TnP - Venkata Subbu.B

Training & PlacementPondicherry Engineering College Puducherry Preparatory Material 1. Numerical Reasoning 2. Analytical Reasoning 3. Logical Reasoning 4. Verbal Reasoning Training & Placement 1. Numerical Reasoning 1.1 Problems on Numbers 1.1.1. Concept: Numbers are broadly classified into Real and Imaginary Numbers. The Real Numbers consist of Integers, Whole Numbers and Natural Numbers. Integers are the set of all Non–Fractional Numbers lying between –∞ and ∞ i.e., –∞... –4, –3, –2, –1, 0, 1, 2, ∞. Natural Numbers are the set of all Non–Fractional Numbers from 1 to infinity i.e. 1, 2, 3, … ∞. Whole Numbers are the set of all Natural Numbers and Zero 0, 1, 2, … ∞. The Imaginary Numbers are generally written in the form of “a + bi”, where a and b are Real Numbers and „i‟ is the Imaginary Number and its value is –1. Operation on Complex Numbers is similar to that on Real Numbers. Let a + bi and c + di be two Complex Numbers. Then the various arithmetic operations are: Addition : (a + bi) + (c + di) = [(a + c) + (b + d) i] Example : (3 + 4i) + (4 – 3i) = [(4 + 3) + (4 – 3) i] = 7 + i Subtraction : (a + bi) – (c + di) = [(a – c) + (b – d) i] Example : (3 + 4i) – (4 – 3i) = [(3 – 4) + (4 – 3) i] = –1 + 7i Multiplication : (a + bi) * (c + di) = ac + adi + cbi+ bdi2 = (ac – bd) + (ad + bc) i Example : (3 + 4i) (4 – 3i) = (12 – (–12)) + (–9+16) i = 24 + 7i The Numbers that have two factors “Only and Exactly” – the number itself and 1 are all Prime Numbers. Natural Numbers greater than 1, which are not Prime, are known as Composite Numbers. A Common Factor of two or more numbers is a number which divides each of them exactly. For example, 4 is a Common Factor of 8 and 12. H.C.F. (Highest Common Factor) is the greatest number that divides each one of them exactly. For example, 6 is the Highest Common Factor of 12, 18 and 24. It is also called Greatest Common Divisor (G.C.D.) The Least Common Multiple (L.C.M.) of two or more given numbers is the least or lowest number which is exactly divisible by each of them. For example, consider the two numbers 12 and 18. Multiples of 12 are 12, 24, 36, 48, 72 … and Multiples of 18 are 18, 36, 54, 72, …. Common Multiples are 36 and 72. Therefore, Least Common Multiple of 12 and 18 is 36. Arithmetic Progression (A.P.) is a sequence of numbers in which the difference between any two successive numbers is always constant. If the Ratio of any two successive terms is equal throughout the sequence, then the sequence is said to be in a Geometric Progression (G.P.) Pondicherry Engineering College 1 Training & Placement Tips: 1. To square any number ending with 5, let the number be „A5‟. (A5)2 = A (A + 1)25. Ex: 252 = (2 *(2 + 1)) 25 = 625. (A = 2, join 25 instead of multiplying). 2. A number is divisible by 3, if the Sum of the digits in the number is divisible by3. Ex: 3792 is divisible by 3 since 3+7+9+2 = 21, which is divisible by 3. 3. A number is divisible by 4, if the number formed by the last two digits is divisible by 4 or both the last two digits are zero. Ex: The number 2616 is divisible by 4 since “16”, the number formed by the last two digits are divisible by 4. 4. A number is divisible by 6, if the number is even and Sum of its digits is divisible by 3. Ex: The number 4518 is divisible by 6 since it is even and Sum of its digits 4+5+1+8 = 18, is divisible by 3. 5. A number is divisible by 8, if the number formed by the last 3 digits is divisible by 8. Ex: The number 41784 is divisible by 8 as the number formed by the last 3 digits is Divisible by 8. 6. A number is divisible by 9 if the Sum of its digits is divisible by 9. Ex: The number 819 is divisible by 9 as the sum of the digits, 8+1+9 = 18, is divisible by 9 7. A number is divisible by 12 if it is divisible by both 3 and 4. Ex: The number 24 is divisible by 3 and 4 is also divisible by 12 8. In Comparison of Fractions, if the Denominators of the Fractions are same, the largest fraction is the one whose Numerator is the largest. Ex: 7/8 > 5/8 > 3/8. 9. If the Numerators of the Fractions are same, the largest fraction is the one whose Denominator is the smallest. Ex: 5/2 > 5/7 > 5/9. 10. If a Composite number C, which can be expressed as C = am * bn * cp….where a, b, c…are all Prime Factors and m, n, p are Positive Integers, the number of Factors is equal to (m + 1) (n + 1) (p + 1) … Pondicherry Engineering College 2 M * H. and H.M. of Fractions = (L.C.2. of Fractions = (H. of Numerators) / (H.M. When the same number is divided by 18 what would be the Remainder? Solution: Dividend = (Divisor * Quotient) + Remainder = 342Q + 47 = 19 * 18Q + 18 * 2 + 11 = 18(19Q + 2) + 11 Remainder is 11.C.C.C. to find the nth term. 3. to find the nth term Tn = a rn–1.C. Sn = n / 2 [2a + (n – 1) d] or Sn = n* (a + l)/2 4.F.1. Tn = a + (n – 1) d.C. To find the Sum of n numbers.F. where a is the first term. the number of Factors = (6 + 1)* (2 + 1) = 21. Pondicherry Engineering College 3 Hence n = 150 .M.M.F.C. of Numerators) / (L.C. = Product of the numbers L. In A. To find the L.F. Formulae: 1. 11. Problem 1: 5. Dividend = (Quotient * Divisor) + Remainder 2.C. 114…… 996 Tn = 996.C. Sum Sn = How many three digit numbers are divisible by 6 in all? Solution: Formula: Tn = a + (n – 1) d Three digit number divisible by 6 range from 102. In G.M. d = 6 996 = 102 + (n – 1) 6 Problem 2: A number when divided by 342 gives a Remainder 47. and d is the difference. of Denominators) 1. L. a = 102.F.C.Training & Placement Ex: The total numbers of Factors of 576 can be found by writing 576 = (24 * 24) = (23 * 3) * (23 * 3) = (26 * 32) Therefore.C. = Product of the numbers H.P. of Denominators) H. 108.P. of Fractions: L.F. a(rn – 1) if r >1 –––––––– r–1 6. 1. 1. 8. Two numbers are such that the Ratio between them is 4:7. find the second number. How many numbers between 11 and 90 are divisible by 7? 3. Concept: Ratio is a comparison of two quantities. If 9 is added to the number. Find the numbers. the digits are reversed. Find the numbers. the Remainder is 454. Find the Sum of the numbers. The Sum of the Squares of two numbers is 3341 and the difference of their Squares is 891. Types of Ratios: Duplicate Ratio: The Ratio of the Squares of two numbers is called the Duplicate Ratio of those two numbers. 1. What is the Maximum number of Slices you can obtain by cutting a cake with only 4 cuts? 6.2. What is the 1. the Sum of the digits is 15. The Product of two numbers is 120 and the Sum of their Squares is 289. If the Ratio between first and second is 2:3 and that between second and third is 5:3. In a two-digit number. the Ratio becomes 3:5. Solution: The Greatest 5-digit number is 99999. 7. The Sum of the Squares of three Consecutive Odd numbers is 251. Ex: 32 / 42. 5. 9. The Sum of three numbers is 136. Pondicherry Engineering College 4 . Problems to Solve: 1. Dividing this number by 463.3. the Fraction becomes 4/5.1. If each is increased by 4. Remainder when the same number is divided by 29? 2. Ratio and Proportion 1. The first term is called Antecedent and the second term is called Consequent. The two quantities that are being compared are called terms.2. 4. A certain number when divided by 899 gives a Remainder 63. The Denominator of a Fraction is 3 more than the Numerator. So. It is a relationship between two quantities with respect to magnitude. 10.2.2.Training & Placement Problem 3: Find the Greatest 5-digit number which is exactly divisible by 463. the Ratio of a to b is a/b and is denoted by a : b. Find the Largest number. If 4 is added to both the Numerator and the Denominator. If a and b are two numbers. the required number = 99999 – 454 = 99545. Find the Fraction. Find the number. If a : b = n1 : d1 and b : c = n2 : d2. then their Incomes are given by. b. two liquids A and B are in the Ratio of a : b. If in a Mixture of „x‟ litres. (aS (d – c)) / (ad – bc) and Rs. Ex: 3 / 5 = 4 / 10. Ex: 33 / 43 Sub–Duplicate Ratio: The Ratio of the Square roots of two numbers is called the Sub– Duplicate Ratio of two numbers. c. Compound Ratio: The Ratio of the Product of the antecedents to that of the consequents of two or more given Ratios is called the Compound Ratio.Training & Placement Triplicate Ratio: The Ratio of the Cubes of two numbers is called the Triplicate Ratio of that numbers. 2. If the Ratio of two numbers is a : b. b. x is called the Fourth Proportional of a. (bS (d – c)) / (ad – bc) 6. Inverse Ratio: If the antecedent and consequent of a Ratio interchange their places. If the Saving of each Person be Rs. a.e. If four quantities are in Proportion. 4 / 5 and 5 / 7 are the given Ratios. S. 12 / 35 is the compound Ratio. Ex: If 3 / 5. If the Incomes of two Persons are in the Ratio of a : b and their Expenditures are in the Ratio of c : d. 5. If two numbers are in the Ratio of a : b and the Sum of these numbers is x. then (3*4*5) / (5*5*7) i. The Equality of two Ratios is called Proportion. then the Quantities of Liquids A and B in the Mixture will be ax / (a + b) litres and bx / (a + b) respectively. If a : b :: b : x. Ex: 3 / 5 is the Inverse Ratio of 5 / 3. 4. then the number that should be subtracted from each of the numbers in order to make this Ratio c : d is given by (bc – ad) / (c – d). b. d are known as Extremes and b. c are known as Means. then each term of the Ratio a / b and c / d is called a Proportional. If a / b = c / d. x is called the Third Proportional of a. then a : b : c = (n1*n2) : (d1*n2) : (d1 : d2) Pondicherry Engineering College 5 . 3. Rs. then Product of Means = Product of Extremes If a : x :: x : b. Tips: 1. If a : b :: c : x. x is called the Mean or Second Proportional of a. Ex: 3/4 is the Sub–Duplicate Ratio of 9 / 16. the new Ratio is called the Inverse Ratio of the first. If the Ratio of two numbers is a : b. then these numbers will be ax / (a + b) and bx / (a + b) respectively. then the numbers that should be added to each of the numbers in order to make this Ratio c : d is given by (ad – bc) / (c – d). we write 3 : 5 : : 4 : 10. Here. b = 5. the Ratio becomes 7 : 5. If A : B = 3 : 2. 500 at the end of the Year. The required number = (bc – ad) / (c – d) = [(6*2) – (5*3)] / (2 – 3) = 3. B. then find their Annual Income. Annual Income of A and B is in the Ratio of 5 : 4 and their Annual Expenses bear a Ratio of 4 : 3. Find the number that must be subtracted from the Ratio of 5 : 6 to make it equal to 2 : 3. If 1 / 3A = 1 / 4B = 1 / 5C.If 14 is subtracted from each. Solution: Here. C and D share a Property worth Rs. The first number = (a * x) / (a + b) = (4 * 27) / (4+ 5) = 12 The second number will be = 27 – 12 = 15. Problem 2: Two numbers are in the Ratio 9 : 7. then A: B: C =? 5. How many Boys were there? 2. Find Share of B. Five Bananas and Four Apples cost as much as Three Bananas and Seven Apples. Rs. 49 was divided among 150 children. 77.3. Pondicherry Engineering College 6 . Find the Ratio of the cost of One Banana to that of One Apple. 500. 1. and x = 27. 9x – 14 = 7 7x – 14 5 5(9x – 14) = 7(7x – 14) => 45x – 70 => 98 – 70 => x Problem 3: = = = 49x – 98 49x – 45x 7 Hence the numbers are 9x = 63 and 7x = 49. A. a = 4. 3. Solution: Let the numbers be 9x and 7x.Find the two numbers.2. B: C = 5 : 4 and C : D = 3 : 7. Each Girl got 50 paise and each Boy 25 paise. Problems to Solve: 1. Find the numbers. 4. If each of them saves Rs. Solution: We have a : b = 5 : 6 and c : d = 2 : 3.Training & Placement Problem 1: Two numbers are in the Ratio of 4 : 5 and the Sum of these numbers is 27. Training & Placement 6. A Man spends Rs 500 in buying 12 items of Tables and Chairs. The cost of one table is Rs. 50 and that of one chair is Rs. 40. What is the Ratio of the numbers of Chairs and Tables purchased? 7. An Alloy contains Copper and Zinc in the Ratio 7 : 3. If the alloy contains 10.5 kg Zinc, find the Quantity of Copper in the Alloy. 8. If Income of A, B and C are in the Ratio of 9 : 3 : 7 and Income of C exceeds the Income of B by Rs. 1200 then find the Income of A. 9. Find the Age of Amit and Puneet which are in the Ratio 2 : 3. After 12 years, their Ages will be in the Ratio 11 : 15. What is the age of Puneet? 10. Two numbers A and B are such that the Sum of 5% of A and 4% of B is 2/3 rd of the Sum of 6% of A and 8% of B. Find the Ratio of A : B. 1.3. Alligation or Mixture 1. 3.1. Concept: Alligation means “Linking”. It is a Rule to find the Ratio in which two or more Ingredients at their Respective Prices should be mixed to give a Mixture at a given Price. By using this Rule, we can also find the Mean or Average Price of a Mixture when the Prices of two or more Ingredients which may be mixed together and the Proportion in which they are mixed are given. Alligation Rule: Suppose, Rs. D per Unit be the Price of first ingredient (superior quality) mixed with another ingredient (cheaper quality) of Price Rs. C per unit to form a Mixture whose Mean Price is Rs. M per unit, then the two ingredients must be mixed in the Ratio: Quantity of Cheaper = C.P. (Cost Price) of Costlier – Mean Quantity of Costlier Mean – C.P. of Cheaper The two Ingredients are to be mixed in the Inverse Ratio of the Differences of their Prices and the Mean Price. The above Rule may be represented schematically as Cost of Cheaper Cost of Costlier C D M (D–M) Ratio = (D – M): (M – C) (M–C) Pondicherry Engineering College 7 Training & Placement Problem 1: Sugar at Rs. 15 per kg is mixed with Sugar at Rs. 20 per kg in the Ratio 2 : 3. Find the Price per kg of the Mixture. Solution: c (Rs.15) d (Rs.20) m (Rs.x) d – m = 20 – x From the given data, m – c = x – 15 20 – x = 2 . x – 15 3 x = 18 Price of the Mixture = Rs. 18 per kg. Problem 2: A merchant has 100 kg of salt, part of which he sells at 7% Profit and the rest at 17% Profit. He Gains 10% on the whole. Find the Ratio. Solution: 7 17 10 (17 – 10) Ratio = (17 – 10): (10 – 7) Therefore, the Ratio is 7: 3. 1.3.2. Problems to Solve: 1. In what ratio is ghee worth of Rs. 30 per Kg and ghee worth of Rs. 45 per Kg be mixed so as to get the mixture worth of Rs. 40 per Kg? 2. How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a Gain of 10% by selling the Mixture at Rs. 9.24 per kg? 3. In what Ratio two varieties of Tea, one costing Rs. 27 per Kg and the other costing Rs. 32 per Kg should be blended to produce a blended variety of Tea worth Rs. 30 per Kg? How much should be the quantity of Second Variety of Tea, if the First Variety is 60 Kg? Pondicherry Engineering College 8 (10 – 7) Training & Placement 4. How much Water is to be added to 14 litres of Milk worth Rs. 5.40 a litre so that the value of the Mixture may be Rs. 4.20 a litre? 5. In what Ratio must Water be added to Spirit to Gain 10% by selling it at the cost Price? 6. Two vessels A and B contain Milk and Water mixed in the Ratio 4 : 3 and 2 : 3. In what Ratio must these Mixtures be mixed to form a new Mixture containing half Milk and half Water? 7. In an Examination out of 480 students 85% of the Girls and 70% of the Boys passed. How many Boys appeared in the Examination if the total Pass Percentage was 75%? 8. Some amount out of Rs. 7000 was lent at 6% p.a. and the remaining at 4% p.a. If the Total Simple Interest from both the Fractions in 5 Years was Rs. 1600, find the Sum lent at 6% p.a. 9. Mira‟s Expenditure and Savings are in the Ratio 3 : 2. Her Income increases by 10%. Her Expenditure also increases by 12%. By what % does her saving increase? 10. The Average Weekly Salary of all Employees (Supervisors and Laborers) is Rs. 100. The Average Weekly Salary of all the supervisors is Rs.600 while the Average weekly Salary of all the laborers is Rs. 75. Find the number of supervisors in the factory if there are 840 laborers in it 1.4. Chain Rule 1.4.1. Concept: There are two Types of Variations 1) Direct Variation 2) Indirect Variation Direct Variation: Apples Cost (Rs.) 20 140 12 x If number of Apples is less, cost is also less. If the Variation of one thing, while increasing or decreasing, affects the other thing in the same way, then it is Direct Variation. 20 / 12 = 140 / x x = 140*12 20 = 84 Indirect Variation: Men 14 24 Days 8 x Pondicherry Engineering College 9 what will be the weight of 6 m of the same rod? Solution: Length (m) weight (kg) 11. In how many Days will 30 Persons working 6 Hours a day. 75 11. complete the same work? Solution: Men Days Hours 39 30 12 x 5 6 More Hours less Days (Inverse Proportion) Less Men more Days (Inverse Proportion) x 12 x x = 5 * 39 6 30 (5 *39 * 12) / (6 * 30) 13 Days. 24 14 = 8 x 24* x = 8* 14 x = 8 * 14 24 x = 14 3 Problem 1: If 11.25 m of a Uniform Iron Rod weighs 42.75 11.25 x = 6 * 42. then the other value decreases.8 kg x Problem 2: = 39 Persons can repair a road in 12 Days.75 kg. the Days required for the Completion of the work will be less.75 6 x Since it is a Direct Proportion.25 42. = = Pondicherry Engineering College 10 .Training & Placement If number of Men increase. It is called Indirect Variation. working 5 Hours a day.25 22. When one value increases. x _ = 6 42. In this. Simple Partnership is one in which the capital of each Partner is in the Business for same time. Pondicherry Engineering College 11 . After 33 Days. If 45 Men can do a Piece of Work in 20 hrs.4.5. For his work. how much will 55 Toys cost? 2. then find the wages of 18 Men for 7 Days.5. If 20 Men can build a wall of 56 meters long in 6 Days. 234. a Partner can be a working Partner or Sleeping Partner. In how many Days will one cow eat one bag of husk? 8. 4/7 of the work is completed. each man now working 9 Hours a day? 9.840 per week. A contract is to be completed in 46 Days and 117 Men were asked to work 8 Hours a day. Concept: In Partnership two or more Persons carry on a Business and Share the Profits of the Business at an agreed Proportion. 20 Men complete one–third of work in 20 Days. If 3 Men or 6 Boys can do a Piece of Work in 10 Days working 7 Hours a day. what length of a similar wall can be built by 35 Men in 3 Days? 5. Working Partner besides investing capital takes part in running the Business. In a dairy farm. If 12 Men working 9 Hours a day can reap a field in 16 Days. If the Wages of 9 Men for 14 Days is Rs. then how much will 9 Men working 6 Hours a day earn per week? 10. Partnership 1. Problems to Solve: 1.2. 2100. he is either paid some Salary or given a certain percent of Profit in addition. Persons who have entered into Partnership with one another are individually called Partners and collectively called a Firm and the name under which their Business is carried on is called the Firm Name. 40 cows eat 40 bags of husk in 40 Days. How many more Men should be employed to finish the rest of the work in 25 more Days? 7. If 5 toys cost Rs. how many Days will it take to complete a Piece of Work twice as large with 6 Men and 2 Boys working together for 8 Hours a day? 1.1. How many additional Men should be employed so that the work could be completed in time. 4. The Partnership may be Simple or Compound. If 6 Men working 8 Hours a day earn Rs. in how many Hours can 15 Men do it? 3. in how many Days will 18 Men reap the field working 8 Hours a day? 6. Compound Partnership is one in which the capitals of Partners are invested for different periods.Training & Placement 1. Sleeping Partner is one who invests the capital in the Business but does not actively participate in the conduction of Business. After Five months. Sweetha. x and B invested Rs.C2 for the periods t1 and t2 respectively. C1 and Rs.35000. P. then Profit of A / Profit of B = (C1*t1) / (C2*t2) 2. If A invested Rs. 40500. x and after 3 months B invested Rs. they earn a Profit of Rs.2380. Sweetha Sumathy if the annual Profit is Rs. 10500. Solution: Ratio Sona‟s Share Sweetha‟s Share Sumathy‟s Share Problem 2: Ramani started a Business investing Rs. then what will be the Share of Raj in the Profit? Solution: Ratio of their Share Raj‟s Share = = = = 9000*12: 8000*7 27 : 14 Rs. y then the Share is in the Ratio A : B = (x * 12) : (y * 9).5.Training & Placement 1. Tips: 1. Formulae: 1. If at the end of the Year. If the Capitals of two Partners is Rs. y then Ratio A : B = X : Y 2. and the total Profit is Rs. 45000 and Rs. 2. If the timing of their investment is in the Ratio t1: t2: t3 and their Profits are in the Ratio P1 : P2 : P3. = = = = 35000: 45000: 55000 40500 * 7 / 27 40500 * 9 / 27 40500*11 / 27 = = = = 7: 9: 11 Rs. If A invested Rs.C1 and Rs. 16500. ((C1 * P) / (C1 + C2)) and Rs. 13500. Rs. Find the respective Shares of Sona. (6970*14 / 41) Rs. C2. 