck. For these problems. there are 52 x 52 = 2704 different outcomes. A successful outcome: The two cards are picture cards (jacks. Therefore. A second card is then drawn. or kings not necessarily the same). Since we are replacing the first card. there will still be 52 possibilities for the second pick. it states: The probability of an event is equal to the product of the probabilities of being successful at each stage of the experiment. Therefore. What is the probability of getting five heads? Solution 13 The probability of getting a head on one toss is liz. Problems involving probabilities sometimes involve data from real-world situations. queens. we have The experiment: Pick two cards. The probability is used as 133 . We can also solve the previous problem by a different use of the Counting Principle. our answer is simply liz x liz X Ij2 X liz X liz = (liz )s = 1 '/zS 5 = 1132. The probability of picking a picture card at each stage is 12/52 or 3/13. we think of the probability as the fraction or percentage that represents the number of times we are successful. Note how this helps with expressing the answer in reduced form. the probability of picking two picture cards is found simply by 3/13 x 3/13 = 9/169. there are 12 picture cards and the number of successful outcomes is 12 x 12 = 144. Example 13 A fair coin is tossed 5 times. the probability is 144/2704. Using the modified Counting Principle. In the deck. which deals with probabilities. Specifically. 3 3.1 3. How many months would you expect rainfall between 2 and 3 in over a 5-year period? Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Inches 2.1 3. the average spectator attendance at the nine high school basketball games is 135. the probability of between 2 and 3 in of rain in any month is 3/12 or 25%. So far this season. Since 3 months out of 12 meet the condition. 134 .9 Solution 14 The solution is quite simple. Each ticket costs $1.50.5 3. What was the average amount for all sixl 2. outcomes. There is no real difference between this and the previous computations for probability. Jordan and Danny averaged $120. Therefore.) city for the past year is given in the table below. determine whether you will use permutations or combinations. 3.3 4. except that we usually have exact counts to work with.2 4.9 2. there will be 25% of 60 or 15 months in which the rainfall is between 2 and 3 in.S.3 4.5 3. Example 14 The monthly rainfall for a Midwestern (U. Be sure to identify all situations carefully and determine exactly what it is you are counting. we expect that over a 5-year period which is 60 months. 2. What is the expected average attendance for the remaining three playoff gamesl Six people were raising money for a charity. Summary I. and successful outcomes in order to understand the situation.the basis of a prediction or expectation. Additional Problems I. Describe the experiment. that is. Michelle and Lindsey averaged $105.3 2. and Mark and Evan averaged $92. Determine whether you are or are not concerned with the specific order.8 4. and the school expects the total revenue for the year to be $2610. From a regular deck of 52 playing cards. On the return trip. How many possible different combinations of courses can a student take if the school offers 6 math courses. 10. Five consecutive odd integers are such that the sum of the first three is 60 more than the sum of the last. 6.3. 4 science courses. second. 12. what is the probability of picking 5 cards and winding up with a flush? Stacy had to get dressed quickly for school one morning. 5. What is the likelihood that two different English classes will read the same selections? A biased coin is weighted so that the probability of landing heads is 2 in 3. What is the mean and median of these numbers? Theresa ordered a combo pizza with three different toppings. and third place? The Senate Committee on Armed Services has I I Republicans and 9 Democrats. She decides to guess at each question. 4 of 10 short stories. a student has to take 2 math courses. there are six different airlines that will take you nonstop to Dallas. She has three pairs 135 . how many possible different combo pizzas did she have to choose from? Windstar Airlines flies from Newark to three different cities. how many different itineraries can you plan starting and returning with Windstar at Newark? Janice is completely unprepared for the quiz on the novel that was read in English class. If three coins are tossed simultaneously many times. How many different answer sheets could Janice possibly produce? There are 12 runners in a race. 13. She reached into her closet for pants and a blouse. 1 I. 7. a "flush" is a 5-card hand consisting of cards all of the same suit. the English classes must read 3 of 5 novels. They decide to form a subcommittee consisting of 4 Republicans and 3 Democrats. How many different subcommittees are possible? During the first 2 years of college at the State University. a science course. what percent of the time will they all land tails? In the game of poker. The quiz consists of 10 multiple-choice questions with four choices per question. 8. if you do not take the same route. If there are six toppings to choose from and one of her toppings is mushrooms. and 5 of 9 poems. From each of these cities. I of 3 Shakespearian plays. and 10 social science courses for freshmen and sophomores? During the junior year of high school. 9. 