9000. Rs. then the Ratio of their capitals invested is (P1 / t1) : (P2 / t2) : (P3 / t3). then Shares of the Partners in the Profits are Rs. If three Partners invested their capitals in a Business. Sumathy entered into a Partnership investing Rs. 8000. ((C2 * P) / (C1 + C2)). 6970. Pondicherry Engineering College 12 . If Capitals of two Partners are Rs. Raj joined with a capital of Rs. 55000 respectively. Problem 1: Sona. Rs.2. Gupta withdraws his capital. 2333 Rs.000 90000 : 180000 1:2 7000 * 1 / 3 7000 * 2 / 3 = = Rs. if the total Profit is Rs. 4600. 60000 respectively. find the value of x. What is Y‟s Share? 5. 8500 respectively in a Business. Pondicherry Engineering College 13 .300 more than the other in the Profit. Rahul started a Business with a capital of Rs. Rs. At the end of 8 months. 12100 which was divided in Proportion to investments made. If they receive the Profits in the Ratio of 5: 9. find how long Bansal‟s Capital was used. Find the amount of Profit each Partner A. 80. 4000. Y and Z so that X‟s Share is Rs. 20000. After 6 months. A and B enter into a Partnership with Rs.3. A and B started a Business investing Rs. The net Profit for the Year was Rs. 50000 and Rs. 20 less than Z‟s Share. B and C will get if they invest Rs. 90000 15000 *12 Rs. 7000. 20 more than Y‟s Share and Rs. 3.5. B withdrew half of his investment. 2. 8000. what would be B‟s Share in it? 4. 70000 and B leaves x months before the end of the Year. If one Partner gets Rs. After Six months. how much did Sanjay invest in the Business? 6.Training & Placement Problem 3: A and B start a Business with the amount of Rs. C joins them after 6 months contributing Rs. Rs. A withdraws Rs. 12000. Two Partners invest Rs. At the end of the Year they get the Profit of Rs. After a Year. 2000 and after 6 months again he withdraws Rs. 40000 respectively in a Business. 4667 1. 1. 120 is divided among X.15000 respectively.10000 and Rs. Sanjay joined him with Investment of some Capital. 50000 and Rs. what is the Total Profit? 7. If they Share the Profit in the Ratio of 20 : 18 : 21. What will be their Share according to the Investment? Solution: A‟s investment = = = B‟s investment = = A: B = = A‟s Share B‟s Share = = (10000 * 3) + (8000 * 6) + (4000 * 3) 30000 + 48000 + 12000 Rs. 12500 and Rs. Gupta and Bansal enter into a Partnership with their capitals in the Ratio 5 : 6. 9000 and Rs. After 3 months. If at the end of the Year each of them gets Equal Amount as Profit. Problems to Solve: 1. 500? 1. 3000 for 4 months and Y invested Rs. If their Business yields them total of Rs.Training & Placement 8. It is measured in cubic units. The Perimeter of a Geometrical Figure is the Total Length of the sides enclosing the Figure. A Trapezium is a Quadrilateral in which any two opposite sides are parallel. It has Three Dimensions namely Length. A Rectangle is a Quadrilateral with opposite sides equal and all the four angles equal to 90 o. Area and Volume 1. This is called the Axis of the Cylinder. The fixed point is known as the Centre and the fixed distance is called the Radius. A Triangle having one of its angles equal to 90 degrees is called a Right–Angled Triangle. Distance between parallel sides is called its Height. A Quadrilateral in which opposite sides are equal and parallel is called a Parallelogram. what will be the Gain of each of them? 9. 8000. Square is a Quadrilateral with all sides equal and all the Four Angles equal to 90o. The Volume of any solid figure is the amount of space enclosed within its bounding spaces.e. A Triangle in which all sides are equal is called an Equilateral Triangle. 2000 for 6 months. A and B jointly Invest Rs. Circle is the path traveled by a point which moves in such a way that its distance from a fixed point remains constant. A Solid is a figure bounded by one or more surfaces.6. How much should X be paid out of a total Profit of Rs. Cuboid is a Solid Figure which has Six Rectangular Faces. Concept: The Area of any figure is the amount of surface enclosed within its bounding lines. 6000 and was joined afterwards by B with Rs. A Right Circular Cone is a solid obtained by rotating a Right–Angled Triangle around its height. A Rhombus is a Quadrilateral in which all sides are equal. X invested Rs. A Right Circular Cylinder is a solid with circular ends of equal radius and the line joining their centers perpendicular to them.1. 1040 as Profit. A Triangle whose two sides are equal is an Isosceles Triangle. It is also called as Rectangular Parallelopiped. A is an Active Partner and hence he gets 25% of Profit separately. A began Business with Rs. X and Y started a Business. A Closed Figure bounded by four sides is called a Quadrilateral. Pondicherry Engineering College 14 . Breadth or Width and Height. A Closed Figure bounded by three sides is called a Triangle. This can be best illustrated by rolling a rectangular paper either lengthwise or breadth-wise in a round way. Area is always expressed in square units.6. The Length of the Axis is called the Height of the Cylinder. When did B join? 10. 3100 respectively in a Firm. At the end of the Year B got one–third of the total Profit. a cylinder is generated by rotating a rectangle by fixing one of its sides. 2100 and Rs. i. 6. AC2 = AB2 + AC2 1. Area of the largest Circle that can be inscribed in a Square of side „a‟ is a2. Total surface area of Cube = 6 * a2 14. a. Volume of Cone = 1/3 * ( r2h) Where h is height and r is Radius of Base. Circumference of the Circle = 2 r 9. Area of equilateral Triangle = 3 a2. 19. Area of Square = (Side) * (Side) = (Side)2 Perimeter of Square = 4*(Side) 2. b and c are sides of the Triangle. 4 3 a. the Square of the Hypotenuse is equal to the Sum of the Squares of the other two sides. 4 2 20. 2 7. 12. Volume of the Cube = (edge) 3 = a3. Area of the total surface = Area of the curved surface + Area of two circular ends. Area of a Trapezium = [½ * (a + b) * h] Where a and b are parallel sides and h is the height. 18. 8. 2 5. Area of a circle = r2 . 10. Area of a Square inscribed in a Circle of Radius r is 2r and the side of the Square is 2 r. Altitude of Equilateral Triangle = Pondicherry Engineering College 15 . Diagonal of the Cube = a* 3 15.: In a Right–angled Triangle ABC Right angled at B. The Area of Largest Triangle inscribed in a Semi Circle of Radius r is r2 . Area of Triangle = (s*(s – a)* (s – b)* (s – c)) where S = a + b + c .6. where „a‟ is side of the Triangle. Area of Rectangle = Length * Breadth Perimeter of Rectangle = 2 * (Length + Breadth) 3. Volume of cylinder = Area of the base * height 16.2. 13. Area of Triangle = ½ * Base * Height 4. where „r‟ is radius of the circle. E. Formulae: 1. Area of Rhombus = ½* d1 * d2 where d1 and d2 are diagonals. 21. Area of the curved surface = Circumference of the base * height 17.g. where „a‟ is side of the Triangle.Training & Placement Pythagoras Theorem: In a Right Angled Triangle. Area of a Parallelogram = b*h 11. 9. 448. each will be in the Ratio a : b. If the Ratio of the areas of two Squares be a : b. then the required Length of the Carpet is (l * b)/ w.F (l.8 m2 4. If all the sides of a Quadrilateral are increased (or decreased) by x%. b)). 2. If the Length and the Breadth of a Rectangle are increased by x% and y% respectively. then its breadth will have to be decreased by (100x / (100 + x)) % in order to maintain the same area of rectangle. 4. The Area of the Path is given by 2w*(l + b + 2w) sq. If all three measuring dimensions of a sphere. = = = = ½ * 3. its diagonals also increase (or decrease) by x%. then the volume of the figure will increase or decrease by (x + y + z + ((xy + yz + xyz) / 100) + xyz/100)% 10.8 m and Height 3. then the least number of Square tiles required to cover the floor = (l*b) / (H.Training & Placement Tips: 1.2 m at the rate of Rs. then the Area of Rectangle will increase by (x + y + xy / 100) %. If one of its Diagonal is 34 cm. y% and z% respectively. If the Length of a rectangle is increased by x%. cylinder or cone are increased or decreased by x%. Problem 1: What is the Cost of planting the field in the form of the Triangle whose Base is 2. 8. 3.48 m2 Rs. If the Length and Breadth of a room are l and b respectively. Pondicherry Engineering College 16 . If the Length and Breadth of a room are l and b respectively. cuboid. 5.100 / m2? Solution: Area of Triangular field Cost Problem 2: Area of a Rhombus is 850 cm2.m. find the Length of the other Diagonal. cube. its area changes by x*(2 + x/100) %. 7. A Rectangular garden l m long and b m broad is surrounded by a path w m wide. and a Carpet of Width „w‟ is used to cover the floor.C. If each of the defining dimensions or sides of any two–dimensional figure is changed by x%. 6.100 * 4.2 * 2. then the Ratio of their sides. Ratio of their Perimeters and the Ratio of their diagonals. then the number of small Spheres = Big Sphere‟s Volume / Small Sphere‟s Volume (R / r)3. If a Sphere of Radius R is melted to form smaller Spheres each of Radius r.48 Rs. find the Distance traveled by it. 4. and a Perimeter of 46 m.m. Find the Base of the Parallelogram.cm and its Height is 12 cm. The Area of the Parallelogram is 72 sq.3. 8. If the Base of an Isosceles Triangle is 10 cm and the Length of equal sides is 13 cm. How many tiles of 20 cm by 10 cm will be needed to cover the floor of a room 25 m long and 16 m wide? 3. 2. The Diameter of a Wheel is 2 cm.6. Find the Percentage increase in Surface Area. d2 Problem 3: The Radius and Height of a Cylinder are increased by 10% and 20% respectively. Percentage increase in Surface Area is = = 1. The Length and Breadth of a Rectangle are increased by 20% and 5% respectively. 7. Find the Area of an Equilateral Triangle each of whose sides measure 6 cm. find its area. What is the Diameter of a Circle? 6. Find the Area of a Trapezium having Parallel sides 65 m and 44 m and their separation being 20 m. Find the Percentage increase in its area. 9. Find the Length of its Diagonal. Pondicherry Engineering College 17 (x + y + (xy / 100)) % (10 + 20 + ((10*20) / 100)) % = 32 % = = 850 / 17 50 cm = = = ½ * d1 * d2 ½ * 34 * d2 17 * d2 . If it rolls forward covering 10 Revolutions.Training & Placement Solution: Area of a Rhombus 850 Therefore. A Rectangular Carpet has an Area of 120 sq. Solution: Let the increase in Radius x be 10% and the increase in Height y be 20%. Problems to Solve: 1. The Diameter of a Circle is 105 cm less than the Circumference. 5. If the occurrence of an event precludes or rules out the happening of all the other Events in the same experiment. In the case of Compound Events we consider the occurrence of two or more Events. dice throwing and playing cards.7. 1. It is expressed quantitatively. where P (A) is the Probability of Event A and P (B) is the Probability of Event B. P (E) = n(E) n(S) E – Event S – Sample space 2. P (A B) = P (A) + P (B) – P (A B). Probability 1. Probability is the chance of occurring of a certain Event. P (A B) is the Probability that A and B occur simultaneously. 3. Pondicherry Engineering College 18 . Find the Length of the Wire. then P (A B) is the Probability that either A or B or both the Events occur. Tips: 1. In Simple Events. The number of combinations of n objects taken r at a time (r <= n) is denoted by C (n. coin tossing. Events are generally denoted by capital letters. because 3.7.2 Formulae: 1. then P (A P(A B) = 0 for Mutually Exclusive Events. we consider the Probability of happening or non happening of single events. Events are said to be “Equally likely” if there is no reason to expect anyone in preference to other. P (E) + P (E1) = 1 2. Probability has its origin in the problems dealing with games of chance such as gambling.7. The possible outcomes of a trial are called Events. Independent Events P (A B) = P (A) *P (B) B) = P (A) + P (B).A Copper Sphere of Diameter 18 cm is drawn into a Wire of Diameter 4 mm. it is called Mutually Exclusive Event.1. r) or nCr and is denoted as nCr = n! / (r! * (n – r)!) If A and B are two Events associated with sample space S. If the Events are Mutually Exclusive. Probability of Occurrence of an Event is always positive. 1.Training & Placement 10. Two Events A and B are said to be Independent if the occurrence of one does not affect the Probability of the occurrence of the other. Concept: Probability is a concept which measures Uncertainties. Probability of Occurrence of an impossible Event is 0. 1) (5. 5.3. P(A B) = P(A) + P(B) – P (A B) = (13 / 52) + (4 / 52) – 1 / 52 = 4 / 13. Find the Probability of getting an Odd number on one Dice and a multiple of three on the other Pondicherry Engineering College 19 . 6) Hence the Probability = 5 / 36. (5. P(B) = Getting a Queen Card.7.Training & Placement 4. P(A) = 13 / 52. 1. Problem 2: A bag contains 2 Green. 4) (5. P(B) = 4 / 52. Required Probability. the Probability that the ball drawn is not black is 8C1 / 11C1 = 8 / 11. Hence. Two Dice are thrown.. the number of favorable cases are 5 viz. Solution: P(A) = Getting a Heart Card. 2) (5. Problem 1: If two Dice are thrown simultaneously. Probability of Occurrence of an Event is from zero to one. Problem 3: From a well shuffled pack of 52 Cards. P (A B) = 1 / 52. 6 Blue. and 3 Black balls. Problems to Solve: 1. what is the Probability that the first Dice shows up 5 and the second Dice does not show up 5? Solution: We want only 5 on first Dice and any number other than 5 on the second Dice. 3) (5. If a Ball is drawn at random. A ball other than a black ball (2 + 6 = 8) can be drawn in 8C1 ways. what is the Probability that it is not a Black ball? Solution: One ball can be drawn out of the 11 balls from the bag in 11C1 ways. a Card is drawn at random. Find the Probability that it is either a Heart or a Queen. One ball is drawn from each bag. Time and Work 1. 4. The Probability that a man will be alive for 25 Years and hence is 0. 10. Find the Probability that both are White. A. Five Letters are placed at random in Five Addressed Envelopes. the more number of Men involved. B and C can do a Piece of Work in T1. A Card is drawn from a pack of 52 cards.8. 9. D2 and D3 Days respectively.Training & Placement 2. A man and his wife appear for an interview for two vacancies for the same post. 1. i. T2 and T3 Days. What is the Probability that only one of them is selected? 3. B and C together) = = = = = D1 / T1 D2 / T2 D3 / T3 D1 / T1 + D2 / T2 + D3 / T3 1 (if the work is complete) Pondicherry Engineering College 20 .3 and the Probability that his wife will be alive for 25 Years and hence is 0.8. A Coin is tossed Three times. Find the Probability of drawing a number which is a Square. Concept: The Time required for the Completion and the number of Men engaged for a Project is Inversely Proportional to each other.e. then Amount of Work done by A Amount of Work done by B Amount of Work done by C Also. 5. One bag contains 4 White and 2 Black balls. 8. respectively. the lesser is the time required to finish a job. Find the chance that Head and Tail show alternatively. Find the Probability of getting a King of Club or a Queen of Heart. Find the Probability that 25 Years hence both will be alive. „A‟ will finish 1 / nth work in 1 day. Two Dice are thrown. 7. A Card is drawn from pack of 100 Cards numbered 1 to 100. Find the Probability that all the Letters are not dispatched in the Right Envelopes. Another bag contains 3 White and 5 Black balls.1. Find the Probability that the total score is a Prime Number. Find the Probability of getting a total of 7 or 11 in a simultaneous throw of two Dice.4. then at a Uniform Rate of Working. The Probability of husband‟s selection is 1/7 and the Probability of wife‟s selection is 1/ 5. If „A‟ can do a Piece of Work in n Days. 6. If they have worked for D1. Amount of Work done (A. Training & Placement Tips: 1. If A can complete a / b part of Work in X Days. Problem 1: Worker „A‟ takes 15 hrs to complete a job. Problem 2: A can do a Piece of Work in 6 Days. A‟s 1 Hour‟s work B‟s 1 Hour‟s work (A + B)‟s 1 Hour‟s work Pondicherry Engineering College 21 . 5. can complete a Piece of Work in X Days. then B. B and C while working alone. How long should it take both A and B. If A can do a Piece of Work in „n‟ Days and B can do the same Piece of Work in „m‟ Days then Work will be done in [m*n / (m + n)]Days. Two Persons A and B. If A works Thrice that of B then Ratio of Work done by A and B is 3 : 1. 3. Working alone. to do the same job? Solution: = 1 / 15 = 1 / 30 = (1 / 15) + (1 / 30) = (2 + 1) / 30 = 1 / 10 Both A and B will finish the Work in 10 Days. then „c‟ Men and „d‟ Women can do the work in (n*a*b) / [(b*c) + (a*d)]. Y and Z Days respectively. What time will they require to do it working together? Solution: Part of the work done by A in one day = 1/6 Part of the work done by B in one day = 1 / 12 Part of the work done by A and B in one day = 1 / 6 + 1 / 12 = 1/4 Time required by A and B together to finish the work = 4 Days. working together. If „a‟ Men and „b‟ Women can do a Piece of Work in „n‟ Days. 2. working alone will complete the work in X*Y / (Y – X) Days. can complete a Work in X. if both Work together. 6. If A. and B can do it in 12 Days. If A. 4. can complete the work in Y Days. then they will together complete the Work in X*Y*Z / (XY + YZ + ZX). Worker B takes 30 hrs to do the same job. then c / d part of the work will be done in (b*c*X) / (a*d) Days. working together but independently. A can do a Piece of Work in 8 Days.8. 4 Men and 6 Women can complete a work in 8 Days. while 3 Men and 7 Women can complete it in 10 Days. in how many Days can they complete the work? 2.2.Training & Placement Problem 3: 12 Men or 15 Women can do a work in 14 Days. If 1 man. If A and B Work Together. A can do a piece of work in 9 Days. N = 14 C = 7 and D = 5. A = 12. With the help of B he finishes the work in 6 Days. He works at it for 8 Days and then Y finishes it in 16 Days. In how many Days will 10 Women complete it? 6. 200. Problems to Solve: 1. How long will it take for C alone to complete the work? 7. then in how many Days can B do the same work? 10. A undertook to do it for Rs. X can do a Piece of Work in 40 Days. 1 Woman and 1 child work together. in how many Days will the work get completed? 3. If B is 50% more efficient than A. B worked for 10 Days and left the job. If all the 10 Men and 15 Women work together. With the help of a Boy they finish it in 3 Days. A can finish a work in 18 Days and B can do the same work in 15 Days. How much is the Share of the Boy? Pondicherry Engineering College 22 = = = n*a*b / [(b*c) + (a*d)] 14*12*15 / ((15*7) + (12*5)) 168 / 11 Days. 4 Women can complete the same work in 3 Days and 5 children can do it in 3 Days. Required Number of Days 1. How long will they together take to complete the work? 5. 320. What will be the Share of B? 9. A can do a Piece of Work in 30 Days while B can do it in 40 Days. 10 Men can complete a Piece of Work in 15 Days and 15 Women can complete the same work in 12 Days. B and C can finish it in 8 Days. A and B can finish a job in 12 Days while A. in how many Days will they complete the work? 8. B = 15. In how many Days. In how many Days. 7 Men and 5 Women would complete the work? Solution: Here. A alone can finish the remaining work? 4. One alone can do it in 6 Days and the other in 8 Days. Two Men undertake to do a Piece of Work for Rs. 5 Men can complete a work in 2 Days. . the part of the Tank emptied in 1 Hour is 1 / Y. the part of the Tank filled in 1 Hour is 1 / X. If both the Pipes are opened together. If an outlet can empty the full Tank in Y Hours.1. Concept: Pipes are connected to a Tank or Cistern and are used to fill or empty the Tank. What is the total time taken to fill the Tank completely? Solution: Time taken to fill the half Tank = 3 Hours. All the four taps need 45 minutes more to fill the remaining Tank.Training & Placement 1. A Cistern has a leak which can empty it in X Hours. 5. the part of the Tank filled in 1 Hour is 1 / X. 4. the part of the Tank emptied in 1Hour is 1 / Y. If an inlet can completely fill the empty Tank in X Hours.9. 2.e. The concept is similar to that of Time and work. net part of the Tank filled in 1Hour = 1 / X – 1 / Y. If an inlet can completely fill the Tank in X Hours. After half the Tank is filled three more similar taps are opened. So. If both inlet and outlet are open. A Pipe which allows Y litres of Water per Hour into the Cistern is turned on and now the Cistern is emptied in Z Hours. Problem 1: A tap can fill a Tank in 6 Hours.9. Part filled in 1 Hour = 1/6 Part filled by the four taps in 1 Hour = (4* 1 / 6) =2 / 3 Remaining part (1 – 1 / 2) =1/2 2 / 3: 1 / 2 = 1: x or x = (1 / 2* 1* 3 / 2) = 3 / 4 hrs i. The time taken by C to empty the full Tank is (X*Y*Z) / (XZ + YZ – XY) Hours. The capacity of the Cistern can be found as (X*Y*Z) / (Z – X) litres. If all the three Pipes are opened together. Pipes and Cisterns 1. If an outlet can empty the full Tank in Y Hours. 6. the Tank is full in Z Hours. net part of the Tank filled in 1 Hour = 1 / X – 1 / Y. then the time taken to fill the Cistern is (X*Y) / (X + Y) Hours. Outlet is a Pipe connected to a Tank or Cistern for emptying it. Inlet is a Pipe connected to a Tank or Cistern for filling it. Two Pipes A and B can fill a Cistern in X Hours and Y Hours. If inlet and outlet are open. respectively. Tips: 1. totally 3 Hours 45 minutes is the time taken to fill the Tank completely Pondicherry Engineering College 23 . 3. There is also an outlet C. Two Pipes A and B can fill a Cistern in X and Y Hours respectively while working alone. 2. 12 and 15 Hours respectively while working alone. 1. B and C can fill a Cistern in 6 hrs. Problems to Solve: 1. If the Cistern is full. in what time shall the leak empty it? Solution: Work done in 1 Hour by the leak and the filling Pipe Work done by the leak in 1 Hour = = = Hence the leak can empty it in 90 Hours.5) litres 162 litres 9 litres (162 / 9) = 18 2. After working together for 2 hrs. Amit opens both the taps simultaneously. Three Pipes A. Find whether the Tank will be filled up or emptied first and in how many minutes? A Tap can fill a bath in 20 minutes and another tap can fill it in 30 minutes. find the time taken to fill the Cistern. he finds that the waste Pipe was open. When the bath should have been full.9. 1/9 1 / 9 – 1/10 1 / 90 = = = = (12 x 13. Two Pipes A and B together can fill a Cistern in 24 min and 32 min respectively. How many buckets will be needed to fill the same Tank.5 litres. Find when the first Pipe must be turned off so that the Cistern may be just filled in 16 minutes. Find the time in which the Cistern can be filled by Pipe C. When must the first Pipe be turned off so that the Cistern may be filled in 10 more minutes? Three Pipes A. Pondicherry Engineering College 24 . A Tap can fill a Tank in 25 minutes and another can empty it in 50 minutes. if the capacity of each bucket is 9 litres? Solution: Capacity of Tank Capacity of each bucket Number of buckets needed Problem 3: A Cistern is filled in 9 Hours and it takes 10 Hours when there is a leak in its bottom. Both Pipes are opened. He then closes the waste Pipe and in another 4 minutes the bath is full. If all the three Pipes are opened together. What time will the waste Pipe empty it? A Cistern can be filled by two Pipes in 20 and 30 minutes respectively. C is closed and A and B fill the Cistern in 8 hrs. B and C can fill a Cistern in 10. 5. Both Pipes being open. 4. 3.Training & Placement Problem 2: 12 buckets of Water fill a Tank when the capacity of each Tank is 13. 6. How many litres does the Cistern hold? Two Pipes A and B separately fill a Tank in 6 hrs and 8 hrs respectively. in how many Hours the reservoir will be filled by A alone? 1. I – interest It can be written as R = (I x 100) / (P*N) Pondicherry Engineering College 25 . With this method. A leak in the bottom of a Tank can empty the full Tank in 8 hrs. Interest 1. If both together fill the reservoir in 6 Hours. 1. For example. How many Hours does the faster Pipe take to fill the reservoir? 8. 10. When the Tank is full. An inlet Pipe fills Water at the rate of 6 litres a minute. 9. Simple Interest on Rs.10. 100 at 5% per annum will be Rs. Both the Pipes are opened together. Interest can be of two types: 1.5 each Year. which is expressed as some percent of the principal and is called the Rate of interest for fixed period of time. A reservoir is provided by two Pipes A and B. Simple Interest (S. the Principal grows as the interest is added to it.I) 2. that is. at the end of one Year total amount will be Rs. One Pipe fills the reservoir 10 Hours faster than the other.10. which comprises of the Principal and the Interest is called the Amount. N – Period – Rate of interest.I) When the interest is payable on the principal only. This amount paid by A is called Interest. but 1½ hrs after the start the Pipe A is turned off. The total amount of money borrowed by A from B is called the Principal. 110 and so on. where the interest for each period is added to the Principal before interest is calculated for the next period. At the end of second Year it will be Rs. Formulae: Simple Interest: Amount = P+I I = (P*N*R) / 100 Where P – Principal.10. it is called Simple Interest. Compound Interest is the method. How much time will it take to fill the Tank? If two Pipes function simultaneously. then A has to pay certain amount to B for the use of this money. Concept: When a Person A borrows some money from another Person B.Training & Placement 7. 105. the inlet is opened and due to leak the Tank is empty in 12 Hours. Compound Interest (C. A can fill the reservoir 5 Hours faster than B. The money paid back to B. The Interest is usually charged according to a specified term.1. the reservoir will be filled in 12 Hours.2. Rate of Interest for the whole Sum is R = (P1R1 + P2R2) / (P1 + P2). then P = (A1T2 – A2T1) / (T2 – T1).I – S.I = P*[R / 100] 2 Example: Difference between C.1000 at an Interest Rate of 4% per annum for a period of 4 Years. then Amount = P*[1+ (R1/100)]* [1+ (R2/100)]* [1+ (R3/100)] Problem 1: Find the S.I and S.I for 2 Years C. Pondicherry Engineering College 26 . 4.I – S. If an amount P1 lent at Simple Interest rate of R1% per annum and another amount P2 at Simple Interest rate of R2% per annum.I on a Principal of Rs. A due in T Years at R% per annum is Annual payment = Rs.(100*A) / ((100*T) + (R*T*(T – 1)) / 2) 3.I = P*[(R / 100)3 + 3 (R / 100)2] Tips (Simple Interest): 1.I for 3 Years C. second and third Year respectively. The annual payment that will discharge a debt of Rs. R2. If a debt of Rs. 7. Z = (n*a) + [(R*a) / (100*b)] * [n*(n–1) / 2] 6. When the Rates of Interest are different for different Years. If a certain Sum in T Years at R% per annum amounts to Rs. the Sum will be P = ((100*A) / (100 + (R*T))) 2. If a certain Sum of money becomes n times itself in T Years at a Simple Interest. R3 per cent for first. A. then the time T1 in which it will become m times itself is given by T1 = (m – 1)* T / (n – 1) 5. Z is paid in n number of installments and if the value of each installment is Rs.Training & Placement Compound Interest: Amount A = P* (1 + R / 100) N CI = [Amount – Principal]. Tips (Compound Interest): 8. then the borrowed amount is given by. say R1. If a certain Sum of money P lent out at SI amounts to A1 in T1 Years and A2 in T2 Years.I and S. If a certain Sum of money becomes n times itself in T Years at Simple Interest. then the Rate of Interest per annum is R = 100*(n – 1) / T %. Example: Difference between C. A. If he returned Rs. R = 15% per annum.1350 Rs. Krishna received a loan at 13% per annum S. I = = = = = = P*R*T / 100 3000*15*3 / 100 Rs. A man borrowed Rs. 5200 for 2 Years at 6% per annum.I.12. She pays 5% Simple Interest. Solution: Here.15625 for 3 Years is Rs. After 4 Years he returned the principal and interest. Solution: Here.1350. T = 2 Years and R = 6% Therefore.Training & Placement Solution: The formula for S. A Problems to Solve: 2.4350.5000 at the rate of 8%? Find the rate percent. In how many Years Brinda ought to clear off the debt? 3. if the Compound Interest on Rs.I. T = 3 Years Interest. 9120. and lent the whole Sum to another Person at the rate of 15%. P + I. 6. Rs. 5.1000 to build a hut.I Problem 2: = = = = = PNR / 100 Rs.3000.3000 + Rs. what will be the Principal Amount? Brinda borrowed Rs. Problem 3: Mahesh borrowed Rs.8000 at the rate of 12% S.5200.1200 as S. can we get Rs. Pondicherry Engineering College 27 . Find the Interest and Money returned by Mahesh to Suresh.1000 4% 4 Years 1000*4*4 / 100 = Rs. 3000 from his friend Suresh at 15 per cent per annum for 3 Years. P = Rs. What will be the Gain after 7 Years? In what time.1951? Mr. Amount. 624. She lets the hut to Ramu and receives the rent of Rs. on Rs. P = Rs.I Where P is principal R is rate of interest N is time period S. 4. Simple Interest = P*R*T/100 = 5200*6*2/100 = Rs.5 per month from Ramu.I.160 Find the Simple Interest on Rs. a. If the S. 405. 3757 is to be divided between A and B such that A‟s Share at the end of 7 Years may be equal to B‟s Share at the end of 9 Years.I.I.1.a on C. and the S. for 3 Years is Rs. on a Sum of money at 5% p. The Population of a certain village increases by 5% annually. 1.11. If the Average of First Five Results is 48 and that of the last five is 53. Find the Average daily Expenditure for the Three months. 9. Its present population is 8000. Compound Interest.1500 Find the C.18000 in 2 Years was Rs. Find the Fifth Result. 11. 19 and 23 is [3 + 9 + 11 + 15 + 18 + 19 + 23] / 7 = 98 / 7 = 14. on the same Sum for the same period and at the same Rate? A man borrowed Rs. Sum of Quantities Average 2) Sum of quantities 3) Number of quantities = = Problem 1: The Average of 9 Results is 30.a. Find B‟s Share 11. Pondicherry Engineering College 28 . 15. the Average of 3. 9. If rate percent be 10% p.8000 at 12% per annum on S.15 during July. What will be the population after 3 Years? 1. 10. Find the Rate of Interest.I. 18. He lent the whole Sum at 12% p.14 during June and Rs. accrued on an amount of Rs. Concept: The Average or Mean or Arithmetic Mean of a number of Quantities of the same kind is equal to their Sum divided by the number of those Quantities. For example. Average 1. Solution: Fifth Result = = = (48 * 5 + 53 * 5 – 30 * 9) (240 + 265 – 270) 235 Problem 2: A man‟s Average daily Expenditure is Rs.10 during May. Rs.I.2. The difference between the C.Training & Placement 7. Formulae: 1) Average = Sum of Quantities Number of Quantities Average * Number of Quantities.11.11. What will be his Gain after 2 Years? 8.I.I. Rs. 4. If the Temperature on Monday is 42o. The Mean of 100 Observations was calculated as 40. Problems to Solve: 1.1200. If the Average of the passed candidates was 39 and that of the failed candidates was 15. 3. If the Average of First Five Observations is 58 and that of the last Five is 56.Training & Placement Solution: As there are 31 Days in May.1195 / 92 = Rs.13 approx. 7. The Average Salary of a Staff of 18 Officers and 32 Clerks is Rs. If each number is multiplied by 12 find the Average of new set of numbers. On an 800-mile trip.11. A Cricketer has a certain Average for 9 Innings. The Average of Ten numbers is 7.3. The number of Days The Average daily Expenditure 1. What will be his Average score after the 19th Innings? 6. If the Average Salary of the Officers is Rs. It was found later on that one of the Observations was misread as 83 instead of 53. thereby increasing his Average by 8 runs. 30 Days in June and 31 in July.1195 31 + 30 + 31 = 92 Rs. If the Average Height of the remaining Boys is 165 cm find the Average Height of the whole class (in cm). a Cricketer increases his Average score by 4. Having scored 98 runs in the 19th Innings. a car traveled half the distance at 80 miles per Hour and the other half at 100 miles per Hour. what was it on Friday? 8. 800. Pondicherry Engineering College 29 . then Find the Sixth Observation. 5. 2. Find the correct Mean. What was the Average Speed of the car? 10. The Average Temperature from Monday to Thursday is 48o and from Tuesday to Friday is 52o. The Average of 11 Observations is 60. The Average of marks obtained by 120 candidates was 35. find the number of those Candidates who passed the Examination 9. The total Expenditure = = = = = (10 * 31+14 * 30+15 * 31) rupees (310 + 420 + 465) Rs. Find his new Average. The Average Height of 30 Boys out of a class of 50 is 160 cm. find the Average Salary of the Clerks. In the Tenth Inning he scores 100 runs. then either of the two operations can be performed in (m + n) ways. Each of the different groups or selections which can be made by taking some or all of a number of things (irrespective of other) is called a Combination. Tips: If „n‟ is Even then the greatest value of nCr is nCn/2. Concept: If an operation can be performed in „m‟ different ways. Let r and n be positive integers such that 1 < r < n. which is independent of the first operation. Formulae: n 1. Each of the different arrangements which can be made by taking some or all of given number of things or objects at a time is called Permutation. following which a second operation can be performed in „n‟ different ways.12. Then. Solution: The number of Permutations is 8P6 = = = n! / (n – r)! 8! / 2! 20160 30 n = n! / (n–r)! = n! / [r! (n–r)!] = nCn = 1 . then the Greatest value of nCr is nCn+1/2 or nCn–1/2. then the two operations in succession can be performed in „n*m‟ ways. nC1 = n Cr nCo Pondicherry Engineering College . Factorial: The continued Product of first n natural numbers is called n! n! = 1*2*3*4*…… (n–1) * n 0! = 1 1. 3. Pr 2. The number of combinations of n different things taken r at a time is denoted by nCr. is denoted by the symbol nPr. Permutations and Combinations 1. taken r at a time. If an operation can be performed in „m‟ different ways and another operation.2.12.1. can be performed in „n‟ different ways.12. the number of Permutations of n different things. Problem 1: Find the number of Permutations and combinations which can be made by taking 6 items at a time from Eight given Distinct items without Repetition.Training & Placement 1. If „n‟ is Odd. If there are 6 periods in each working day of a school. and 1I. in how many ways can one arrange 5 subjects such that each subject is allowed at least one period? Pondicherry Engineering College 31 . How may different words can be formed with the letters of the word „BHARAT‟. In how many ways they are made to sit in a row if the candidates in Mathematics cannot sit next to each other? 4. the number of Permutations = 10! / 4! 2! 2! = 37800 Problems to Solve: 1 There are 15 buses running between Delhi and Mumbai. 1N. can be formed from the letters of the word INVOLUTE? 7. Hence. and if each two of them shake hands with each other. In how many different ways can the letters of the word „TRAINER‟ be arranged so that the vowels always come together? 6. A select group of 4 is to be performed from 8 men 6 women in such a way that the group must have atleast one woman. 2T‟s. In how many ways can this be done? 10. In how many ways can a man go to Mumbai and return by a different bus? 2. In how many different ways can the letters of the word „ALLAHABAD‟ be permuted? 5.Training & Placement The number of Combinations is 8C6 = = = n! / r! (n – r)! 8! / 6! 2! 28 Problem 2: How many different Permutations can be made out of the letters of the word. how many handshakes happen in the party? 3. There are three different rings to be worn in four fingers with at the most one in each finger. „ASSISTANTS‟ have taken all together? Solution: In this word. each of 3 vowels and 2 consonants. there are 10 letters composed of 4S‟s. 2A‟s. In How many of these B and H are never together? 8. How many words. In how many different ways can it be done? 9. There are 8 students appearing in an examination of which 3 have to appear in a Mathematics paper and the remaining 5 in different subjects. If there are 12 Persons in a party. The symbol % is often used for the term percent.1. If A is x% more than B. A Fraction whose Denominator is 100 is called a Percentage and the Numerator of the Fraction is called rate percent.13. Problems to Solve: 1. 2. In an Examination 75 % of the candidates passed in English. Percentage 1.13. multiply by 100 and put % sign. To convert a Fraction into a percent. It is the Abbreviation of the Latin phrase „per centum‟. Solution: Increase in Length Decrease in Breath Percentage = = = = 20% –10% X + Y + (XY / 100) % 20 – 10 – (20*10 / 100) = 8% Problem 2: The Price of a Hindi book in 1987 was Rs.c. Concept: The term Percent Means per Hundred or for every Hundred. The term percent is sometimes abbreviated as p. 100 but due to devaluation of the rupee it has risen to Rs. 2. 80. 250. Tips: 1.13. A rise of 20% in the Price of rice compels a Person to buy 4 kg less for Rs. Find the increased Price per kg. then B is more than A by [X / (100 – x) *100] % Problem 1: Find the Percentage Change in Area if the Length is increased by 20% and Breadth is decreased by 10% of a Rectangle. Example: 5% = 5 / 100 5 parts out of every hundred parts.2. Pondicherry Engineering College 32 . then B is less than A by [X / (100 + x) * 100] % 3. If A is x% less than B. What is the Percentage increase in its Price? Solution: Percentage Increase = = = [(Final Price – Initial Price) / Initial Price]*100% [(250 – 100) / 100]*100% 150 % 1.Training & Placement 1. Find the Pass Percentage. 65% in Mathematics and 27% failed in both subjects. 25% on house rent. A reduction of 12. Formulae: If the speed of the boat be x kmph and the speed of the stream be y kmph then (a) Speed of the boat down stream (a) = ( X + Y) kmph (b) Speed of the boat upstream (b) = = = = ( X – Y) kmph (Downstream speed + upstream speed) 2 ½ (a + b). If Water of the river is moving. A house owner was having his house painted. If a boat moves along the stream that is in the direction of the stream. If her saving at the end of a month is Rs. (c) Speed of the boat in still Water (d) Speed of the stream Pondicherry Engineering College 33 .14. A reduction of 20 percent in the Price of mangoes enables a man to buy 25 mangoes more for Rs. Concept: If a boat moves against the stream that is in the Opposite Direction of the stream. Boats and Streams 1. it is called Still Water. it is called a Stream. it is called Upstream. 7. 8. 4375. He was advised that he would require 25 kg of paint. If the speed of the Water in the river is zero. At an election a candidate who gets 35% of the votes is defeated by a majority of 150 votes. Find the reduced Price of the basket containing 200 mangoes. ½ (a – b). find the percentage of students who are not adults.1. what would be the cost of paint purchased.1200. Find the original Price (in Rs) of the table.2.14. 15% on entertainment and 5% on conveyance. 6. A candidate must get 33% marks to pass.5% in the Price of a dining table brought down its Price to Rs. 4. He gets 220 marks and fails by 11 marks. The petrol Prices are reduced by 10%. If 20% of the Boys and 25% of the Girls are adults. 