4. and 3 social science courses. How many different outcomes are there for first. The wheel is divided into 20 regions. When emptying her pocket one night. On a popular TV game show. to the nearest hundredth. On a circular dartboard whose diameter is 18 in. Gayle found that she had eight nickels. and two pairs of brown pants. What is the probability that these papers will have scores greater than the average of the class~ Scores SS 62 6S Frequencies 3 3 S 1 3 9 4 2 78 83 88 93 100 15. The next 136 . If the ratios of the distinct angles formed by the radii are I:3:5: I I. What is the probability of a thrown dart landing in the bull's-eye~ 18. If these regions are each one-fifth the size of the others. five dimes. If there are five bananas. of blue pants. On his way out the door to go to school in the morning. what is the probability that he will take two bananas and a pear~ 19. While going over the test with the class. what is the probability of the spinner landing in one of the two smaller regions~ 16. A circular spinner is divided into four regions of different colors by four radii. All these regions are equal in area except for two which award the player the grand prize.14. and two pears. three oranges. the contestant has to spin a wheel. Brandon's mother tells him to take three different pieces of fruit from the bin. two pairs of black pants. She also has three white blouses. What is the probability that she picked blue pants and a blue blouse~ The test scores of a class are given in the table below. what is the probability that the contestant will win the grand prize~ 17. five quarters. the teacher randomly picks two test papers out of the pile simultaneously to use as models. the bull's-eye in the center has a diameter of 6 in. and two pennies. and one red blouse. four blue blouses. 20. Therefore.6 = $105. and 39. she reached in and took out two coins. Jordan and Danny raised $120 x 2 = $240. The others are x + 2.$1. our equation is I.3 = 175. At lunchtime. The Counting Principle applies to this problem with three slots to fill. 35.50 per ticket = 1740 tickets sold..67. We need to compute the total amount of money raised by all six people.50 per ticket. What are the chances that she took out less than 35q! ~ In an envelope are 10 straws each measuring a different whole-numbered length from 2 to I I in. At $1. day she put these coins in her pocket. 3. x + 6. the school needs to sell 1740 . Therefore. Let x be the first integer. the average for each person is $634 -.. 3x + 6 = x + 8 + 60 -+ 2x = 62-+ 3x x + 6 = x + 68 -+ = 31 4. 2.1215 = 525 more tickets. The sum of the first three is x + x + 2 + x + 4 = 3x + 6. The mean and median are the same. the others are 33. The average attendance for the remaining three games would be 525 -. Since we know that one of Theresa's toppings is mushrooms.. the second slot has five choices and the third has four. When taking three different straws out of the envelope without looking. namely 35. There are I x 5 x 4 = 20 different combo pizzas that she can create. The total is $240 + $210 + $184 = $634. The remaining toppings are different. and Mark and Evan raised $92 x 2 = $184. we know that the total attendance of the nine games played so far is 9 x 135 = 1215. the expected total attendance is $2610-. Michelle and Lindsey raised $1 05 x 2 = $210. what is the probability that the three straws could form a triangle~ Solutions to Additional Problems We need to work backward from the expected revenue and compute the expected total attendance for the 12 games. 137 . Therefore. Using the fact that the sum of the attendance figures equals the average attendance times the number of games. Therefore. and x + 8. 37. If the first odd integer is 31. the first slot has only one choice. x + 4. We need to use some algebra for this. S. Therefore. That is. a very small chance!) This is a permutation problem since we are interested in the arrangement of the three people who finish in the first three positions. The number of different subcommittees is 330 x 84 = 27. The number of groups of Republicans is C II 4 = II P4 = 4! II x 10 x 9 x 8 4 x 3x 2x I = 330 The number of groups of Democrats is C 9 3 = 9 P3 = 9 x 8 x 7 = 84 3! 3x 2x I 9. 7. Therefore. Since you cannot take the same Windstar routes or the same airline to and from Dallas. the number of different itineraries is the product 3 x 6 x 5 x 2 = 180.576 or approximately 0.2 x I - 138 . each possible group of Republicans can be joined with any of the possible groups of Democrats. (Note that the probability that Janice will answer all the questions correctly is 1/1.048.048. The answer is 12 P3 = 12 x II x 10 = 1320 possible outcomes for the race.720. The Counting Principle is applied with 10 slots.000000095.x airline airline to to Newark from Dallas with Windstar Dallas 6. There are four slots to fill in this problem using the Counting Principle. the number of different answer sheets is the very large number 4 10 = 1.576. each representing one question with four possible choices for each question. From Newark with Windstar x -. the number of different subcommittees will be the product of the two numbers of ways we can pick groups of each. The Counting Principle can be applied with the slots as -----x ------x Math courses There are Science courses Social science courses c _ 6 2 - 6 P2 _ 6 x 5 _ 15 2! . 8.x -. The solution to this problem involves both the Counting Principle and combinations. 7% of the time. 10C. 12. there are lOx 210 x 3 x 126 = 793. 10.000001259 or there is almost no likelihood that this will happen! 1 I.037.different ways to choose math courses. Therefore. A key fact about probability is that the sum of all possible outcomes is I. A "flush" is a group of cards of the same suit in no specific arrangement.793. the probability of landing tails is 1/3. therefore Number of possible different selections There are sC 3 = 10 different groups of 3 novels. In other words. Since the only outcomes for tossing a coin are heads and tails and we are given the probability that landing heads is 2/3. Therefore. The probability is I -7. the numerator and denominator of the probability fraction must be equal. 3 different Shakespearian plays. if any outcome is considered to be successful. The word likelihood is a synonym for probability. We use the modified Counting Principle for probabilities and find the probability of three tails on one toss ofthe three coins to be 1/3 x 1/3 x 1/3 = 1/27 or 0. Two classes reading the same selections is the same as already knowing the selection that one class has made and asking. we can expect this to occur 3. the number of different 139 .800 different selections to choose from. Therefore. There are IOC3= IOP3 = 3! IOx9x8 3x 2x I = 120 different ways to choose social science courses. using the Counting Principle. "What is the probability of that being the selection for another class~" The probability is.800 = . and 9C 5 = 126 different groups of poems. The problem needs to be restated in a way that identifies the event. Therefore. there are 15 x 4 x 120 = 7200 different combinations of these courses that a freshman or sophomore can take. There are different ways to choose a science course. for anyone suit. = 210 different groups of short stories. 