40. 16? Swati spends 40% of her Salary on food.14. 10. 5. it is called Downstream. Allowing for 15% wastage and assuming that the paint is available in 2 kg cans.Training & Placement 3. 1. if one can costs Rs. The Boys and Girls in a college are in the Ratio 3:2. find her Salary per month. 9. What is the maximum number of marks? 1. Find by how much a user must increase the consumption of petrol so as not to decrease his expenditure on petrol. Find the total number of votes recorded. 4.14. A man can row upstream at 12 kmph and downstream at 17 kmph. 5. Problems to Solve: 1. Find the total time taken by him.Training & Placement Problem 1: A man can row 2 / 7th of a kilometer upstream in 25 minutes and return in 10minutes. Solution: Upstream speed Down stream speed Speed in still Water = = = Problem 2: A man can row upstream at 8 kmph and downstream at 10 kmph. The Speed of a boat in still Water is 9 kmph and the Speed of the Stream is 1. If a man can swim downstream at 8 kmph and upstream at 4 kmph. A man rows to a place 48 km far and comes back within 14 hours. 6. If the Speed of the Stream is 4 kmph. Solution: Speed in still Water = ½ (10 + 8) kmph = 9 kmph Speed of current = ½ (10 – 8) kmph = 1 kmph. 3. find the Speed in still Water. Find the velocity of Water.2 kmph = = = [(2 / 7) / (25 / 60)] 24 / 35 kmph [(2 / 7) / (10 / 60)] =12 / 7 kmph Pondicherry Engineering College 34 .3. 7. Find the Velocity of the Stream.5 km and comes back to the starting point. find the Speed of the Stream. A man can row a boat at 6 kmph in still Water. [(24 / 35) + (12 / 7)] / 2 [(84 / 35) x (1 / 2)] 1. A man rows upstream 12 km and downstream 28 km taking 5 hours each time. Find the Speed of the man in still Water. Find the man‟s speed in still Water and the speed of current. find the time taken to row a distance of 90 km down the Stream. A man rows 13 km upstream in 5 Hours and 28 km downstream in 5 Hours. Find speed of the Stream. He finds that he can row 4 km with the stream in the same time as 3 km against the stream.5 kmph. A man rows to a place at a distance of 10. 1. 2. 1.15. If A goes from X to Y at s1 kmph. Average Speed = 2s1*s2 / (s1 + s2) Pondicherry Engineering College 35 . a motor boat goes 10 km upstream and back again to the starting point in 55 min.Training & Placement 8. Distance 2. then the Average speed during the whole journey is given by.1.2. 9. and comes back from Y to X at s2 kmph. 3.15. find the speed of the current in m per min. 10. then the Average speed during the whole journey is given by. Find the rate of the current and speed of the man in still Water. It is obtained by dividing the distance covered by the object. Concept: The terms „Time‟ and „Distance‟ are related to the speed of a moving object.15. Time Tips: 1. Also. If a Person A covers a distance d1 km at s1 kmph and then d2 km at s2 kmph. The Speed of an object is defined as the distance covered by it in a unit time of interval. For converting kmph into m/sec multiply by 5/18 2. Time and Distance 1. For converting m/sec to kmph multiply by 18/5. A man can row 30 km upstream and 44 km downstream in 10 hr. Speed 3. Average speed = [s1*s2*(d1 + d2)] / [(s1*d2) + (s2*d1)] 5. Find the speed of the motorboat in still Water. If the difference between these two times is 10 min. In a stream running at 2 km/hr. Formulae: 1. he can row 40 km upstream and 55 km downstream in 13 Hours. A man who can swim 48 m/min in still Water swims 200 m against the current and 200 m with the current. 1. by the time it takes to cover that distance. If two Persons A and B start at the same time from two points P and Q towards each other and after crossing they take t1 and t2 Hours in reaching Q and P respectively then A‟s speed B‟s speed = t2 t1 = = = Speed * Time Distance Time Distance Speed 4. 3. and travels towards B at 60 km / hr. Vikas can cover a certain Distance in 1 hr 24 min by covering two third of the distance at 4 kmph and the rest at 5 kmph.15.m.78 = 9. If he takes T Hours in all. Let the Trains be P and Q respectively Distance by P in „x‟ hrs + Distance by Q in (x –1) hrs 60x + 75 (x –1) x They meet at 11 A. and travels towards A at 75 kmph. Another train starts from B at 9 A.78 m/sec Distance / Speed 530 / 57. Problem 3: Find the Length of the Bridge which a train 120 m long traveling at 54 kmph can cross in 30 seconds. Problem 2: The distance between two cities A and B is 330 km.17 Sec. 250m? Solution: Distance covered Speed Time = = = = = 280 + 250 530m 208*5 / 18 = 57.Training & Placement 6. If a Person goes certain distance (A to B) at a speed of s1 kmph and returns back at a speed of s2 kmph.M. Problems to Solve: 1. What is the total distance? Pondicherry Engineering College 36 54 kmph 15 m/sec 15 *30 450m Distance covered – Length of the Train 450 – 120 330m = = = 330 330 3 . A train starts from A at 8 A.M. Solution: Speed of the Train = 54 * 5 / 18 = Distance covered in 30 seconds = = Length of the Bridge = = = 1. At what time do they meet? Solution: The Trains may meet x Hours after 8 a. then the distance between A and B is = T*[s1*s2] / (s1 + s2) Problem 1: In how many seconds does a 280 m long train moving at 208 kmph cross a platform of length.M. Find the distance. A bullock cart has to cover a distance of 80 km in 10 Hours.1. Age some Years hence. 3. Find their Present Ages.16. How far I have to walk? 4. Problem 1: The Ages of two Persons differ by 10 Years. How many Hours will it take to cover a distance of 385 km? 10. I had to be at a certain place at a certain time and I found that I shall be 40 min too late. For how many minutes does the train stop per Hour? 9.Training & Placement 2. if he walks at the rate of 3 kmph. However. the speed of the train is 45 kmph and including stoppages. Present age. If a man runs at 3 m per sec. 1. There may be three situations:– 1. Simple linear equations are to be framed and the solution can be obtained. if I walk 3 km an hr and 30 min too soon. The distance traveled in both the cases being the same. How many kilometers does he run in 1 Hour 40 min? 5. the elder one was 2 times as old as the younger one. Find the length of the train 7. If it covers half the distance in 3/5th of the time. the knowledge of linear equations is essential. A man travels uphill with an Average speed of 24 kmph and comes down with an Average speed of 36 kmph. The former arrives half an Hour before the latter. Age some Years ago. What is the distance of the school from his house? 3. Problems on Ages 1. he reaches the school 10 min earlier than the schedule time. one at 4 kmph and another at 3 kmph. Two Men start to walk together to a certain distance. walking along the line in the opposite direction at the rate of 4 km an Hour and passes him in 5 sec. if I walk 4 km an hr. Excluding stoppages. If 5 Years ago. Concept: To solve these Problems. 2. what is the Average speed during the journey? 6. he reaches 10 min late.16. A train running at the rate of 40 km an Hour meets a Person. Pondicherry Engineering College 37 . A car starts with the initial speed of 40 kmph. with its speed increasing every Hour by 5 kmph. What should be its speed to cover the remaining distance in the time left? 8. it is 36 kmph. If a Boy walks from his house to school at the rate of 4 kmph. The Average age of five members of a family is 21 Years. Find the age of the grandfather 3. Seven years hence the Sum of their united ages will be 83 years. The ages of Ram and Shyam differ by 16 years.16. The Ratio of Laxmis age to the age of her mother is 3:11. Three years ago. The difference between the ages of two Persons is 10 Years. Mother‟s Present Age = (4x + 4) Years Four years hence (4x + 4 + 4) = 2(x + 4) + 6 4x + 8 = 2x + 14 x = 3. mother‟s age will be 6 Years more than twice the age of the Daughter. Find their present age? 5. 15 Years ago. Mother‟s Present Age = (4x + 4) = (4*3 + 4) = 16 Years. the elder one was twice as old as the younger. One Year ago.2. X‟s age was double of Y‟s. their Present Ages are 15 Years and 25 Years. Four Years hence. find the Ratio of their ages after 3 years? 7. Find the Present Age of the mother? Solution: Let the Daughter‟s present age be x Years. Find the present age of elder 4. Problem 2: The Present Age of a mother is 4 Years more than Four Times the age of her Daughter. a father was four times as old as his son. Find the Ratio of the present ages of Promila and her daughter 6. Mohan‟s age was thrice as that of Rams. Promila‟s age will exceed her daughter‟s age by 9 Years. If the age of the grandfather is included. Hence. Six Years hence. Promila was four times as old as her daughter. In 6 Years time his age will exceed twice his son‟s age by 9 Years. the Average is increased by 9 Years. 1. find the age of X today Pondicherry Engineering College 38 . The difference of their ages is 24 years.Training & Placement Solution: Let the age of the Younger Person be x Years. One Year ago. Problems to Solve: 1. Find the Ratio of their present ages 2. Then. Six years ago. age of the Elder Person = (x + 10) Years 2(x–5) = (x + 10 – 5) 2x – 10 = x+5 x = 15 Hence. Then. P. 4. Find her Gain percent. Problem 1: S.60 – 17. In How many years. (18. [100 / 100 – Loss%] * S.28% Pondicherry Engineering College 39 .17. C.P. 8. 100 + P% [(100 – Loss %) /100] * C. It is abbreviated as S. 100 [ 100 ]* S.60 So.P.P. – S.50 and sells it for Rs. he will be 3 times her age. Narmatha‟s father is now four times her age. Three Years ago. the Average age of a family of 5 members was 17 Years. 17. The Cost Price of an article is the Price at which an article has been purchased. The Ratio of father‟s ages to son‟s age is 3:1. 3.1.50 S.P. the Average of the family is the same today. It is abbreviated as C. Solution: C.P C.1.P = Rs.P)* 100 % [100 + P%] * C. = = A Girl buys an article for Rs.10 / 17. Formulae: 1. 1.60. The Selling Price of an article of an article is the Price at which an article has been sold. there is a Gain or Profit. What is the age of the baby? 1.P)* 100 % (L /C.50) = Rs. The Product of their ages is 243. will he be twice her age? 10. 2.P. Gain = Rs. – C.P. 6.17. Profit %( P %) = Loss % S.2.P.P (P /C.17.10 Gain% = ((1. Profit and Loss 1. A baby having been born. Profit (P) Loss (L) = = S. In 5 years.P.17.50)*100) % = 6.P = Rs. The terms that describe Profit is Cost Price and Selling Price. C. it is defined as Loss.P. If the cost Price of an article is greater than the selling Price. Concept: The aim of every Business is to earn Profit.P.P. What will the Ratio of their ages be 3 years hence? 9. If the selling Price of an article is more than the cost Price. 5. = = = 7.18. 18.Training & Placement 8. Training & Placement Problem 2: An Article is bought for Rs. 80 and sold at a Gain of 20%. What is Selling Price of the book? Solution: Gain = Rs. 5, so Gain % 1.17.3. Problems to Solve: 1. A Man bought 5 apples for Rs. 3 and sold each for Rs. 2. What did he Gain or Loss? 2. A man sold two pens at Rs.20 each. He sold one at a loss of 10% and the other at a Gain of 10%. Find his Goss or Gain %. 3. If Cost Price of 8 articles is equal to the selling Price of 10 articles. Find the Gain or Loss. 4. Vikas bought paper sheets for Rs. 7200 and spent Rs. 200 on transport. Paying Rs. 600, he made 330 boxes, which he sold at Rs. 28 each. Find his Profit percentage. 5. A sells a bicycle to B at a Profit of 20% and B sells it to C at a Profit of 25%. If C pays Rs.1500, What did A pay for it? 6. The profit earned after selling an article for Rs.625 is the same as loss incurred after selling the article for Rs. 435. What is the Loss percent? 7. A machine is sold at a Profit of 10%. Had it been sold for Rs. 80 less, there would have been a loss of 10%. Find the C.P of the machine. 8. A cloth merchant says that due to slump in the market, he sells the cloth at 10% loss but he uses a false metre scale and actually Gains 15%. Find the actual length of the scale. 9. If a Person makes a Profit of 10% on 1/4th of the quantity sold and a loss of 20% on the rest, then what is his Average percent Profit or Loss? 10. When a producer allows 36% commission on retail Price of his Product, he earns a Profit of 8.8%. What would be his Profit percent, if commission is reduced to 24%? = (5/20) * 100 =25% 1.18. Calendar 1.18.1. Concept: The Process of Finding the day of a given date depends upon the number of Odd Days present. The Days more than the complete number of weeks in a given period is called Odd Days. The Year (except century) which is divisible by 4 is called a Leap Year, where as century is a leap Year by itself when it is divisible by 400. Leap Year has 366 Days, i.e. 52 Weeks and 2 Odd Days. An Ordinary Year has 365 Days, i.e. 52 weeks and 1 Odd day. Pondicherry Engineering College 40 Training & Placement Tips: 1. A Century has 76 Ordinary Years and 24 Leap Years i.e. 76 + 48 Odd Days = 124 = 17 weeks and 5 Odd Days. 200 Years contain 3 Odd Days. 400 Years contain 1 Odd day. Since the order is Cyclic (i.e. every Century has 0, 1, 3 or 5 Odd Days) The First Day of a Century must either be Monday, Tuesday, Thursday or Saturday. Last Day of a Century cannot either be Tuesday, Thursday or Saturday. To find the Day of the Week on a particular date when reference date given: Find the net number of Odd Days for the period between the reference date and the given date. To find the Day of the Week on a Particular Date when no Reference Day is given: Count the net number of Odd Days on the given date and write the day corresponding to the Odd day. 2. 3. 4. 5. 6. 7. 8. Odd Days: Sunday Monday Tuesday Wednesday Thursday Friday Saturday =0 =1 =2 =3 =4 =5 =6 Month code: Ordinary Year has the Odd Days as follows January = 0 Odd Days February March =3 April May =1 June June =6 August September =5 October November =3 December =3 =6 =4 =2 =0 =5 Month code for leap Year adds 1 with ordinary Year’s corresponding month code. Problem 1: 11th January 1997 was a Sunday. What day of the Week was on 7th January 2000? Pondicherry Engineering College 41 Training & Placement Solution: Total number of Days between 11th January 1997 and 7th January 2000 = (365 – 11 Days) in 1997 + (2*365 Days) in 1998 and 1999 + 7 Days in 2000. = (50weeks + 4 Odd Days) + 2*(52 weeks and 1 Odd day) + 0 Odd day = 6 Odd Days Hence, 7th January 2000 would be 6 Days ahead of Sunday, i.e. it was on Sunday. Problem 2: What Day of the Week was 5th June 1999? Solution: (A+B+C+D)-2, Take the Remainder 7 A = Divide the year by 7 and take the Remainder 1985/7, Remainder is 4 B = Divide the year by 4. Take the Quotient and divide it by 7. Take the Remainder. 1985/4 = 496; 496/7, Remainder is 6 C = Divide the Date by 7 and Take the Remainder. 2/7, Remainder is 2 D = Month Code June Code is 4 (4+6+2+4)-2 = (14-2)/7 = 5 7 th Hence, 5 June 1999 was Saturday. 1.18.2. Problems to Solve: 1. 2 nd July 1985 was Wednesday. What Day of the Week was 2nd July 1984? 2. India got Independence on 15th August 1947. What was the Day of the Week? 3. The First Republic Day of India was celebrated on 26th January 1950 What was the Day of the Week on that Date? 4. Smt. Indira Gandhi died on 31st October 1984.What was the Day of the Week? 5. What was the Day of the Week on 26th June, 2002? Pondicherry Engineering College 42 1. Clocks 1. The Two Hands of the Clock will be at Right Angles between H and (H + 1) o‟ Clock at (5H 15)12 / 11 Minutes past H o‟ Clock.1. Problem 2: How many times do the Hands of a Clock Coincide in a Day? Solution: The Hands of a Clock Coincide 11 times in every 12 Hours The Hands Coincide 22 times in a Day. Solution: Angle between the two hands at 7: 20 = 30h – (11m / 2) = 30x7 – (11*20 / 2) = 100 Degrees. The Hour hand (or short hand) indicates time in Hours and the Minute Hand (or the long hand) indicates time in Minutes. where h denotes Hour and m denotes Minutes.2. 3. 4. Formulae: To find the Angle between the hands = 30 h – 11m or 11m – 30 h. The Clock has two hands – the Hour Hand and the Minute Hand. The Minute Hand moves 12 times as fast as the Hour Hand. Problem 1: Find the Angle between the Minute Hand and Hour Hand of a Clock at 07:20.19.19.19. In every Hour the hands are twice at Right angles.Training & Placement 1. In every Hour both the hands coincide once. If the Minute Hand of a Clock overtakes the Hour Hand at intervals of M minutes of correct time. the Hour hand covers 5 minute spaces while the minute hand covers 60 minute spaces. 2 2 Tips: 1. 2. Pondicherry Engineering College 43 . Thus in One Hour or 60 minutes. The Clock gains or loses in a Day by [(720 / 11) – M]*(60*24 / M) minutes. The Two Hands of the Clock will be together between H and (H + 1) o „clock at (60H / 11) minutes past H o‟clock. the minute hand gains 55 minute spaces over the Hour hand. In an Hour. Concept: The Circumference of a Dial of a Clock is divided into 60 Equal Parts called Minute Spaces. Training & Placement 1. 2.1 Concept: Deduction Means reducing two statements into one or three statements into one. Read the statements very thoroughly and repeatedly to clearly understand the meaning (known or unknown) which they carry.30 and 6 o‟clock will the hands of a Watch be 90 ? How much does a Watch Lose per day. 2. Deduction 2. Tips for Solving Analytical Reasoning questions: 1.1. 6. the approach for solving analytical reasoning questions varies from question to question as analytical reasoning questions are of different types. Problems to Solve: 1.The other approach is based on the specific type of the questions. There are two ways of solving analytical reasoning questions. Find the Angle between the Hour Hand and the Minute Hand of a Clock when the time is 3. This method consists of a standard approach which must be followed in all the analytical reasoning questions (based on the need). Syllogism is a noun which means a form of reasoning in which a conclusion is drawn from two statements. 3. 5. Analytical Reasoning Concept: As it is very much evident from the word „analytical‟ itself that this type of reasoning is based on the analysis of the statements which are there in the question. preferably. 2. In more clear terms Syllogism is a mediate deductive inference in which two propositions are given in such an order that they jointly or collectively imply the third. Maps.1. i. Use notations. 4. Deductive Reasoning. Do not jump from one statement to the other before completely understanding the statements. if its hands coincide every 64 Minutes? 2.3.19. The first way is the one which consists of the general method for solving any type of analytical reasoning questions. 3.e. It is also known as Syllogism. or Tree.. Thus Syllogism can be defined Pondicherry Engineering College 44 . symbols and abbreviations wherever required. 5. Do not make unnecessary assumptions. Mark the keywords which are present in the statements.25? At what Time between 4 and 5 o‟clock will the Hands of a Watch Point in Opposite Directions? At what Time between 9 and 10 o‟clock will the hands of a Watch be together? At what time between 5. 4. Organize the information given in the question in the form of suitable Tables. e. No –––– Universal Statement Some. The child is a human being. Types: The types of deduction are: Deduction in two statements and Deduction in three statements.Training & Placement as „a form of reasoning in which the conclusion establishes a relation between two terms on the basis of both terms being related to the same third term as derived in the premises. √ –DISTRIBUTED Subject Universal Affirmative Universal Negative Particular Affirmative Particular Negative Rules for Deductions:  The middle term must be distributed at least once. Universal Negative: If the statement starts with „NO‟ it is Universal Negative. Pondicherry Engineering College 45 √ √ X X X –UNDISTRIBUTED Predicate X √ X √ . 1. If the statement starts with „ALL‟ and if the statement has the word „Not‟ it is Universal Negative. with both subject (children) and the predicate (mortal). All. The child is mortal.‟ For example. Particular Affirmative: If the statement starts with „SOME‟ or „MANY‟ it is Particular Affirmative. 3. Therefore. Many–––– Particular Statement No.  Two statements should have only Three Distinct terms. If the statement starts with.. „human being‟. in a Syllogism two premises are necessary to arrive at a conclusion. Not–––– Negative Statement Universal Affirmative: If the statement starts with „ALL‟ or „NO‟ it is Universal Affirmative. Particular Negative: If the statement starts with „SOME‟ or „MANY‟ and if the statement has the word „Not‟ it is Particular Negative. All human beings are mortal. i. The conclusion is reached through the medium of a middle term. 2. One Conclusion. Two Conclusions. All pots are jacks Solution: Statement 1: Universal affirmative Subject (pens) – Distributed Predicate (pots) – Undistributed Statement 2: Universal affirmative Subject (pots) – Distributed Predicate (jacks) – Undistributed Conclusion: “All pens are jacks” 2.2 Problems to Solve 1. Two Conclusions. One Conclusion. – – – – Some Many No Conclu. the Conclusion would be Particular If one Statement is Negative.1. All All All No No No Conclu. All Some Some All No No Some No Some. All dogs are Parrots No train is dog Some buses are trains Pondicherry Engineering College 46 . the Conclusion would be Negative. If one Statement is Particular. 1 Stat.Training & Placement      The Middle term should not come in the conclusion If both the statements are „Particular‟ there would be No Conclusion If both the statements are „Negative‟ there would be No Conclusion. All bats are keys All locks are keys No keys are doors 2. 2 Conclusion Universal Affirmative Universal Negative Particular Affirmative Particular Negative Problem 1: All pens are pots. All horses are fruits Some boxes are trees Some trees are horses 3. Not Stat. 2 Linear Sequencing 2. The direct condition should be found first and then the question can be solved. arranging it in single row manner. __ __ __ __ C A is to the immediate left of C. All sports are enthusiastic Cricket is not a sport 9. i. and E) in a room who are seated in a single row in the following manner: A does not sit immediately next to E. __ __ __ A C E does not sit next to A.2.e. C is at the extreme right. A is immediately to the left of C. Shyam is bachelor All bachelors are intelligent 8. E and D sit immediately next to each other. There may be 4–5 questions asked from a single linear pattern given. because the remaining conditions are based on this direct condition.e. All poles are Guns Some Boats are not Poles 2. Concept: Arranging the items in a linear fashion is called Linear Sequencing. C. i.e.1. __ E __ A C or E __ __ A C D sits next to E.e. Some fools are professionals All engineers are fools 6. E takes 2 seats: i. Problem 1: There are five Persons (A. Who must sit in the middle seat of the row? Solution: First draw linear arrangement as given below. either DEBAC or BEDAC E takes 1 seat: EDBAC Hence Middle Seat: B or D ––––– Pondicherry Engineering College 47 . i. D. C is at the extreme right end of the row.Training & Placement 4. B.e. Some guns are Pistols All Pistols are Bombs All Crackers are Pistols 5. No magazines are caps All caps are cameras 10. Some dogs are cats None of the cats are cows 7. So occupy 1 or 2 seat i. story books. I. who is at one end. Fiat. CHDE b. I. lying on the table one above the other. who is second to the right of C. B. Which of the following groups of friends are sitting to the right of G? a. Fiat. Mercedes. which was third to the left of Ambassador Car. ICHD 2. was at one end. there are novels. Who is sitting in the middle of the row? II. Problems to Solve: 1. IBJA d. Which of the following was the correct position for the Mercedes? a) Immediate right of Cardilac b) Immediate left of Bedford c) Between Bedford and Fargo d) Fourth to the right of Maruthi II. Which cars are on the either side of the Cardilac cars? a) Ambassador and Maruthi b) Maruthi and Fiat c) Fiat and Mercedes d) none of these 3. I.2. Education and Accountancy. Sociology is on top of all the books. facing the east such that: Cadillac car was to the immediate right of Fargo. J is the immediate neighbor of A and B is third to the left of G. If Sociology and English. In a Car Exhibition. Eleven Students A. C.2. every story–book has a comic next to it and there Is no story–book next to a novel. CHDF c. H is to the immediate left of D and third to the right of I. A is second to the right of E. D is to the immediate left of F. Hindi.Training & Placement 2. J and K are sitting in the first row of the class facing the teacher. Fargo was fourth to the right of Fiat. which book will be between Psychology and Sociology? a) Accountancy b) Psychology c) Hindi d) Economics 4. Every novel has a drama next to it. Ambassador. Accountancy and Hindi and Education and Psychology interchange their positions. F. English. E. what would be the order of the books in the pile? Pondicherry Engineering College 48 . Economics is immediately above Psychology but not in the middle. Bedford and Fargo were displayed in a row. There are seven books one each on Psychology. If there be a novel at the top and the number of Books is 40. Economics. seven cars of seven different companies such as Cadillac. Maruthi car was between Ambassador and Bedford. Sociology. G. I. dramas and comics. Economics is in–between which of the following books? a) Accountancy and Education c) English and Psychology b) Psychology and Hindi d) Psychology and Sociology II. D. H. Hindi is immediately below psychology. Maruthi. In a pile of reading material. Accountancy is immediately below Education which is immediately below Sociology. A sits in between E and D. M is to the right of L. F. D sits in one extreme of the row. H2. 5 people A. B. N. B sits in the fourth seat from the left end and D sits in the third seat from right end. A yellow bus is standing to the right of green. H. C. H5 in a row H1 is to the right of H2 & H5 is to the left of H3 and right of H1. If the seven Persons sit facing forward. and E read a novel. P. L is to the right of N. N. H3. R & S are 7 Persons who are sitting in a row. M is to the immediate left of N and there are exactly three Persons between M and S. E. There are 5 houses – H1. There are 6 Persons (A. One of the 2 buses at the extreme end is red and the other blue. One who finishes first gives it to C & one who reads last had taken it from A. The seats are numbered from 1–9 from left to right as per the following restrictions. and R is to the immediate right of P. Orange bus stands at ________ Position from right. C. There are two places in between B and E.Training & Placement a) NSCD b) NDSC c) CSDN d) DNCS 5. I are seated in a row. An orange bus is to the left of blue bus & Green bus stands between the red and yellow buses. If G sits to the immediate left of D then who will sit in the centre seat of the row? If F sits between C and I. O sits to the immediate left of S. Who is sitting second from your right side? Pondicherry Engineering College 49 . and P are sitting in a line. H2 is to the right of H4. then with whom should he exchange the seat? How many Persons are there between R and Q? 10. Nine Persons A. Q. O. M is to the left of P. E has not read it first or lastly two people read between B & A. E & F) in a Team who are seated in a linear manner. E is to the immediate right of H. C. N is to the right of O. H4. then in how many ways altogether can all of them be seated in the row? Which of the following is at one end of the row? 11. Which bus stands at the centre? Yellow bus stands to the left of _____? 7. B. B. Five Persons L. M. C. E is not between B and D. Which is to the house in the middle? 6. D. M. The buses stand in a row. F. viewing from front. O. I sit together. N has equal number of people on either side of him. then who sits to the immediate right of Q? If Q wants to sit to the immediate left of N. Whom did B pass the novel to? Who read the novel last? 8. D. G. How many arrangements are possible? If C takes the 2nd spot and E shifts the place to B How many people are sitting in between E and D? 9. D. A.2. R. W are 8 employees allocated with eight different lockers numbered 1–8. Problem 1: P. We have to form a table. 2.Training & Placement Identify the Persons sitting on the extremities. V. C and D are law reports while the rest are gazetteers Which is the red colored new law report volume? Which two old gazetteers have blue covers? Pondicherry Engineering College 50 . P has been given locker 1 while v has been allocated locker 8. Fitting the data in the table results in Left 1 4 6 7 Right 2 3 5 8 P S/U S/U W 1) 2) 3) 4) T Q R V Employee Q is allocated locker 3 There are 2 arrangements possible W would take the top row and P would take the bottom row If S is given locker 4 there is only 1 solution possible. Problems to Solve 1) On the shelf are placed six volumes side by side labeled A. Which locker is allocated to employee Q? 3. D and F are new volumes while the rest are old volumes. 2. 3 volumes C. B. The lockers are arranged in four rows with two lockers in each row.3. D. Draw the table? 2. There will be more than one question per passage. If W and P are interchanged what does the table look like? 5. Double Lineup 2. If S is given locker 4 how many solutions are possible? Solution: 4 Rows and 4 Columns are given.3. How many arrangements are possible? 4. T has locker just above that of Q which is just above that of R whereas W locker is in bottom row. T. Lockers 1 and 2 in top row from left to right and lockers 7 and 8 in bottom row from left to right.3. B and E have blue covers while the rest have red covers. U. Concept: The question will consist of at least two different variables. C. E and F. Lockers 3 and 4 are in the second row from top arranged from right to left and so do 5 and 6.1. 1. S. Q. C. Volleyball. Green. White. Lakhan and Kishan are a doctor. B. Pondicherry Engineering College 51 .Q. Who plays Volley ball? Which colored car F owns? 5) 6 friends Ramesh. Company T does not Sponsor Purple or Yellow colored tie and Lokesh works in Company P. E has white car and plays Baseball. B. C. No postman works in two post offices. S. B. A works neither in post office P nor in S. Which Colour Tie is sponsored by company R? Which of the following Colour Company Person combination is correct? a) Green – R – Nilesh b) Blue – S – Lokesh c) Red – T – Dinesh d) Yellow – R – Shaliesh Which of the following is true? a) Company U sponsored Green tie c) Nilesh works in T b) Shailesh wears Red tie d) Red Colour is sponsored by T 6) Among Ram. T & U and each one wears company sponsored different coloured tie such as Blue. C & D) sit and have dinner in a room. Shaliesh & Hitesh work in different companies namely P. Blue and Red. Hitesh does not work in company R.R. Ram and Gita are having a love affair secretly. E & F. A plays Carrom and he has black car. Each of their wives is working. TT and Polo. Carrom. Baseball. C has Z. Purple & Red though not necessarily in the same order. If A has Y. Nilesh does not work in Company T and purple colour tie is not sponsored by Company R.T and U are 6 post offices. They are married to Radha. Shaliesh works in company U and neither Nilesh nor Dinesh works in company S. D. X. Black. Pink. Every one is good in one of the following Games. Each owns a different colored car Yellow. E and F workers. What are the different arrangements possible? 4) There were 6 friends A. The various dishes provided are (W. Sita and Gita not in that order. C does not play TT & Hockey and owns neither blue nor yellow car. D. Nilesh. Radha is married to the Engineer. Q. The one wearing Blue tie works in Company S and the one wearing Green tie works in Company P. Green. C works neither in post office Q nor in R but D works in T and A works in Q. If B works in post office P and C does not work in post office U then E works in (a) S (b) U (c) R (d) R or U 3) 4 people (A. Sita and Gita saw the film Ram Lakhan. Dinesh. Yellow.Training & Placement 2) P. In each of which one is a post man out of A. R. B cannot have W. while B works neither in post office T nor in U. Teacher‟s wife teaches in St. Bangalore. The day Ram and Lakhan saw the film “Sita aur Gita”. B does not play TT and has red car.S. Kishans wife is an artist and regularly holds exhibitions at Venkatappa art gallery. Lokesh. Hockey. Ramesh wears Pink tie and works in Company Q. Y & Z). Each of them has one dish each as follows. Joseph‟s Convent. D plays Polo and has yellow car. a teacher and an engineer. A can speak Spanish and Hindi Mr. I like the following 5 Places in India A. Pondicherry Engineering College 52 . E is a peaceful city.5 and 6. E. B e) any of the other three executives.C converses in English and Hindi. B understands Spanish and English Mr.3. There are six display windows number 1. B and E have Seashores. Mr. One Product is to be put in one window. B. Sunil. Who is the student of commerce? Who does not play Football? Who plays cricket? Who studies Maths and plays Cricket? 9. One of them studies Commerce and plays Golf and Table tennis. A or Mr. A b) Only Mr.5? There are four friends Dinesh. can also speak French. B. W must be placed in which window? If U is placed in window no. Dinesh and Sunil study Maths.Training & Placement Sita is a good cook and one can find her recipes in every issue of Women‟s Era magazine. Which of the following can act as an interpreter when Mr. All the friends play two games each and study one subject each. a native Indian. W must be immediately to the left of X. Mr. D & A are metropolitan. Y and Z – are display windows of a shop. Rahul is a student of Physics. Six Products – U. C and Mr. If X is placed in window no. D. B c) Only Mr. D wish to converse ? a) Only Mr. Which city has seashore and historic significance? Which city is both a metropolitan and has seashore? Which city is not peaceful but has historic significance? 8.3. C. Five executives of CHOGM hold a conference in Delhi. Z cannot be in window number 6. Both the Maths students play Football. Rahul and Atul. Moreover U cannot be immediately to the left or immediately to the right of V. 2.3. Who is the Doctor? a) Ram b) Lakhan c) Kishan d) Can‟t Say Who is Ram‟s wife? a) Gita b) Sita c) Radha d) Data insufficient Who is the Artist? a) Gita b) Sita c) Radha d) Can‟t Say Who is Radha‟s Husband? a) Ram b) Lakhan c) Kishan d) Can‟t Say Who is married to the teacher? a) Radha b) Sita c) Gita d) Can‟t Say Who is the Engineer? a) Ram b) Lakhan c) Kishan d) Data Insufficient 7. A & E are is of historic significance. immediately to the right of X which Product must be placed in window no. Dinesh plays cricket. E because of the following Characteristics.4. 10. X. Mr. The Physics student plays Football and Badminton. V. E d) Mr. W. D speaks French and understands Spanish quite well. L. Hence: K is the tallest. according to the condition. Since P Eliminated. Concept: Items are to be arranged based on the conditions given. Problem: K. M. Types: Ordering: If More than one Variable is given in the problem. B and Mr. A and Mr. L. O is as tall as P but taller than M. less than. O is not. O and P are 6 Men. Remaining: K L O Statement 3: L is shorter than K So. height. A and Mr.4. B c) Mr. B and Mr. age. use ordering method according to the condition. Statement 1: Neither M nor N are tallest So.1. Neither M nor N is the tallest. E Of the languages spoken at this conference. M. rank etc. then use elimination or ordering method. Which is the tallest? Solution: The Persons are K. Here the terms like greater than. L is taller than P but shorter than K. Elimination: If the Problem contains only one variable. Remaining: K L O P Statement 2: L is taller than P So. The questions will be based on the Person‟s weight. N. N. Remaining: K O Statement 4: O is as tall as P. Ordering 2. O and P.Training & Placement Which of them can not converse without an interpreter? a) Mr. E b) Mr. C d) Mr. Pondicherry Engineering College 53 . A and Mr.4. choose the least common language: a) English and Spanish b) English and French c) Hindi and Spanish d) English and Tamil e) French and Spanish 2. not equal to will come into play. D e) Mr. How much does Banti have? a) 12. Find which is lightest. D. and N are five Boys in a class. B2 is 4½ times heavier than B3. B3. Chetan has half as much money as Dolly. Pondicherry Engineering College 54 . and is taller to D and E. J. Jugal is younger than Sanju and older than Mughal. B2. whereas M is the cleverest of all but shorter than J. F is heavier than D. 4. L is not as clever as J. K is taller than N. C is younger to A. Arti has more money than Banti. Arti has Rs. D is heavier than B and taller than C. D is elder to A but is shorter in the Group. Manju is younger than Sita and older than Jugal.3 d) None of these Who has the second highest amount? a) Arti b) Chetan c) Dolly d) Banti 2. A belongs to the age group of five while H belongs to the age group of six. Who is the third tallest? How many people are shorter than K? If L is the third cleverest then who is the dullest? 6. Banti. B was 2 months old.2. but younger to E. but ½ of B5‟s weight. when Sheila was born. B4 & B5 all of different weights. No two Persons got the same ranks in any parameters. Who among them is the oldest? 7. B. K. Anju is younger than Manju and older than Sanju. but not as tall as C. Raksha is younger than Saksha and older than Sita. E. Arti and Banti together have as much money as Chetan and Dolly. B5 is less than B1 but more than B3 in weight. A is shorter than E but taller than F. Chetan and Dolly have Rs. B1 is twice as much as B2. Who among them is the Tallest? Who is the third tallest from the top when arranged in descending order? Who among them is the lightest? 3.Training & Placement 2.03 b) 11. Which of the following pair is elder to D? a) BA b) BC c) BE d) EA e) None of these If another friend F is taller than C how many of them will be between F and E according to height? a) None b) one c) two d) three e) None of these 5. Heena is older than Raksha. B. A. They are ranked in the order of their heights from tallest to shortest and in the order of cleverness from cleverest to dullest. C. M. but not as clever as J and L. Four People Arti. Problems to Solve 1. E is shorter than D but taller than F. Saksha is younger than Beena and older than Heena. 5 more than Dolly. I have 5 balls B1. A. B4 is twice of B3. D and E are five friends. B and C are shorter than F but heavier than A. H is six months younger to Sita while B is three months younger to A.6 c) 13. While L is shorter than M but taller than K. A is taller to D. and F are six students in a class. C. L. 100 with them.4. B is elder to E. S and T are 5 people who sit at a round table.5.Either P is sitting when viewed anticlockwise or T and R when viewed clockwise. If S is not sitting between Q and R then who is sitting between Q and S? Solution: 1. If S is between Q and R then the arrangement is QSRTP. Mohit. Mohan is older than Raju but not as old as Lalit. 1. B. Q. Mostly it is in the form of a circle. Anuj is little shorter than Kunal and little taller than Sachin? Who is the Shortest? Who is the second tallest? 9. A Blacksmith has five iron articles A. etc. We have to draw the Shape which is given in the question and arrange them according to the conditions given. Sachin is Shorter than Kunal but taller than Rohan.Two different arrangements are possible. Triangle. Pondicherry Engineering College 55 . 