960. Score 55 62 65 78 Frequencies 3 3 5 1 Products 165 186 325 78 83 88 93 3 9 4 2 249 792 372 100 Sums 30 200 2367 140 . 14. Therefore. and a successful outcome--picking any of three pairs of blue pants and any of four blue blouses. 3/7x 1/2=3/14.13. Identifying the components of the problem. flushes is 13C5 = 1287. we have the probability of picking blue pants = 3/7 and the probability of picking a blue blouse = 4/8 = 1/2.00198. there are 4 x 1287 = 5 148 different flushes. and a successful outcome--picking a pair greater than the average. Using combinations. To find the average score. The answer is. there is an extremely small chance that this will occur!) Identifying the components of the problem.960 = 0. the probability is 5148 -:. which states that the probability of a successful outcome is the probability of picking blue pants x the probability of picking a blue blouse.598.2. we have to get the sum of the scores by mUltiplying each score by the number of times it appears in the set of 30 and then add the products. This is easiest to do by adding a third column to the table which will contain the products and adding a row at the bottom that will contain the sums of the second and third columns. we have the experiment-picking two papers without replacement (each pick represents a different stage of the experiment). We can use the modified Counting Principle for Probabilities. we have the experiment-picking pants and picking a blouse (each pick represents a different stage of the experiment). The total number of different five-card hands is 52C 5 = 2. (Since this number is nearly 0. an outcome--picking any pair of papers. Since there are four different suits.598. an outcome--picking any of seven pairs of pants and any of eight blouses. therefore. The probability of a spinner landing in a region is the ratio of the area of the region to the area of the circle.-----30 x 29 30 x 29 which indicates that the problem could have been solved by finding the ratio of (a) the number of ways to select 2 from the 18 scores that are greater than the average to (b) the number of ways of selecting 2 scores from the entire set of 30.35. let x be the multiple and let the angles be x. Since the second paper is pulled out at the same time as the first. The frequencies indicate that there are 18 of these scores. and I Ix.93. Using the algebra of ratios. it is helpful in understanding the situation. 5x. Their sum is 20x.The average is 2367 -:. the probability of the first paper having a score greater than the average is 18/30. Therefore. its probability would be 17/29.30 = 78. the two smaller angles are 18 and 54 Together this is 72/360 of the 0 . and 100 are all greater than the average. The modified Counting Principle for Probabilities states that the probability of both papers having scores greater than the average = 18/30x 17/29 = 0. 15. rounded.88. 3x. we lose one possibility from those greater than the average and from the total pile. While it is not necessary to compute the angles. Therefore. 0 0 • 141 . Since the entire rotation about the center of the circle is 360 we have 20x = 360 and x = 18.9. The scores of 83. Note that this answer is equivalent to (18 x 17)/(2 x I) (30 x 29)/(2 x I) = 18xl7 2x I x 2xl 18 x 17 . The ratio of the areas of the regions is directly proportional to the ratio of these central angles. Therefore. the probability of landing in either of the smaller regions is 2x /92x = 1/46 or. 15 and 16 as it involves finding the ratio of areas. Therefore. 72/360 = 2/ 10 or 0. 0. using combinations. 8. (This. in fact.) The probability of the dart landing in the bull's-eye is the ratio of the area of the bull's-eye to the area of the dartboard. Such problems are referred to as problems of geometric probability. will be true of any problem involving interior regions of a geometric shape. The probability is the ratio of the number of ways Brandon can pick 2 bananas and I pear to the number of ways he can pick any 3 fruits from the collection of 10 fruits. Note that this is the square ofthe ratios ofthe radii: (3/9)2 = (1/3)2 = 1/9. Since the smaller regions are one-fifth the size of the others. The area of the bUll's-eye is 9n and the area of the entire dartboard is 81 n. therefore. The total area of the regions we wish to land in is 2x. The probability of landing in either of the two smaller regions is.16.02. to the nearest hundredth. This problem can be solved using the same idea as in Prob. the probability is 9n 781 n = 1/9.2. We can use algebra to determine this ratio. we let x be the area of each of the smaller regions and Sx be the area of each of the other 18 regions. and the total area of the wheel is sx x 18 + 2x = 92x. 17. The "bull's-eye" is a circle whose center is that of the larger circular dartboard. The 142 . This will always be true about the ratio of areas of two circles. This problem is similar to Probs. The probability of a spinner landing in a region is the ratio of the area of the region to the area of the circle. we know that the radii are 3 and 9 in. In this problem. rotation. Therefore. 18. Since we know the diameters of each circle. we are given information about the areas of the region. 