3. Mohit is the tallest. Mughal in age? a) Manju b) Sita c) Anju d) Raksha 8. Jugal. Neelesh is younger than Raju but not the youngest. Lalit is Older than Neelesh and Kabir. R. Anuj and Rohan. Seating Arrangement 2. D and E. P sits two tables to the left of R and Q sits two tables to the right of R. There are five friends – Sachin. C. A weighs twice as much as B B weighs four and a half times as much as C C weighs half as much as D D weighs half as much as E E weighs less than A but more than C Which is the lightest in weight? Arrange the articles in descending order of their weights. How many different arrangements are possible? 3. each having a different weight. Concept: It is the manner in which the people are seated in Place. Kunal. Problem 1: P. If S is in between Q and R what is the arrangement? 2. 10. Based on the question. Among 5 friends. Who is the fourth in the descending order of age? 2.5. 2.1.Training & Placement Who is the Youngest? a) Anju b) Sanju c) Jugal d) Mughal Who precedes Sanju. draw the shape like Square. D. he doesn't sit either opposite or beside C. If D is adjacent to F then who is adjacent to C? (a) E and B (b) D and A (c) D and B (d) none 10. R. C. is not the neighbor of R. 6 people A. C. D sits to the immediate left of A and B sits two places to the right of A. B. E is to the right of A but on the left of C. N is second to the left of S . Six Persons A. Who is immediate right of S? 9. Q. E is to the left of D. P. 6. E and F sit at a round table with the following conditions given how does the arrangement look like: A sits opposite to E. whose right hand neighbor is G. Six friends A. E and F are seated in a closed circle facing the centre. B enjoys having H to his left and F to his right. S. B. C. D. M. B. F is to the left of D.Training & Placement 2. N. B is sitting between G and D. C is between A and B. G and H are sitting in a circle facing the centre. F. L. Since A does not like C. There are five different houses A to E in a row. 4. R. R and S are sitting in a circle and playing cards. N who is the neighbor of P. V and W are sitting around a circle facing the center. B. who is the neighbor of M. A group of eight members sit in a circle. E and F sit around a table for dinner. Which of the houses is in the middle? a) A b) B c) D e) E 2. B. C. Q. What is the formation like? How many possibilities are there? 3.5. T has two seats in between himself and U. Eight friends A. Problems to Solve: 1. Who is between A and F? 8. Who is to the left of B? 7.2. Who is third to the left of D? 5. U is to the immediate left of P. Q. D. H is third to the left of B and second to the right of A. Who is third to the left of T? Pondicherry Engineering College 56 . Find the member who is diagonally opposite to A. S. If 6 people A. C. E. C is sitting between A & G and B & E are not sitting opposite to each other. Q is second to the left of R. P. T and U are seated in a circle with P opposite to S. F is between E and A. D. B is to the right of D. E and F are standing in a circle. B and F always like to sit opposite each other. R is third to the right of V who is second to the right of P. A is to the right of B and E is to left of C and right of A. P. T. V is sitting between S and W. T is the second to the left of Q who is second to the left of W. B is between F and C. D. A is between E and D. D is between A and F and is opposite to G. 1. D and Girls are X. Y cannot come with W. If S selects C then J does not select L. Among Girls are Priti. M. If T is selected then Z cannot be there. C. Bhel. Chaitanya. L is selected P cannot be selected.6.Training & Placement 2. A and C will not be selecteed which Means B ≠ A.6. C Problem: Three students J. Y and X come together. If L is selected then P cannot be selected. In a class of 7 members they are going to select 4 members 2 from Boys and 2 from Girls. Y and Z. If A is selected who are the remaining members? If B is selected who are the remaining members? Who is the Person who will be always there in the team? 3. (given) So C can‟t be selected Answer: P and C 2. C. Pondicherry Engineering College 57 . The selection criteria are such that Amit and Sonia have to be together.  Y & C can be together  A& X are always together  C & D cannot be together but C & B are always together. Four Students are to be selected from T. Boys are Amit. X. V. If B selects Y then S cannot select C and if J selects P then S selects H and L.6. Write the condition in a shortcut like. Y. W. H. Selections 2. Select the item from the whole according to the condition and start to solve from the direct condition. Z. B. A team of 5 is to be selected for a Basketball tournament. The good players among. S have to select 3 subjects out of 6 subjects P. Infer from the statements without Assumptions. Find the number of possibilities If T comes in the team who are its members? 2. Kiran and Sonia. Y. The Boys are A. (given) P can‟t be selected If B selects Y then S cannot select C. Problems to Solve: 1. U should be present in the team. A and B together Means A = B If A come and B will not come Means A ≠ B If B is selected.2. Dilip & Imran. and L. B. Concept: Selecting a part out of whole. V cannot be a part of the team. U. If S selects L and B selects Y then S cannot select? Solution: S selects L. B selects Y. Rani. Of these A. 3 from Junior and 2 from Senior. B. B. N. and however Chaithanya and Imran have to be together. S.  Debaters Y & A can‟t work together. B. O. H. P and Q are females and the rest are males. Y & Z who are seniors and A. who would be the others? Now let two members be Girls and Dilip is one of the members too. B. C. there will be no female doctor. S is not selected  C and D are always together. who are the members the team will consist of? If debater A is in the team. If P is selected the remaining members are: a) ACDS b) RBAB c) CDAQ d) QRBA 6. The school requires that they should be two seniors and two seniors in the team. City high school must put together a debating team consisting of four debaters. There are candidates of equal ability X. The formation of teams is subjected to the following conditions:  Wherever there is a male doctor. there will be no female teacher. D and E. three female teachers and two engineers. which other debaters must be in the team as well? If Y & Z are selected.Training & Placement Priti however cannot be with Rani. C. which of the debaters must be in the team with them? 5. Q. K and L and six teachers M. similarly we cannot have Rani with Bhel or Dilip with Kiran. C and D. In senior team the members are P. If the team consists of two doctors. R. four engineers G. the team who will be the members of the team? Now if one of the members is Rani.  Debaters Z & C can‟t work together. If two of the members have to be Boys. G. It is also necessary that all the debaters be able to work with one another. From amongst five doctors A.  Wherever there is a male engineer.  Debaters A & B can‟t work together. If debater B is selected and debater Y is rejected.  There should not be more than two male teachers in any team. P. in junior team the members are A. Q and R some teams are to be selected. O.who are the members of the team ? a) A B O P Q G H b) C D K L O P Q c) C D O P Q G H d) D E G H O P Q If the team consists of two doctors. all the following teams are Possible except? a) A B G M N O P b) A B H M O P Q c) A B H M R P Q Pondicherry Engineering College 58 . In a college a new football team was formed consisting of five members. D who are juniors. then the who would be the members of the team ? 4. H. one engineer and four teachers. The selection of the team is based on the following conditions:  P does not go along with R  Q and R work together  If B is selected. and two engineers. B. O. C and D work together  If H is selected then A cannot be selected. C. A school decided to go for NSS camp for a week. who will be the members of the team? a) P Q D A B b) P S D B C c) Q S D C B d) R T D A B If R be one of the members. all the following teams are possible except : a) A B G H O Q b) A B G H P Q c) A B K L P Q d) O P G H A B 7. Boys team consists of A. P. R. C and D.Training & Placement d) A B K N R P Q If the team consists of two doctors. X. A team of five is to be selected from amongst five Boys P. E) and Executives (W. D. There was a need of five member team of 3 Managers and 2 Executives. N. Unless otherwise stated. Q. two female teachers. these criteria are applicable to all the questions below: If the two of the members have to be Boys. D. A Company wanted to start a new project. Some criteria for selection are:  P and D have to be together. B. Six member team of 3 Girls and 3 Boys were to be selected with the conditions as  A and B cannot be together  M and P cannot be together  P and B go together  D and E go together  M and N go together Pondicherry Engineering College 59 .  S and B cannot go together. B. S and T and four Girls A. Z) were to be selected with the following conditions:  A and C work together  X and Y work together but X does not work with Z  H does not work with X. the other members of the team are? e) A D P S f) BDPS g) B D R T h) D P R T 8.  R and T have to be together. If B is selected then who are the remaining Persons? 9. C.  A cannot be put with C.  C cannot be put with Q. The managers (A. E and Girls team consists of M. Y. How many males and females are present? Solution:  Denotes Female  Denotes Male I. and F who are related to each other. These three attributes are often termed as „reasoning‟. This particular way of relationship between the elements. Why and How its validity can be judged. B is F‟s daughter–in–law. D is A‟s only grand child. objects or subjects leads our mind towards the formation of thoughts which are expressed in language. E is father of C. P were to be selected for a competition. one male and one female. D. C. B is F‟s daughter-in-law F Father/Mother-in-law B Daughter-in-law 60 Pondicherry Engineering College . E and 5 Girls L. B. 3. A has only 2 children F and C. Then. The selected team consists of 3 Boys and 2 Girls with the conditions as: From Boys side B and Girls side M should be selected  A does not go with B  A and E go together  If B is not selected then L is selected  M does not go with O  N and P do not go together How many arrangements are possible? 3. Concept: Deals with Hierarchical structure of a family.1. So logic can be defined as the science of thought. D. C. Family Tree 3. B. O. M.Training & Placement 10. The perception and conception of the objects or elements become knowledge for us if our mind conceives them in a systematic order so that certain relationships can be established between the objects or elements. It is also called as Blood relation problem. Problem 1: There are six members in a family A. In a class there are 5 Boys A. Reasoning: When a thought is associated with the three attributes of What. C is D‟s only uncle. Logical Reasoning Concept: Logic: Our mind always seeks a rational explanation of the objects which we come in to contact with our day-to-day life. N. Reasoning is the function of the mind passing from known to unknown by establishing a systematic relationship between the elements. Knowledge of a subject is said to be systematic and scientific when the different parts of knowledge are related together in a particular way to make a system.Who is the grand mother of D? 2.1.1. 1. This process is known as logic. E. 2. There are 3 males and 3 females in the family. E is the father of C B A x E C F D x B Daughter-in-law 1. C is D‟s only uncle A C C ? IV. A has only 2 children F&C.Training & Placement II. one male and one female A C F D B V. A is the grand mother of D. Pondicherry Engineering College 61 . D is A‟s only grand daughter A ? D III. one mother and 1 daughter. C. d) Data Inadequate d) None Looking at a portrait of a man. E is brother of F. a) Find which are the brothers? b) Who is E‟s husband? Mohan is son of Arun‟s father‟s sister.1. R and L are of different sexes. What is the man to the lady? F‟s grandfather is A. B is Brother of D. who is S‟s uncle? A man was going with a Girl. How is F related to E? 7. What is the relationship between the man and the Girl? Out of A.” At whose portrait was Sanjay looking? If B‟s mother was A‟s mother‟s daughter. how many males are there d) Who is the daughter–in–law of Vivek ? e) If N is Vivek‟s sister. Somebody asked his relationship with the Girl.2. C is sister of F. L is the spouse of one of his child. Purab is father of Neha and grandfather of Mona. “My paternal uncle is the paternal uncle of her paternal uncles”. Rekha is Purab‟s wife. „Her mother is the only daughter of my mother–in– law‟. M is the son and Q is the daughter. Prakash is son of Rekha who is mother of Vikki and grandmother of Arun. 3 brothers. 4. a) Who is the father of S and R? b) Whose spouse is L? c) If all the above characters belong to one family. How is Rohit related to Anil? Pondicherry Engineering College 62 . C is married. Sanjay said. D. N and his/her spouse W. How is Mohan related to Rekha? a) Grandson b) Son c) Nephew How is Vikki „s wife related to Neha? a) Sister b) Niece c) Sister in law 6. E and F. have two children S and R who are of the same sex as W. “the son of the only brother of his father‟s wife”. D has no brothers but is the father of A (male) and grandfather of F. 10. We have 2 fathers.Training & Placement 3. 3. 8. N is of the same sex as Q. B. “His mother is the wife of my father‟s son. He replied. Anil introduces Rohit as. 9. 2. Brothers and sisters I have none. Problems to Solve: 1. a man said. 5. D is sister of E. B is brother of E‟s husband. How is F related to B? Vivek is the father of M and Q. How was A related to B? Introducing a lady. A has two sons – E and C. The arrangement can be any one of the following: 1. certain code values are assigned to a word or a group of words and the original words should be found out. TRIPPLE is written as SQHOOKD How would DISPOSE be written in that code? Pondicherry Engineering College 63 . Problems to Solve: 1. Codes may be numerical or alphabets. C D E (Gaps in Between arrangement) So SNOW can be coded as QLMU. If in a certain language TEACHER is coded as VGCEJGT. Codes 3. then how will SNOW be coded as? Solution: The term FIRE is coded as DGPC.1.2. Hence. Reversed order 2. P Q R. Gaps in between 3. G H I. in first.e. „ASTROLOGY‟ would be written in this code? Solution: In this Code. Decoding is the ability to decipher a certain code. if we note it we find that the code is two alphabets preceding the original word. The answer is “ZTSSNMNHX”. In a certain code. Problem 2: If NUMERICAL is written as MVLFQJBBK. 3. then how.2.2. how would „DULLARD‟ be coded in that code? 2. That is D E F. fifth. the letters in Odd place. i. In these types of questions.2.Training & Placement 3. Tree format Problem 1: The term FIRE is coded as DGPC. third. Concept: Coding is a method of transmitting a message from one place to another. +1 and–1 format 4. seventh and ninth place letters have been coded to their preceding letters and the remaining ones have next letters as code. It is easy to decode if we are able to find the format of arrangement. Which word would be coded as FSQFCE? 3.E. Condition 1: If A occurs then B also will occur. In a certain code. In a certain code RIPPLE is written as 613382 and LIFE as 8192.C. Based on them the decisions are made.B.3. blue is called Water. How will PEARL be written in that code? 7. flower is called cooler. In a certain code COMPUTER is written as RFUVQNPC How would MEDICINE be is coded as 4. If in a certain language CARROM is written as BZQQNL. Concept: Conditions are given based on which the actions take place. There are four basic conditions. If room is called bed. white is called rain. Problem 1: From a group of 6 Boys A. Condition 4: If A has not occurred then B also will not occur. CALCUTTA is coded as GEPGYXXE. PALE is written as 2134 and EARTH as 41590. How would HOUSE be written in that code language? 5. then on what would a man sleep? 9. Based on these conditions the groups are formed. Condition 2: If A occurs then B will not occur.F and 5 Girls L. green is called air. window is called flower.N.3. rain is called green.M.P a group of 6 is to be selected based on the following conditions: O and P have to be together C cannot go with O Pondicherry Engineering College 64 . air is called blue. How is 23549 written in that code? 8.Training & Placement 3. How should PILLER be written in that code? 6. Conditionality and Grouping 3. bed is called window. In a certain code 15789 is written as EGKPT and 2346 is written as ALUR.D.1. If in a certain language. Where will the bird fly? 10. Condition 3: If A has not occurred then B will occur.O. If cloud is called white. From among 6 Boys A.R. Six member team of 3 Girls and 3 Boys was to be selected with the conditions as:  P and Q cannot be together  W and Z cannot be together  Z and Q go together  S and T go together  W and X go together 3. how many arrangements are possible? 2. Therefore M is selected. If A is selected the remaining members are: (a) PRSD (b) CBPR (c) RSPB (d) BCQP If B and C are selected. and F and Five Girls P.3. In a college a new football team was formed consisting of five members. In senior team the members are A. So choice is (iii) BFLNOP If B is selected 2 Girls M and N have to be selected. C. E. S. D. In junior team the members are P. 3 from Junior and 2 from Senior.2. The selection of the team is based on the following conditions:  A does not go along with C  B and C work together  If Q is selected. X. B. S. B. R. D is not selected  R and S are always together. Q. 3. O and P D and A cannot be there. The Girls are L. Q. T and Girls‟ team consists of W. Z.Training & Placement A and D have to be together D cannot go with L C and M have to be together Band N have to be together B and E cannot be together If the team consists of four Girls the members of the team are: i) BELNOP ii) EFLNOP iii) BFLNOP If the team consist of 5 Boys and 1 Girl that Girl would be: i) L ii) M iii) N iv) O Solution: O and P will be there so C cannot be there as well as M. Problems to Solve: 1. Y. So B is rejected. Boys‟ team consists of P. R. A school decided to go for NSS camp for a week. C and D.Q. N.S and T a team of six is to be selected under the following conditions:  A and D have to be together  C cannot go with S  S and T have to be together  B cannot be teamed with E Pondicherry Engineering College 65 . C. C.Training & Placement  D cannot go with P  B and R have to be together  C and Q have to be together If four members have to be Girls. F. A college selection committee sat to finalize Cricket Team of five members 2 from team X and 3 from team Z. If H and G are sitting in the same car. B. N. then find the members of the team? 4. G and Z team consist of 5 members B. Team X consist of 4 members A. Eight students A. E and 5 Girls L. 5 Boys A. D. B.  B and C can‟t sit in the same car in which D is sitting  F will sit in the car of four people only along with A and E but certainly not with G.  A will sit in the same car in which D is sitting and H is not in the same car. D. The selected team consist of 3 Boys and 2 Girls with the conditions as. J. The conditions given were:  A and E go together but cannot go with B  D and F go together but cannot play with H  H and J go together but cannot play with E  E does not play with D 5. H. E. O. which of the following statements is true? (a) Five students are sitting in the same car (b) B is sitting in the same car (c) F is not sitting in the same car (d) G is not sitting in the same car (e) None of these Which of the following statement is superfluous for the above seating arrangements? (a) Only (i) (b) Only (ii) (c) Only (iii) Pondicherry Engineering College 66 . From Boys side B and Girls side M should be selected  A does not go with B  A and E go together  If B is not selected then L is selected  M does not go with O  N and P do not go together How many arrangements are possible? 