6 and 18 in. 20 = 380 filled cells (note that this is 20 x 19). Of these.82. Note that the diagonal contains no values since we cannot use the same coin twice. and a successful outcome-picking two coins whose total value is less than 35¢. p p X p 2 X 6 6 6 6 6 6 6 6 II II II II II P 2 n 6 6 6 6 6 6 6 6 II II II II II n 6 6 X 10 10 10 10 n 6 6 10 X 10 10 10 n 6 6 10 10 X 10 10 n 6 6 10 10 10 X 10 n 6 6 10 10 10 10 X n 6 6 10 10 10 10 10 n n 6 6 6 6 10 10 10 10 10 10 10 10 10 10 d d d d d q q q q q II II II II II 26 26 26 26 26 II II II II II 26 26 26 26 26 15 15 15 15 15 30 30 30 30 30 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 n n n n n n n d d 10 10 10 10 10 X 10 10 15 15 15 15 15 30 30 30 30 30 10 10 10 10 10 10 X 10 15 15 15 15 15 30 30 30 30 30 10 10 10 10 10 10 10 X 15 15 15 15 15 30 30 30 30 30 15 15 15 15 15 15 15 15 X 20 20 20 20 35 35 35 35 35 15 15 15 15 15 15 15 15 20 X 20 20 20 35 35 35 35 35 15 15 15 15 15 15 15 15 20 20 X 20 20 35 35 35 35 35 15 15 15 15 15 15 15 15 20 20 20 X 20 35 35 35 35 35 15 15 15 15 15 15 15 15 20 20 20 20 X 35 35 35 35 35 d d d qUUmmmmmmmm3535HHHXmmmm qUUmmmmmmmm3535HHHmXmmm qUUmmmmmmmmHH35HHmmXmm qUUmmmmmmmm35H35HHmmm q 26 26 30 30 30 30 30 30 30 30 35 35 35 35 35 50 50 50 L--·~- There are 20 x 20 . Identifying the components for this problem. the probability is 310/380 = 31/38. an outcome-pick any two coins totaling any amount. therefore. In this case a table whose column and row headings indicating the different coins can be used. The total number of 143 . When all else fails. Each cell contains the total value of the two coins.answer is 10 x 2 120 20 120 I 6 19. or the chance that Gayle took out less than 35¢ is approximately 0. a probability problem can be solved by listing all possible outcomes. we have the experiment-picking two coins. 70 are not successful. We move along through the 2s until 2.35)/190 = 155/190 = 31/38. and a successful outcome--the three lengths can represent the sides of a triangle. For example. Note that there are loe 3 = 120 possible different triplets that can result and we should list the possibilities in a way that makes it easy to see that we have considered all possibilities. which would add no additional successful triplets. ways to do this. our probability is (190 . it would be reasonable to find a systematic way to list the combinations and pick out the ones that work.) It should be clear from the problem that there are considerably fewer ways to pick two coins unsuccessfully than successfully. (Note that we are not considering the order in which we pick the coins. Therefore. We can find the number of ways to do this and subtract. Therefore.4 in the row above. Since there is more to consider than just counting the number of different groups of 3. The experiment-picking three different straws. is not true! A basic fact about triangles is that the sum of any two sides must be greater than the third side. Any pick of two quarters is unsuccessful and there are SC2 = -2! = -x. you might say that the probability should be I. but anything higher would not be less than 2 + 3. When moving down the column. however. This. 144 . where we consider the pairs.= 10 2 I SP2 5x 4 20.4. and there are 5 x 5 = 25 ways of doing this. Any pick of a quarter and a dime will be unsuccessful. A possible third side is 3. and then to the next column. thinking that you can form a triangle from any three lengths. but we already listed the triplet 2. We can have a side of 4. we go to 2. I I. we can start a table and include in the cells only those possible third sides which were not considered previously. an outcome--three different lengths.3. The only other remaining possibility is 5. that is why this is half of the number we saw in that table. we can start with supposing that our first two picks are 2 and 3.outcomes is c _ 20 2 - 20 P2 _ 2! - 20 x 19 = 190 2x I which will be the denominator of the probability ratio. including a 3. At first thought. 11 11 X 10.11 6.10.10 9.7 4.6 6.11 11 X 5.11 11 X 4.7 3.9 4.10 7.First Possible First Possible First Possible First Possible First Possible First Possible First Possible First Possible First Possible two third two third two third two third two third two third two third two third two third 2.4 3.7 5.4 2.8 8.11 9.10 4.11 11 X 9.7.11 10.6 2.10.5 2.9.9 9.6 5.11 5.9.10 5.10.9.9 6.11 9.8 5.9 8.11 10.8 4.9 8.5 4.8.10.9 5.8 7.11 7.8 7.6 3.10.11 10.9.10 6.5 3.7 7.11 11 X 8.10 3.8 6.10 9.10 10.7 2.10 8.11 11 X 6.11 8.11 X VI ~ .10.8 2.7 6.9 2.11 10.11 11 X 7.9 3.3 2.10 2.11 4 5 6 7 8 9 10 11 X 3.11 10.8 3.9 7.10 8.6 4.11 9.8. 146 .We can now count the number of entries in the "Possible third" column 70. Therefore. the probability of picking three straws that could form a triangle is 70/120 = 7/12 or approximately 0.58. The answers are given at the end. concepts.7S/ft. If the grass covers approximately 3550 ft2. he weighed himself again and found that he then weighed 142 lb. in 15 minutes.S0/ft. how much more would Jim be willing to pay for molding than Donna wouldl 2. you can look back at the solutions to the examples and additional problems in the chapter mentioned before each set of problems. How many pounds did Danny gain or lose while in ltalyl 147 . how far is her house from the storel 4. he weighed himself and found that he weighed 63 kg. and the entire diamond (which is really a square) is covered with grass except for the pitcher's circle. Jim likes the molding that costs $2. what is the diameter of the pitcher's circle to the nearest footl 3. Good luck! Problems Similar to Those in Chap. If the rectangular den is 15 ft x 12 ft. I I. Laura walked to the store from her house in 20 minutes and then from the store to her friend's house. while Donna prefers the molding that costs $1. When Danny first arrived in Italy on his trip. which is dirt.Chapter 6 Miscellaneous Problem Drill This collection of problems will give you an opportunity to use the skills. If she walked at the same rate on both trips. Jim and Donna would like to put a fancy molding around their den. The distance from homeplate to first base at the local schoolyard is 60 ft. and strategies discussed in the previous chapters. When he returned home from his trip. If you need help. which was I mi away. The sum of the squares of the first and last of three consecutive positive integers is 164. Today Allen is 8 years older than Harvey.99. The owner of the store creates a 50-lb mixture of the two that sells for 148 . This year she invested all the principal and interest from that fund into another fund that earned only 6 percent. 12. The larger of two numbers is 8 less than 3 times the smaller. However. A second particle is projected 5 seconds later along the same path at a speed of 1200 m/second. The tens digit of a two-digit number is 3 less than the units digit. If 3 is added to both the numerator and the denominator. The number is 3 more than 4 times the sum of the digits. The prices of two kinds of coffee are $1. the new fraction has a value of 0. Given that the sum of the number:s is 88. 