6. D. In a class. F. and P were to be selected for a competition. C. who are the other two students sitting in the same car? (a) B and C (b) C and D (c) B and D (d) E and B (e) None of these If E and A are sitting in the same car. M. G and H are planning to enjoy car racing. E. There are only two cars and following are the conditions  One car can accommodate maximum five and minimum four students.  If Q is to be operated on a day. T. R and S Some criteria for selection are:  A and S have to be together. S. Pondicherry Engineering College 67 A. Q. V. A team of five is to be selected from amongst five Boys C. S. S. The program operating must also satisfy the following conditions:  If Program P is to be operated on a day. U. T must be one of the programs to be operated after Q. V (c) Q. T. V. S. U. V. On any one day. Which of the following could be the set of programs to be operated on the first day of a month? (a) V. R. V (b) Q. T.  The last program to be operated on any day must be either S or U. At an Electronic Data Processing Unit. only three of the program sets must be the ones that were operated on the previous day. R. except for the first day of a month. T.  P cannot be put with R. S (d) Q. V (d) T.  If R is to be operated on a day. T. W (c) T.Training & Placement (d) Only (IV) (e) None of these 7. V 8. S (b) U. S. U Which of the following is true of any day‟s valid program set operation? (a) P cannot be operated at third place (b) Q cannot be operated at third place (c) R cannot be operated at fourth place (d) T cannot be operated at third place (e) U cannot be operated at fourth place If R is operated at third place in a sequence. V must be one of the programs to be operated after R. R. five out of the eight problem sets P. U (e) T. B. S. Q. which of the following could be the other programs on that day? (a) P. R. R. Q. U (e) P.  D and Q cannot go together. . V cannot be operated on that day. S. Q. which of the following cannot be the second program in that sequence? (a) S (b) T (c) U (d) W If the program sets R and W are to be operated on the first day. D and E and four Girls P. V and W are to be operated daily. four engineers G. one engineer and four teachers. P and Q are females and the rest are males. all the following teams are possible except: (a) ABGMNOP Pondicherry Engineering College 68 . D and E. B. K and L and six teachers M. If the team consists of two doctors. some teams are to be selected. the members of the team are? (a) (b) (c) (d) ABOPQGH CDKLOPQ CDOPQGH DEGHOPQ If the team consists of two doctors. P.  R cannot be put with B. Of these A. the members of the team other than D are? i) P Q B C j) P Q C E k) P S A B l) P S C E If A and C are members. there will be no female doctor. H. Q and R. H. three female teachers and two engineers. N. there will be no female teacher. G.Training & Placement  C and E have to be together.  There shall not be more than two male teachers in any team. the other members of the team are? e) P S A D f) Q S A D g) Q S C E h) S A C E If two of the members are Girls and D is one of the members.  Wherever there is a male engineer. O. these criteria are applicable to all the questions below: If the two of the members have to be Boys. From amongst five doctors A. C. B. The formation of teams is subject to the following conditions:  Wherever there is a male doctor. O. the other members of the team cannot be? m) B E S n) D E S o) E S P p) P Q E 9. Unless otherwise stated. the team will consist of? a) A B S P Q b) A D S Q R c) B D S R Q d) C E S P Q If R be one of the members. all the following teams are possible except? (a) (b) (c) (d) ABGHOP ABGHMN CEKLNR CDKLOP 10. 2. two female teachers. and two engineers. F above d) Both B and C e) None of the Which may occur as a result of cause not Mentioned. who could be the members of the team? (a) (b) (c) (d) ABCKLMR BCDKLNR CDEKLMN CDEKLPR If the team consists of two doctors. two engineers and two teachers. 3 Pondicherry Engineering College 69 . H or both  H occurs if E occurs  G occurs if F occurs If B occurs which must occur? a) D b) D & G c) G & H d) F & G e) J If J occurs which must have occurred? a) E b) either B or C c) both E.Training & Placement (b) (c) (d) ABHMOPQ ABHMRPQ ABKNRPQ If the team consists of two doctors. A causes B or C. all the following teams are possible except : (a) (b) (c) (d) ABGHOQ ABGHPQ ABKLPQ OPGHAB If the team consists of three doctors. but not both  F occurs only if B occurs  D occurs if B or C occurs  E occurs only C occurs  J occurs only if E or F occurs  D causes G. two male engineers and two teachers. 1) D 2) A 3) E a) 1 only b) 2 only c) 1 & 2 d) 2 & 3 e) 1. A. 92 So the solution is 72 = 49. 32. Problems to Solve: 1.5. 1. 9. V. 45. 4. 9. 1.5. T.5. W. 0. 13. 3. S. 24. The sequence can be any of the following. J. M. U. R. 13. _. Concept: Series involves the numbers or letters that are arranged in a particular sequence. 16.__ ? 3. Direction Sense 3.1. 52. Solution: As given in the problem it consists of Squares of digits with a digit left in between It is 12. X. Problem 1: „A‟ starts from his office and walks 3 km towards north. Concept: These problems test our sense of Direction. 1. Problem 1: Fill in the series 1. G. 33. __? 5. C.2. 1. 9. 4.5. _. __? 9. 81. 1. D. __ ? 6. 17. Z. 2. 22. 121. The best way of solving these questions is to follow the instructions given in the question carefully and make a diagram accordingly with the help of which the question can be solved. Gap 4. Y. 25. 5.4. Descending 3. 25. +/– order series.5. 6. Series 3. ___ ? 7.4. H. 225. He then turns right and walks 2 km and then again turns right and walks 2 km. He then turns right and walks 2 km and then turns right and walks 5 km. ___ ? 8. Q. __ ? ___ ? 4.4. M 3. __? __? 2. ?. A.5. 361.1. 29. Squares and cubes 5. 6. In which direction is he from the starting point? a) South b) North–east c) South–East d) He is at the starting point Pondicherry Engineering College 70 . __ ? 10.Training & Placement 3. F. Ascending 2. 5. Finally. Problem 2: Deepa moved a distance of 75 metres towards the north. She was finally moving in the direction DE i. the answer is (e). turned left and walked 25 m upto C.Training & Placement 3 5 5 Solution: He is at the starting point. How many kilometers will he have to cycle to reach his home straight? Pondicherry Engineering College 71 . turned towards right and walked for a distance of 10 km. She then turned to the left and walking for about 25 metres. Problems to Solve: 1. Turning to the right at an angle of 450 . she turned to the right at an angle of 450.2. turned left again and walked 80 metres. One day Kannan left home and cycled 10 km southwards. turned right and cycled 5 km and turned right and cycled 10 km and turned left and cycled 10 km. moved 75 m upto B.. Arman walked towards east for a distance of 5 km. Then again turned to his right and walked 15 km. 3. In which direction is Sathish with reference to Amir? 3. Next he turned to his left and walked 5 km. In which direction was she moving finally? a) North-east b) North-west c) South d) South-east e) South-west Solution: Deepa started from A. Hence. Rajesh is standing to the west of Amir and north of Ruchir and Sathish is to the west of Ruchir and south of Salman. She then turned left again and moved 80 m upto D. In which direction is he now with reference to his starting point? 2.e. South-west. After moving a distance of 20 m. He then turns towards East and walks 15 m.6. He again took a left turn walking 20 metres. Further. Statement Logics 3. and then I turn left and go 8 km. A walks southwards then right. a) If conclusion 1 follows b) If conclusion 2 follows c) If neither 1 nor 2 follows d) Both 1 and 2 follows Example: Statement: Quality has a Price tag. then turn right and go 8 km. I walk from my back door 100 m. What is the straight distance in meters between his initial and final positions? 5. He then turns left and walks 40 m. How far is he from his original position? 3. he turns towards South and walks 10 m. The Door of my house faces East.Training & Placement 4. then left and then right.6. Concept: The ability of Analysis and interpretation of data logically is tested in these problems. in what direction is A with respect to C? 9. then right and walk 100m and turn left and walk 50 m and reach a point X. A man leaves for his office from his house. Then I turn left and go 5 km. I go 5 km east. He again turns left and walks 20 m. India is allocating lot of funds for education. Gaurav walks 20 m towards North. In which direction is he from the starting point? 7. Then he walks 35 m towards the West and further 5 m towards the North. he moves 20 m after turning to the right. In which direction is I from the starting point? 10. He walks towards East. Ram started walking towards South. Logically apply the Pondicherry Engineering College 72 . Which direction is he facing now? 6. At what distance am I from the starting point now? 8. conclusions and give answers.1. Types: 1) statement– conclusions 2) statement– assumptions 3) statement– arguments Conclusions In each questions there is a statement and two conclusions. If A is to the South of B and C is to the east of B. He took a right turn after walking 10 metres. Training & Placement Conclusions: 1) Quality of education in India would improve soon 2) Funding alone can enhance Quality of education Arguments Each question given below has 1 statement and two arguments. Give answer: a) If only 1 implicit b) If only 2 implicit c) Both 1 and 2 implicit d) Neither 1 nor 2 implicit Answer: Example: Statement: The pen is mightier than the sword Assumptions: 1) The pen is made up of stronger metal than the sword 2) The power of the mind is much stronger than brute physical power Answer: (1) (a) Pondicherry Engineering College 73 . It is risky to put them in private hands Assumptions Each question contains 1 statement and 2 assumptions. We should decide which of the arguments is strong Give answer: a) If only 1 is strong b) If only 2 is strong c) If neither 1 or 2 is strong d) If both 1 and 2 is strong Answer (a) Example: Should Private sector be permitted to operate telephone services? Arguments: 1) Yes. Check the data and answer. they are operated in advanced western countries 2) No. II) Evening walks are harmful. Statements: Should military service be made compulsory in our country? Argument: 1) No. Statements: Should there be internal assessment in colleges? Argument: 1. Statement: The Best way to escape from a problem is to solve it Conclusion: I) Your life will be dull if you don‟t face a problem. 2. it will enable the teachers to have a better control over the Students.Training & Placement 3. 2. Every citizen should protect his country. No. II) Then Price of the Product must be reasonable. 3.6. Conclusion: I) all healthy people go for morning walks. Statements: Should there be world Government? Argument: 1) Yes. 7. Go ahead. Statement: Company X has marketed the Product. Conclusions: I) The Products must be good in quality. II) To escape from problems you should have solutions 4. It is against the policy of non violence. purchase it if Price and quality are your considerations. Statement: Morning walks are good for health. Problems to Solve: Statement Conclusion a) Only conclusion I follow b) Only conclusion II follows c) Either I or II follows d) If neither I nor II follows e) If both I and II follows 1. 2) Yes.2. Statements: Vegetable Prices are soaring in the Market Conclusions: I) Vegetables are becoming a rare commodity II) People cannot eat Vegetable Statement Argument a) If only 1 is strong b) If only 2 is strong c) If either 1 or 2 is strong d) If neither 1 nor 2 is strong e) If both 1 and 2 is strong 5. 6. Yes. It will help in eliminating tensions among the nations Pondicherry Engineering College 74 . since it will encourage favoritism among teachers. how well they do comprehend the idea conveyed in the given passage and how well candidates can express the given idea. 2) Recent news is broadcasted only on radio 9.00 pm news on radio– candidate tells the interview board. Verbal Ability falls under following categories: Pondicherry Engineering College 75 . A language is a practical medium of expression of our ideas and thoughts. Statements: To keep myself up to date . Verbal Reasoning Concept: It is the general notion that English is a foreign language and so it is very difficult.Training & Placement 2) No. Assumption: 1) The Assurance is not genuine. 2) People want their money to grow. The aim is to find how well candidates do understand the grammatical rules and their usages. They are only interested in the grammatical structure of the language as a means of getting things done. Statements: Double your money in five months – An advertisement. But this notion is far from truth. Then only the developed countries will dominate the Government Statement Assumption a) If only 1 implicit b) If only 2 implicit c) Either 1 or 2 implicit d) Neither 1 nor 2 implicit e) Both 1 and 2 implicit 8. I always listen to 9. People who study and use a language are mainly interested in how they can do things with the language. The questions in English Language are designed to test candidate‟s understanding of English and its usage. A student must understand that English is not a tough language. Assumption: 1) Telephone facility is not available. Statements: I cannot contact you on phone. 10. get attention to their problems and interests. Verbal Ability is to understand the prose and to work with specialized and Technical Vocabulary. Grammar puts together the patterns of the languages. Assumption: 1) The candidate doesn‟t read newspaper. 2) Now a Days it is difficult to contact on phone 4.Acquiring a simple. how they can make meanings. direct and forceful knowledge in writing and reading calls for constant practice. influence their friends and colleagues and create a rich social life for themselves. make smaller) e) Electrify ( thrill. If you have trouble with a definition. Look for cognates from French. or Italian if you recognize them. try using the word in the sentence. Example: MAGNIFY a) Forgive b) Comprehend c) Extract d) Diminish e) Electrify Solution: Magnify (enlarge. Use the Latin roots. it is most likely correct. Try to predict an answer before looking at the answer choices. pullout) d) Diminish (reduce. 4. 2.Training & Placement Synonyms Antonyms Analogy Sentence Completions Idioms Spotting errors Reading Comprehension 4. 5. absolve) b) Comprehend ( understand. know. Tips: 1. Pondicherry Engineering College 76 .1 .1. Before you look at the answer choices. excite. Spanish. and suffixes to figure out what hard words mean. Consider slight variations in the meaning of each word. The Antonym for the word „Old‟ is „New‟.1 Antonyms 4. expand) a) Forgive ( excuse. If an answer choice matches your predicted answer.Concept: The word „Antonym‟ means a word that is opposite in meaning to another word. remove. realize) c) Extract ( take out. Always consider all the answer choices before you select an answer. prefixes. Remember that you are looking for the answer choice that has a meaning opposite to that of a given word. amaze) The answer is Diminish. try to clearly define the given word. For example. 3. DEPRECATE a) Praise b) Approve c) Eulogize d) Depreciate 5. ABATE a) Aggravate b) Amplify c) Enlarge d) Enhance 2. BUCOLIC a) Urban b) City like c) Slovenly d) Polished 3. PROVIDENTIAL a) Ill–starred b) Unfortunate c) Unlucky d) Simulation Pondicherry Engineering College 77 .1.2 Questions to Practice: 1. ILLUSORY a) True b) Elusive c) Real d) Factual 8. EMOLUMENT a) Penalty b) Punishment c) Monument d) Retribution 6.Training & Placement 4. CATACLYSM a) Steadfastness b) Privilege c) Benefit d) Blessing 4. FRACTIOUS a) Complaisant b) Genial c) Sagacious d) Accommodating 7. complete) Noisy: (loud. VARIEGATE a) Set type b) Multi–colour c) Differ d) Reject e) Reply in Kind 2. piercing.2. Example: ASTUTE a) Sheer b) Noisy c) Astral d) Unusual e) Clever Astute: (smart. „Shut‟ and „Close‟ are synonyms. FILCH a) Pretend b) Dirty c) Embarrass d) Steal e) Honour Pondicherry Engineering College 78 . For example.2 Synonyms: 4.Training & Placement 9. TURBULENT a) Quiet b) Quiescent c) Calm d) Truculent 10.) Astral: (planetary.2 Questions to Practice: 1. bright.2. curious) Clever: (smart. intelligent. intelligent) Answer – Clever 4. VENAL a) Honourable b) Likeable c) Incorruptible d) Patriotic 4. Odd. lunar) Unusual: (strange.1 Concept: The word „Synonym‟ means a word or phrase with the same or nearly the same meaning as another in the same language. shrewd) Sheer: (pure. absolute. INFINITE a) Verbal b) Indefinite c) Endless d) Strange e) Vague 4. DISCONCERT a) Sing in harmony b) Pretend c) Cancel Program d) Confuse e) Interrupt 10. DEMISE a) Residence b) Dismissal c) Accident d) Act e) Death 5. FRUGALITY a) Extravagance b) Ripening c) Thrift d) Miserliness e) recurrent 6. GARRULOUS Pondicherry Engineering College 79 . UNEQUALED a) Outstanding b) Different c) Praised d) Resentment e) Miserliness 7. FASTIDIOUS a) Speedy b) Precise c) Squeamish d) Hungry e) Slow 9.Training & Placement 3. ADVERSITY a) Opponent b) Hardship c) Opening d) Public announcement e) Agency 8. Class and members 4.3 Questions to Answer: Pondicherry Engineering College 80 . Symbol Example Book: Library:: Cannon: a) Artillery b) Powder c) Shell d) War Solution: Library is the place to keep books. Synonyms 6. Part to Whole 7.3. look for a narrower approach Consider secondary Meaning as well primary Meaning. Antonyms 5. Function 8. The relationship between the words in the original pair will always be specific and precise.1 Concept: Analogy questions ask you to determine the relationship between the two words in a pair and then to recognize a similar or parallel relationship between the members of a different pair of words. Tactics: State the relationship between the capitalized words in a clear sentence. Worker and article created 11.2 Analogy Types 1. Definition 2.Training & Placement a) b) c) d) e) Laconic Strangling Ecstatic Frozen Wordy 4. Action & its significance 10. Worker and action 13.3. If there is more than one answer. 