9. Josh wanted to buy a football helmet that cost $49.5.7. I I. which are 21 mi apart. how fast was Jeff traveling if they met each other in 45 minutes~ 15. If each soup bowl holds ~ cup of soup. How much money does she have now~ 7. A nuclear reaction projects an atomic particle along a straight line at 900 m per second (m/second). 10. What should Josh do in order to pay the least amount for a helmed Problems Similar to Those in Chap. 1 8. find the larger number. Jeff and Steve left their homes. how many second servings of soup could the chef serve. at the same time and bicycled along the same straight road toward each other.40/Ib. In November. 13. there was another helmet in the store that sold for $42. In 6 years he will be twice as old as Harvey will be then. In how many seconds into its trip will the second particle pass the firsd 16. Find the three integers. Last year Gabriella invested $3500 in a mutual fund that had an annual return of 12 percent. The denominator of a fraction is 3 more than the numerator. Find the original fraction.99. Find four consecutive odd integers whose sum is 96. if the pot has a 3-gal capacity~ 6. A chef had to prepare soup for 50 people and used the entire capacity of the pot. The salesperson told him that after the football season in February the helmet would sell for 15 percent less. How old is Harvey now~ 14.75/Ib and $3. If Jeff's speed was 4 mph faster than Steve's. Find the number. How much time did it take for the other pump to complete filling the tankl Phyllis invested part of her $10. The ratio of two positive numbers is 4: I. A recipe in a cookbook calls for ~ cup of sugar and 3 cups of flour to make a cake that serves eight people.000 inheritance in a fixed annuity earning 7 percent annually. but actually earned 12 percent over the year. Joan was expecting more than eight people and used 10 cups of flour. If she has $1. $2. 18.85 in change. If 2600 votes were cast.25. much did she invest in both situations if her earnings for the year were $1 050l Problems Similar to Those in Chap. by how many votes did Nicole win the electionl 26. how long will it take for the job to be completedl If two pumps are used to fill a tank at the same time. how much. how many students are in the schooll 22. 19. The ratio of votes received respectively were 2:5:6. Lindsey. Separate 144 into two parts so that the ratio of the two numbers is 5:7. In 10 years. and the sum of their squares is 153. Melissa usually does the same job in 30 minutes. to the nearest pound. This morning both pumps were turned on at the same time and left to fill the tank. Assuming that all 50 Ib are sold. After I ~ hours.17. If they work together. how many dimes does she havel Carole is asked to do a job that ordinarily takes her 45 minutes to complete. Kristen. She invested the rest of her money in various stocks that were expected to earn 8 percent. How many cups of sugar did she have to usel 149 . Find the numbers. The faster pump fills the tank by itself in 5 hours. Gayle's age will be 22 less than twice Danielle's age. and Nicole ran for student body president. 23. of the cheaper coffee should she use in the mixture to make sure that she makes the same money as if she sold the coffees separatelyl Amy has twice as many nickels as quarters and two more dimes than nickels. How old is Gayle nowl 25. A school district boasts that there are three teachers for every 70 students. How. the slower pump stopped working. the tank is filled in 3 hours. If the total number of teachers and students in the school is 4015. 3 21. 20. The ratio of Gayle's age to Danielle's age is 3:2. 24. 4 30. 32. In the time it took for all the papers to be graded. The mixture is heated so that it becomes a 40 percent basic solution by weight. a 5 Y2-ft-tall person casts a shadow of 8 ft. Committee A usually grades 30 more papers in an hour than committee B. How much water must be eliminated in this process to achieve the proper mixturel Problems Similar to Those in Chap. A rigid cable is needed to support a 12-ft antenna on the roof of a building. how many papers can each committee grade in one hour. The chemist knows that the pressure of a gas within a container varies directly with the temperature in the container. and each leg is I0 cm less than the base.. 36. to the nearest tenth of a centimeter. To the nearest foot. If the pressure is measured at 35 Ib/in 2 when the temperature is 18°C. The cable is to be attached to the top of the antenna and to a point 7 ft from the base of the antenna. and how many hours did it take for all the papers to be gradedl 28. how long is the tree's shadow at 3 p. What are the measures of all three anglesl The measures of two consecutive angles of a parallelogram are in a ratio of 4:5. In the manufacturing process of a cleaning fluid. committee A graded 350 papers while committee B graded 280 papers. 150 .27. A nearby tree is 21 ft tall. 33. In a triangle. If the area of the trapezoid is 184 cm 2 . must the temperature be raised so that the pressure reaches 40 Ib/in 2 l 29. To the nearest tenth of a foot.M.M. 35. to the nearest tenth of a degree. What are the lengths of the three sides of the trianglel At 3 P.l One of the legs of a right triangle is 17 in more than the other. 34. are the nonparallel sidesl 0 0 31. In June the final exam in math given to all ninth graders is graded by two teacher committees. how much cable must be available to accomplish thisl The parallel sides of an isosceles trapezoid are 16 and 30 cm. Find the measure of the larger angle. how long. how much higher. The perimeter of an isosceles triangle is 52 cm. Find the length of the larger leg. 180 kg of a dry base is added to 500 L of water. On the basis of these results. the measure of the largest angle is 16 more than twice the measure of smallest angle and the third angle is 20 more than the smallest. given that the hypotenuse is 25 in long. to the nearest tenth of a foot. In an experiment. How high. A floor design in the lobby of a building is in the shape of a rhombus. Jennifer has been improving in math. The weights of some of the members of the wrestling team are 172. 183. Her last three test scores were in the ratio of 4:5:7. the angle of elevation of the kite is 32°. How long. What was her lowest score if her test average is 72? 42. of the four interior angles of the trapezoid? 38. and 179 lb. along the side of the building does the top of the ladder reach? 39. 185. 43. and mode of the reaction times. seconds Number of reactions 23 2S 28 31 3S 31 46 S2 61 10 ISI . A 15-ft ladder is leaning against a building at an angle of 39° with the ground. 212. and its diagonals are laid out in red tile. the times of many chemical reactions were recorded and are given in the table below. to the nearest degree. Time. From a point 25 ft from where a kite is directly over the ground. In the trapezoid described in the previous problem.37. If the diagonals measure 6 and 8 m. There are as many floors in the building as there are whole degrees in the smaller angle of the rhombus. is the string attached to the kite? Problems Similar to Those in Chap. 20 I. median. Find the median and mean weights of this group. S 41. what are the measures. to the nearest foot. Find the mean. how many floors are in the building? 40. 49. 50. 48.50. 8. 3 have to be selected to be the officers. Josh should wait because the discounted price on the helmet will be $42. 45. What is the probability that they are both chocolate-chip cookies? Answers to Chap. what is the probability that the next card drawn is not a heart? "Rolling doubles" means that when two dice are tossed. 47. Mike went to the video store to rent some newly released movies to watch over the weekend. eight sugar cookies. If the first three cards are hearts. 2. and the red and yellow regions equally occupy the rest of the area. and six oatmeal cookies. 46. How many times should you expect to lose a turn if you toss the dice 78 times during the game? On a circular spinner whose radius is 6 cm. there are four chocolate-chip cookies.4 lb. the blue region occupies 911" cm 2 .44. I~mi. 5. Steve reaches into the bag without looking and takes out two cookies. the up sides are the same. the green region occupies 1211" cm 2 . $40. 64. 14 bowls of soup.50. 4.20. 3. To the nearest hundredth. 6 I. If he wanted one of each. In a certain game. you lose a turn whenever doubles are rolled. what is the probability that the four people consisted of two boys and two girls? In a bag of cookies. $4155. How many different groups of 3 students can be chosen to be officers? Four cards are drawn from a standard deck of playing cards without replacing a card after it is drawn. Only four people were able to get on the last remaining car on the roller coaster. and three dramas. Danny gained 3. 152 . six action movies. 8 ft. Among the new releases there were four comedies. What is the probability of the spinner stopping in the red region? A group of seven boys and five girls went to the amusement park. 7. how many different possible groups of three movies could he rent? Among the 12 students in the club. 6. the median is 28 seconds. 26.4 ft. Committee A grades 150 papers per hour. 2~ hours. 15 seconds. and 10. 13. the angle is approximately 73. 27. 2 years old. and 88°. 29 ft. 28. 12 and 3. 37. The median weight is 184 Ib. 31. 14ft. 30.6°C. committee B grades 120 papers per hour. 22. 10.9. 47. The mean is 27. II. 18 minutes. 29. 12.131°. All the papers were graded in 2~ hours. 49°.7°. 14. Gayle is 36 years old now. 3850 students. 17. 33. the mean weight is 188. 18. 23. 72. 230 L. 38. 56° . 100°. 2. 73 floors. and the mode is 3 I seconds. 15. 19.9. 14. and 24 cm. 153 .7 lb. I ~ cups of flour. and Lindsey received 400 votes. 30.131°. 21. 25. 24. 16 mph. 10. 14. 35. $3000 in the annuity and $7000 in the stocks. 42.6 cm. 45. 20. 9. 8 dimes. 60 and 84. 4f7. 41. 54. Nicole received 1200 votes. 36. 34. 44. 21. 24 in. 200 votes. 36° . 35 lb. 43. 32. 220. 8. 39. Kristen received 1000 votes. 16.5 ft. 40. and 27.8 seconds.49°. 25. 23. 46. 39/49. 47. 13. 48. 5/24 or 0.208333 .... 49. 0.42 SO. 12/306 or 2/51 or approximately 0.04. 154 Appendix A Brief Review of Solving Equations The information below will refresh your memory about some of the most important facts about solving equations. The two most common types of equations that you need to solve are linear equations and quadratic equations. For each there are several steps to follow that will ensure that you get the correct answer every time. Linear Equations An equation is linear if the highest power of the variable is 1 such as 4x - 5 = 11. In fact, you won't even see the power since we rarely write exponents of 1. The goal is to undo all the complications in the equations and work your way to having the variable appear on only one side, with a numerical term on the other. The following example illustrates the necessary steps involved: Example. Solve the equation 4(2x + 5) + 6x = 8x - 4 - 2x for the value of x. Step 1 Use the distributive property to 8x+20+6x = 8x-4-2x "remove" any parentheses 14x+20 = 6x-4 in the equation. At this point the variable should appear Step 2 Combine like terms on each at most once on each side, and a side of the equation separately. number should appear at most once on each side. 155 Step 3 Remove the variable with the smaller coefficient by adding its signed opposite to both sides of the equation, and combine the variable terms. Step 4 Remove the numerical term from the side of the equation with the variable by adding its signed opposite to both sides of the equation and combine the numerical terms. Step 5 Divide both sides of the equation by the coefficient of the variable and simplify. Step 6 Check your answer in the original equation by using it in place of the variable. 14x + (-6x) + 20 = 6x + (-6x) - 4 8x+20 =-4 8x+20+(-20) = -4+(-20) 8x = -24 8x -.;..8 = -24-.;..8 x =-3 4(2x + 5) + 6x = 8x - 4 - 2x 4(2 x (-3)+5)+6 x (-3) = 8 x (-3) -4-2 x (-3) 4(-6+5)+-18 -24-4+6 4 x (-1) + -18 - 22 -4 + -18 ..