4. Sex 14. Manner 9. Worker and tool 12.3. Cannon Means gun so Artillery is the place to keep gun. Defining Characteristic 3.3 Analogy 4. Familiarize yourself with common analogy types. mortar 4. SYMPHONY: MUSIC (a) Mural (b) Ode (c) Preface (d) Editorial 8. CURTAIN: DRAPERY (a) Cockroach (b) Bed sheet (c) Pillow (d) Mat 4. REVOLVER: HOLSTER (a) Book (b) Eye (c) Juice (d) Nostril 7. CROWN: ROYAL (a) Throne (b) Wrap (c) Pen (d) Crucifix 6. BALANCE: WEIGH (a) Airplane (b) Radar (c) Satellite (d) Television 5. TRAITOR: DISLOYALTY (a) Executioner (b) Rebel (c) Manager (d) Hope 9.Training & Placement 1. EXPLOSION: DESTRUCTION (a) Talk (b) Girl (c) Success (d) Engagement 2. TEXTILE: MILL (a)Eggs ` (b)Coal (c)Food (d)Brick : : : : Exaggeration Woman Failure Marriage : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : Function Ticket Scale Article Insect Bed Cushion Floor Height Detection Revolution Picture Regal Ermine Author Religion Bag Eyelid Glass Nose Painting Prose Book Journal Reliability Defiance Administration Pessimism Hen Mine Agriculture Kiln 81 Pondicherry Engineering College . AGENDA: MEETING (a) Program (b) Performance (c) Map (d) Foot note 3. poetically or prosaically formally or informally etc. prepositional phrases.4. Eclectic. An author‟s style depends on such details as word choice. Grammatical complexity : They include entire range of grammatical possibilities. MODESTY: ARROGANCE (a) (b) (c) (d) Passion Practice Cause Debility : : : : Emotion Perfection Purpose Strength 4.e. ornately on sparely.1 Concept: It tests your a) b) Ability to use vocabulary Ability to recognize logical consistency among the elements in a sentence What makes Sentence Completion difficult is A. gerunds etc. Thought extenders – Continue the idea Thought contrasters – Reverse the idea Signal Words: Additionally Also And As well Besides Further more Although But Despite nevertheless In spite of On the contrary Pondicherry Engineering College 82 . Sentence Completion 4. ironic. D. Use of clauses. playful. imagery repetition. rhythm. sentence structures and length. in a convoluted manner C. Style: Ideas maybe expressed in different manners. Tone: Writer‟s attitude towards the subject matter i. Harbingers.4. somber etc.Training & Placement 10. Vocabulary: Use of words like nonplussed. Example: Economy – restrain Toy – Dally B. skeptical. some plastic six–pack rings. In a revolutionary development in technology. abstract (b) theoretical …… challenging (c) fraudulent …… deceptive (d) Derivative …….. (a) Harden (b) Astagnate Pondicherry Engineering College 83 . Avaricious (Greediness) “Altruistic” (unselfish. testing the first word in each choice. Opposite of self seeking (selfishness).4.2 Questions to Answer: 1. several manufacturers now make biodegradable forms of plastic. subject to error.. In double blank. a complex activity. In one shocking instance of _______ research. a) Redundant b) Frivolous c) Inexpensive d) Ephemeral e) Altruistic Solution: The sentence presents simple case of cause and effect. Organized (d) Innovatively ……. decided (b) erratically …… analyzed (c) Systematically ……. Check whether the Metaphor controls the choice of words. (a) Mistakenly …. authoritative (e) Erroneous ……. Example: Because experience had convinced her that he was both self–seeking and avaricious. go through the answer. Impartial 2. Look at all possible answers. misunderstood 3. like any other human Endeavour. Measurement is. one of the nation‟s most influential researchers in the field of genetics reported on experiments that were never carried out and published deliberately ________ scientific papers on his nonexistent work. for example. selfless) 4. refined (e) Properly ……. (a) Comprehensive…. Watch for single words that link one part of the sentence to another.. she rejected the likelihood that his donation had been ______. not always used ________ and frequently misinterpreted and _________.. gradually ________ when exposed to sunlight.. Break down complex sentences into simpler components.Training & Placement More over Still Too While Likewise Which Tactics: Read the sentence and think of a word that makes sense. Job failure Means being fired off a job. being asked to resign. who began systematic astronomical and weather observations shortly after the ancient Egyptians. thought to be _______ the consumer. (a) Wane (b) Moderate Pondicherry Engineering College 84 . becoming ever more severe. (a) Reform (b) Discredit (c) Coerce (d) Intimidate 7.Training & Placement (c) (d) (e) Inflate Propagate Decompose 4. but during certain conditions the storm may _________. because the misdeeds of individuals are often used to _______ the institutions of which they are a part. Normally an individual thunderstorm lasts about 45 minutes. (a) Refinance (b) Documentation (c) Maintenance (d) Domination 6. the study panel recommends that the federal government shift its inspection emphasis from cursory bird–bird visual checks to a more _______ random sampling for bacterial and chemical contamination. The Chinese. (a) Reactive…shielding (b) Stable…blackmailing (c) Excessive…gouging (d) Depressed…cheating 9. for as long as four Hours. An institution concerned about its reputation is at the mercy of its members. To alleviate the problem of contaminated chicken. and because of this. the Price of gasoline was so _______ that suppliers were generally. (a) Rigorous (b) Perfunctory (c) Symbolic (d) Discreet (e) Dubious 5. were assiduous record–keepers. can claim humanity‟s longest continuous _______ of natural events. or leaving _______ to protect yourself because you had very strong evidence that one of the first two was _______ (a) Voluntarily…impending (b) Abruptly…significant (c) Knowingly…operative (d) Eventually…intentional 8. During the widespread fuel shortage. 5 Spotting Errors 4. A B C D E 3. more we want/ in this way/ there is no end to human desires/ No error. More we get. He had ordered you/ to paint it green/ but you painted/ the blue house. / No error D E 5. / No error. Mistakes in preposition and tenses are more common Examples: 1) The car is / almost / the same / like mine./ No error. Little money/ that he was left behind after/ the marriage of his daughter/ was not A B C Sufficient for a big family like his. D E Pondicherry Engineering College 85 .5. A B C D 4. Ashoka the Great / was regarded one/ of the greatest emperor the/ world has ever A B C D Produced / No error E 2. (a) Indifference (b) Suspicion (c) Veneration (d) Recklessness (e) Bewilderment 4.1 Concept: Grammatical errors form the most probable errors in error-spotting exercises.Training & Placement (c) (d) (e) Persist Vacillate Disperse 10.5.2 Questions to Answer: 1. A B C D E 6. Perhaps because something in us instinctively distrusts such displays of natural fluency. some readers approach John Updike‟s fiction with __________. None of the two Boys / who were / present there / came to his help. / No error. Ans: as mine 2) The weather / feels / as / spring Ans: like spring 4. When I arrived / on the station / I saw a man who had robbed me / because he A B C thought I carried the King‟s wallet. 3. No less than/ four thousand people/ lost their lives/ in the earthquake/ no error. „Catch in the act‟ means „Finding someone doing something wrong or bad. Idioms 4. I have read several plays of Shakespeare/ who was one of / the greatest dramatists/ A B C the world has ever produced/ No error.6. Test was about eating a cake. He has paid for the books. She ate cake before the test. A B C D E 9. Nobody in their senses/ would have/ uttered/ such silly remarks/ No error. my friends and I hit the slopes. He wished me/ to dine with him that evening/ but which/ I declined/ No error.Training & Placement 7. For example. A B C D E 8. (a) Hit each other. I don‟t want them on my hands any longer. She was thinking of eating a cake. (a) (b) (c) (d) Mary thought that the test was a piece of cake. I‟m trying to sell off these books. Pondicherry Engineering College 86 . The books are heavy. He wants to get rid of the responsibility. D E 10. I will not carry the books. During our winter break.‟ Example: The manager‟s bark is worse than his bite a) He shouts b) He punishes c) His action is worse than his speech d) His speech is worse than action Solution: The manager‟s bark is worse than his bite d) His speech is worse than action Questions to Answer: 1.1 Concept: A Group of words whose meaning is different from the meanings of the individual words are idioms. The task was relatively easy. (a) (b) (c) (d) 2.6. A B C D E 4. (a) (b) (c) (d) I haven‟t studied at all for the exam tomorrow. The Police found the case a tough nut to crack. (d) Hit the mountains. I lived a stone‟s throw from a popular beach. Jack always managed to keep his eye On the ball (a) He is a cricket player (b) Being focused. Prepare soup Sandy is often too bogged down with her studies to spend time with her friends. (a) (b) (c) (d) 9. (a) Got scared (b) Fell down (c) Falls down from the rock (d) His nerve was cut. (a) (b) (c) (d) My father insisted that I put my nose to the grindstone next semester. (b) People threw stones at his house. Fall into the soup Get into trouble. (c) Police found some nuts in the crime scene. In spite of many different jobs he had to do. (c) Stones were piled on the beach (d) Lived near the beach.Training & Placement (b) Went snow–skiing (c) Fell in the slopes. (d) The case was easy to solve. (a) Police could not crack nuts. Happy Overwhelmed Forgetful Smart 10. I‟m really going to be in soup. but lost his nerve and turned back. 6. (d) He is searching the ball. 4. He had climbed almost to the top of the rock. 8. (b) The case was difficult to solve. Hit my nose if I fail Work really hard Grind the stone Copy during the exam 7. (a) Throwing stones and playing. Pondicherry Engineering College 87 . Drink soup before exam. (c) Hit in the eye with a ball. Throughout the Summer. 5. What is a Deduction? If we can infer a statement either by correlating information or by discovering an implication. they felt was all the greater because British foreign policy still remained a stronghold of the aristocracy. The Questions are generally of the following types: 1. Example: The emotional appeal of imperialism never completely stilled the British conscience. Even in the heyday of colonialism British radicals continued to protest that self-proclaimed imperialists however honorable their motives. 2. enlightened British liberals looked forward to the day when India would stand on its own feet. However this fact did not escape Gandhi. In the first seventy years of the nineteenth century. it assumes that “There has been a Change in the Public Mood”. in defiance of the revealed truths of classical economics. many politicians were conscious of saddling Britain with a heavy burden. it is said to be a deduction. 3. It took the humiliation of the Boer war to teach the British government what it would cost to hold an empire by force.7. Any argument which gives ideas similar to the Supporting Argument strengthens it. the statement becomes incorrect. The only difference is that questions are designed to evaluate the Reasoning Power of the candidates. What is an Assumption? Many statements are made by presuming something to be true. For example if we say.7. The danger. If we don‟t assume it.1 Concept: Questions on reasoning comprehension resemble questions set on English Comprehension. 4. the British officer class. liberal thinkers throughout the nineteenth century argued that democracy was incompatible with the maintenance of authoritarian rule over foreign peoples. Thus when the British government took over responsibility for India from the East India Company in 1858.Training & Placement 4. while that related and persuasive lobby. Which statement Weakens or Strengthens the Argument? Any Statement that goes Contrary to the assumption or to the Supporting Argument weakens it. To think imperially was to think in terms of restrictive and protective measures. “The Change in the Public Mood is Noticeable”. the supreme tactician of the Indian Pondicherry Engineering College 88 . would place faith accomplishment before the country and commit blunders of incalculable consequence. Reading Comprehension 4. also had a vested interest in imperial expansion. What is the assumption of the Passage? Which of the statements weakens the Argument? Which of the statements strengthens the Argument? Whether a Particular statement can be deduced from the passage or not. However. that Britain could not long continue to rule India expect with the co-operation of many sections of its population. Who was the supreme tactician of the Indian liberation movement? (a) Mrs. the foundation of British authority in India would crumble.2 Question to Answer: Read the following passage and pick up correct answers for each of the questions which follow: Pondicherry Engineering College 89 .Training & Placement liberation movement. Once that cooperation was withdrawn. Answer : ( D) 4. What does the term authoritarian rule mean? (a) Rule of the dictionary of law (b) Dictatorial rule of an aristocrat un-accomplished by the rule of law (c) Arbitrary exercise of power by officials (d) Rule having stability (e) None of these Answer : ( B) 2. the Indian nationalist leaders were able to exploit the aversion of the British liberal conscience to methods used by the local colonial rulers in combating Indian non co-operation. 1. namely. What according to you would be the most suitable title for this passage? (a) British imperialism and India (b) British liberals attitude towards imperialism (c) Role of mahatma Gandhi in Indian freedom movement (d) The emotional appeal of British imperialism (e) British as a colonial power. He saw what some perceptive British thinkers had much earlier recognized.7. What according to the author was the attitude of the British liberals towards the British imperialist and colonial policy? (a) One of active co-operation (b) One of only verbal co-operation (c) One of total indifference (d) One of repeated protests (e) One of disagreement Answer : ( D) 3. Furthermore. Annie Besant (b) The enlightened British liberals themselves (c) Lokmanya Balgangadhar Tilak (d) Mahatama Gandhi (e) None of these Answer : ( D) 4. Standardization is the development and adoption of standards. (b) Accomplished by the Person who is best informed about the functions (c) The responsibility of the people who are to apply them. the construction of standards to which the performance of job duties should conform is most often. (a) Work of people responsible for seeing that the duties are properly performed. a processing standard that requires the use of materials that cannot be procured is most likely to be (a) Incomplete (b) Unworkable (c) Inaccurate (d) Unacceptable 2. artistry of Means. According to the above paragraph. they cannot be considered to be satisfactory. Pondicherry Engineering College 90 . which may save many detours and delays: it is the middle way the golden Mean. According to the above paragraph. accurate. when standards call for finer tolerances than those essential to the conduct of successful Production operations the effect of the standards on the improvement of Production operations. Virtue. Standards must not only be correct. (d) Attributable to the efforts of various informed Persons. Passage 2: The chief condition of happiness. The qualities of character can be arranged in triads in each of which they first and the last qualities will be extremes and vices. but they must also be workable in the sense that their usefulness is not nullified by external conditions. According to the above paragraph. When they are formulated.Standards which do not meet certain basic requirements become a hindrance rather than an aid to progress.will depend on clear judgment . and practice in requiring no more and no less than what is needed for satisfactory results. symmetry of desire . it is not the possession of the simple man. 1. If they are not acceptable. standards are not usually the Product of a single Person. barring certain physical prerequisites. So between cowardice and rashness is courage: between stinginess and extravagance is liberality. although they may posses all the other essential characteristics. nor the gift of innocent intent. between sloth and greed is ambition. leavened with the knowledge and information which are currently available . Yet there is a road to it. is the life of reasons – the specific growth and power of man. but the achievement of experience in the fully developed man. self control. and the middle quality a virtue or an excellence. Which one of the following is the most suitable title for the above paragraph? (a) The evaluation of formulated standard (b) The attributes of satisfactory standards (c) The adoption of acceptable standards (d) The use of process or Product standards. but represent the thoughts and ideas of a group. or rather excellence . (a) Negative (b) Negligible (c) Nullified (d) Beneficial 4. Standards should also be acceptable to the people who use them.Training & Placement Passage 1: A standard comprises characteristics attached to an aspect of a process or Product by which it can be evaluated. a guide to excellence. 3. and of a large class of subjects. Which of the following is the most suitable title for the passage? (a) Qualities of Character (b) Chief Condition of happiness (c) Golden Mean (d) None of these Passage 3: To a greater or lesser degree all the civilized communities of the modern world are made up of a small class of rulers. and where there is not at least the considerable degree of non attachment in activity. in ethics or contact is not different from right in mathematics or engineering . (b) Middle path avoids delay in achieving excellence. (c) Courage is the middle path of indecisiveness and impulsiveness (d) None of the above. it Means correct and fit what works best to get the best results. Participation in a social order of this kind makes it very difficult for individuals to achieve that non attachment in the midst of activity. corrupted by too much passive and irresponsible obedience. What is the main idea of passing? (a) The qualities of character are there – extremes and middle (b) In some respects ethics and mathematics resemble (c) Happiness can be achieved by following the middle path (d) None of these 2. between Hamlets‟ indecisiveness and Quixote‟s impulsiveness is self control „Right‟.Training & Placement between humility and pride is modesty. which is the distinguishing mark of the ideally excellent human being. The author has not said: (a) The middle path between humility and pride is modesty. Who of the following is not the writer of either Hamlet or Don Quixote? (a) Ben Johnson (b) Shakespeare (c) Cervantes (d) None of these 6. between moroseness and buffoonery is good humour between quarrelsomeness and flattery is friendship. corrupted by too much power. A bad social order is one that Pondicherry Engineering College 91 . right in ethics means that works to get the best results. What is the implied Meaning of the passage? (a) Happiness depends upon physical and mental qualities (b) Self control is necessary (c) Excellence should be achieved (d) Rational approach lies in following the middle path 3. between secrecy and loquacity is honesty. Which of the following is not the middle path of different qualities? (a) Liberality (b) Ambition (c) Friendship (d) Secrecy 5. 4. the ideal society of the Profits cannot be realized. A desirable social order is one that delivers us from avoidable evils. 1. (b) Power is of no significance. 2. 4. 3. so dear to advance thinkers are not in themselves Sufficient to produce desirable changes in the character of the society and of the individuals composing it unless carried out by the right sort of Means and in the right sort of governmental administrative and educational context. (d) Economics is necessary. (d) Reforms are necessary for freeing us from power and obedience. our system of education and our metaphysical and ethical beliefs 1. Our present Business is to discover what large scale charges are best calculated to deliver us from the evils of too much power and off too much passive and irresponsible obedience. The passage implies (a) Power and obedience corrupt the society. (c) Changes in society are natural. The main idea of the passage is (a) Ideal society is one which is free from evils. The author does not say (a) By participating in this kind of society one cannot remain non–attached. such reforms are either fruitless or actually fruitful evil. Pondicherry Engineering College 92 . What is the inference that one derives? (a) Changes in the character of society are necessary (b) Ideal society is envisage by Profits (c) It is the mark of ideally excellent human beings (d) None of these. Who do you think the writer is? (a) Capitalist (c) Anarchist 5. would never rise. our methods of public administration. we must change our machinery of government. (b) Man‟s participation is necessary (c) No–attachment is the distinguishing mark. In order to create the proper contexts for economic reform.Training & Placement leads us into temptation which. (b) Socialist (d) None of these. if matters were more sensibly arranged. It has been shown that the economic reforms. (b) Subjects indulge in obedience (c) Bad social order leads to temptations (d) None of these.