( -22 Example. Solve the equation 4(y - 7) - 2(8 Step 1 Use the distributive property to "remove" any parentheses in the equation. Step 2 Combine like terms on each side of the equation separately. Step 3 Remove the variable with the smaller coefficient by adding its signed opposite to both sides of the equation, and combine the variable terms. Step 4 Remove the numerical term from the side of the equation with the variable by adding its signed opposite to both sides of the equation, and combine the numerical terms. Step 5 Divide both sides of the equation by the coefficient of the variable and simplify. y) = 6(y + 3) - 3Y + 1. 4y - 28 - 16 + 2y = 6y + 18 - 3y + 1 6y-44=3y+19 6y + (-3y) - 44 = 3y + (-3y) + 19 3y - 44 = 19 3 y - 44 + 44 = 19 + 44 3y = 63 3y -.;.. 3 = 63 -.;.. 3 y = 21 156 14 = 6. Solve the following pair of simultaneous equations for x and y: 2x+ y= 6 y . Example. For example. Simultaneous Linear Equations with Two Variables A pair of equations are said to be a simultaneous pair if you are told to look for the values of each variable that satisfy both of the equations. (x. 157 .14 for y in the first equation gives 2x + 3x .3x = -14 The second equation is easily solved for y by adding 3x to both sides. giving y=3x-14. Combining and solving for x. the pair of numbers x = 7 and y = 2 is the only pair that will satisfy the equations x + y = 9 and x . 2). Only one pair of numbers will work. Simultaneous equations arise frequently in word problems that involve two unknowns. Each side has a value of 82 when y = 21. y). Step 2 Use the expression for the solved variable in its place in the other equation. Step 3 Solve the "new" equation for its variable. In this 'case the solution is (7. we have Sx -14 = 6 Sx = 20 x=4 y-3x = -14 y-3(4) = -14 y-12=-14 y =-2 Step 1 Solve one of the equations for one of the variables. Step 4 Use this variable to find the other variable from the equation you used in Step 1. Substitution is usually used when one equation lets you easily write one variable in terms of the other.y = 5 at the same time. We usually write the solution as an ordered pair.Step 6 Check your answer in the original equation by using it in place of the variable. There are many methods for solving simultaneous pairs of equations. multiply. or divide only if you perform the same operation on each side of the equation. (Try the check yourself. Substituting 3x .) Key Fact: You can add. The two most common methods are by substitution and by combining. subtract. ( Solving by combining is sometimes easier as it immediately eliminates one of the variables.( Example.( 6 In the second equation -2 .b = 20 2a+b = 15 The second equation is easily solved for b by adding -2a to both sides. giving b = -2a + 15. Step 6 Check that your solution satisfies both equations. Step 4 Use this variable to find the other variable from the equation you used in step 1. (4. Step S Write your solution. b).(-2a + 15) = 20. In the first equation 2(4) + (-2)=6 8 + (-2) . Solve the following pair of simultaneous equations for a and b: 3a . In the first equation 3(7) . The solution is the pair x = 4 and y = . Step 2 Use the expression for the solved variable in its place in the other equation. (7. we can add the given equations to eliminate b. Step 6 Check that your solution satisfies both equations. Step 3 Solve the "new" equation for its variable. y). 2a +b = 15 14+b = 15 b = 1 The solution is the pair a = 7 and b = 1 or.Step S Write your solution. Using the last example.( In the second equation 2(7) + 1 = 15 14 + 1 15 . We can continue by using 158 .1 = 20 21-1 20 . -2). as an ordered pair (x. as an ordered pair (a. 3a-b= 20 +2a+b = 15 Sa = 35 and we immediately see that a = 7. 1). we have Sa -15 = 20 Sa = 35 a = 7 Step 1 Solve one of the equations for one of the variables. Substituting -2a + 15 for b in the first equation gives 3a . Combining and solving for x.2 or.3(4) = -14 -2-12 -14 . Sy = 0. giving us the pair 12x + 8y = 144 and 12x . If the equations can't be added to eliminate one of the variables.this value in one of the equations to find the value of the other variable as in step 4 above. Step 5 Write your solution. 6). Adding the two equations gives us 23 y = 138 and y = 138 -. y).lSy = 6. we can "fix" the situation by doing a bit of multiplying first. and solve the resulting equation for the other variable. x =8. Quadratic Equations In a quadratic equation. For example. 23 = 6. the following are quadratic equations: x 2 = 25. Step 6 Check that your solution (Try the check yourself as practice.5Y = 0 has the solutions y = 0 and y = s. we will multiply the first equation by 4 and the second by 3. If we choose to eliminate x. 159 . we two original equations to find have 3x + 12 = 36. The solution is the pair x = 8 and y = 6 or as an ordered pair (x. Multiplying both sides of the second equation by -1 gives us the pair 12x + 8y = 144 and -12x + lSy = -6. y2 .6n + 8 = O. x2 = 25 has the obvious solutions x = 5 and x = -5. and the value of the other. y2 . Example. Solve the following pair of simultaneous equations for x and y: 3x + 2y = 36 4x - Sy = 2 Step 1 Select one of the variables to eliminate and multiply each equation by the coefficient of this variable from the other equation. Key Fact: Every quadratic equation can be solved for two values of the variable. the highest power of the variable is 2.. and n 2 . Step 3 Add the equations to eliminate the variable. 3x = 24. (8. Step 4 Use this value in either of the Using y = 6 in the first equation. Example. Example.) satisfies both equations. Step 2 Multiply both sides of one of these "new" equations by -1. one side is O.5 = 0. x 2 .5(5) = 0 25-25 160 . 0-0 o o ./ Using y = 5. + 5)(x - 5) = o./ original equation. This particular quadratic is a difference of two perfect squares. which have the solutions and solve for the variable. so it factors as expression.5) = O. We have the two equations. and x . Example./ . Example. we have 02 . Solve for all values of x that satisfy x 2 = 25. The next step uses the fact that if the product of two expressions is 0. Both terms of the particular quadratic have Step 2 Factor the quadratic the common factor y. x 2 = 25 when x is +5 or -5. and We have the two equations. . This is accomplished by adding -25 to both sides: x 2 + (-25) = 25 + (-25) or x 2 . We sometimes find that these two solutions are equal. Step 1 Rewrite the equation so that This is already the case: y2 .Example. Example. y = 0 and Step 3 Set each factor equal to 0. Step 2 Factor the quadratic expression.5 = 0. n2 . which factors as (x Step 1 Rewrite the equation so that one side is O. Step 4 Check each solution in the Clearly. Step 4 Check each solution in the Using y = 0.25 = O. (How else could you multiply two numbers to get O?) Step 3 Set each factor equal to 0.5y = O.8x + 16 = 0 has the solutions x = 4 and x = 4. The most common method of solving a quadratic equation is by factoring using the steps described below. then one of the factors must be O. x + 5 = 0 solve for the variable.5y = O. we have 52 . which have the solutions x = -5 and x = +5. y .0(5) = 0 original equation.6n + 8 = 0 has the solutions n = 2 and n = 4. y(y . y = 0 and y = +5. Solve for all values of y that satisfy y2 . The equation x 2 . giving us x 2 .e.4ac = ------------2a 2 (where a. b. you need to apply the following special quadratic formula.. The quadratic expression ax 2 + bx + c = 0 has the two solutions: x -b±Jb . Solve for all values of n that satisfy n2 + 8 = 6n. b = -9. Step 4 Check each solution in the Usingx = 4. This is done by adding -6n to both sides. We have the two equations.9x22 = 0 has the coefficients a = 1. and We have the two equations. Solve for all values of x that satisfy x 2 = 8x .16.16 16 . Step 3 Set each factor equal to 0. 16 32 . and x .. and c = -22. they are the coefficients of the terms).Example. and c represent the numbers that appear in the expression in those specific places. These 161 .6n + 8 = O. Step 4 Check each solution in the original equation. both sides.4)(x . Using n = 4.. Step 3 Set each factor equal to 0. Key Fact: There are quadratic expressions that do not factor. giving us n 2 . Step 2 Factor the quadratic expression. we have4 2= 8(4) ..16 original equation. which have the solutions n = +2 and n = +4. Example.4 = 0 solve for the variable. which have the equal solutions x = 4. Step 2 Factor the quadratic expression.2)(n . This is beyond the scope of this book. x . To solve these. Step 1 Rewrite the equation so that one side is O. we have 22 + 8= 6(2) 4 + 8 12 12 . This particular quadratic factors as (x . Step 1 Rewrite the equation so that This is done by adding -8x and +16 to one side is O. we have 42 + 8= 6(4) 16+ 8 24 24 . and solve for the variable.2 = 0 and n . Using n = 2.4) = O.4 = 0.4 = 0. but the following example will show how it can be used.4) = O. This particular quadratic factors as (n .8x + 16 = O. i. n . .4(1)(-22) x= 2(1) 9±J81+88 222 9 -13 9 + 13 = 22 = 11 = - 9 ± J169 9 ± 13 ---:~ = 2 2 x- 2 -4 .2 . this quadratic.-.9x .22 = 0. x 2 .values are substituted into the formula. can be solved by factoring. and we have -(-9) ± )(-9)2 .2 - You should check these values to verify that they are solutions. and you should do so as practice. 162 . Also. 147 Circumference. 49. 23 Interest. 82. 124. 136. 91-122 Geometry. 126-130 Counting problems.26 Coin problems. table of relationships. 127. 45. 152 Counting principle.Index The index below includes the most relevant pages to the items listed. 128 Fahrenheit. 50.140 Cubic units. 17. Within the statements of problems or their solutions. 13 Circles.44 Counting. 91-94 Angle of elevation. 149 Combinations. 148 Algebraic solutions. 110 Apothem. 101 Cost. 38 Distance measuring. 11. 17. 15 Investment problems. 150 Discount. 117 Area definition. 49. 32.96 Average (mean). 121 Factorial. compound. 18. 134. 123-126 Average problems. 120-122 Interest. 133. angle of depression. 26-27 Fractional equations. 122. 79.90. 2 Distance problem. 14 Distance formula. 69. 148 Converting measurements. there may be ideas related to these items not included here. 135. 9 Diameter. 147 Dry measure.50. 26 Area problems.78. 138 Counting principle.157-159 Equilateral triangle. 10-12. 46. 35-37. 2 Equations with two variables. 120 Consecutive integers. 15. 17 Congruent figures. 7. 131 Exterior angles. 2-6. 138 Complementary angles. 75. 77. 11 English system. 3 Experiment. 4.25 Cosine. 148 Centigrade (Celsius). 51. Adjacent angles. 99. 120 Capacity. 13 Formulas for measurement. 16. 103-105. 33. 111. 120 Cone. simple. 7. 120 Angle measurement. 4 Area formulas. 121 Estimating. 151 Bisector. 120 Age problems. for probability. 7 Direct variation.82. 42 Geometric probability. 26. 129. 14. 142 Geometry. 29-63 Altitude. 148-149 163 . ISO Pentagon. 112. 138 Pi.121. 124. 157-159 Sine. 97-101. 9 Volume problems. 123-126. 18. 16. 95. 101 Trigonometry. 101 Speed. 10 Mass. 121. 120.148 Number problems. 4 Sphere. 51. 70. 151 Measurement. 82. defined. 120 Parallelogram. 150-151 Tree diagram. 29-35. 34. 18 Similar figures. 2. 49. 149 Mode. 68. 147 Permutations. 155-159 liquid measure. 131 Parallel lines. 11 Square units. 110. 12 Weight problem. 109. 65-90 Percentage. 111. 120-122 Rhombus. 13. 120 Tangent. 98 Quadratic equation. 109-110. 149-150 Pythagorean Theorem. 17. 130-146 Probability problems. 2-13 Mixture problems. 81 Median. 73. 41-43.Isosceles triangle. 122. 100. 161 Radius. 121 Pythagorean triples. 127 Triangle problems. 43. defined. 151 Units of measurement. ISS.95. 48. 151 Right triangles.7 Polygons. 123-146 Supplementary angles. 65-90 Proportion. 135-136. 135. 67-69. 123-126. inverse. 139 Linear equations. 27 Weight. 49. 118. 16. 81-82. 82. 97-109 Right triangle problems. 78 Perimeter formula. 94-97 Principal. special. 129. 137 Triangle. 121 Simultaneous equations. 107 Trigonometry problems. 112. ISO Likelihood. 122. 148-149 Outcome. 107-109 Rounding decimals. ISO Triangles. with proportion. 120 Volume. 14 Scale model. 80-82. 149-151 Regular polygon. IS Probability. 119. sum of the angles. 109. 148 Volume formulas. 101 Temperature problems. 1-13 Measurement problems. ISO Similar triangles. 151 Money-related formulas. 109-110. 71. 141. 91. 27 Motion problems. 7 Rate. 5 Statistics. ISO Right triangles. 82 Shopping problems. 149 164 . 49. 120-126. 94. 100. 93. 2 Vertical angles. 122. 93 Trigonometric ratios. 148 Work problems. 110 Percentage. 113. defined. Ill. 112. 91-92. 26 Perimeter problems. 12 Mean (average). 72 Proportion problems. 152 Proportion. 43-46. 74. 151 Metric system. isosceles. 37-41. 101-119. 38-42 Ratio. 17 Transversal. 1 Sales tax. 159-161 Quadratic formula. 114. 82. 113. 120 Trapezoid. 118. 65-90 Ratio problems. 110. 121-122 Trigonometry.