Quantitative AptitudeTime and Distance Importance in CAT • Time and Distance is broadly classified under Arithmetic. • It is one of the more important chapters with respect to CAT as well as other MBA entrances. • Over the years, average weightage of Time and Distance in CAT has been 10% speed and distance Average speed Acceleration/deceleration Relative speed Concept of Clock gaining and losing time Speed-time graphs. .Key Concepts • • • • • • Relation amongst time. Problems based on gaining and losing time • Problems based on escalators • Problems based on graphs .Problems related to motion of minute hand and second hand in clocks -.Relative speed in linear motion -.Types of Questions frequently asked • Calculation of average speed/time/total distance • Problems using concept of relative speed -.Questions based on race (contest of speed) • Clocks -.Problems related to trains -.Relative speed in circular motion -.Problems related to speed of boats and streams -. -.Important formulae/properties – Time and Distance • Speed = distance time speed time total distance travelled = total time taken • Acceleration = • Average speed • Savg = • Savg = 𝑛 1 1 1 1 : : : … : s1 s2 s3 s𝑛 (time for different distances is not constant) (time constant) s1 :s2 : …:s𝑛 𝑛 • Relative Speed: For two bodies A and B are travelling with speeds a and b resp.moving towards each other.moving away from each other. relative speed is |a – b| . relative speed is a + b -. then A can cover y metres in x minutes.r.when moving in opposite direction = circumference/(a + b) • Race If A beats B by x minutes or y metres.t hour hand = 5½° per min .Speed of minute hand = 6° per minute -.Speed of hour hand = ½ ° per minute -. time taken to meet for the 1st time -. • Clocks: The time taken by the minute hand to gain x minute spaces over the hour hand = (60/55)x = (12/11)x minutes -.Relative speed of minute hand w.when moving in the same direction = circumference/(|a – b|) -.Important formulae/properties – Time and Distance If A and B are moving on a circular track. -.If a speed time graph is a straight line horizontal to the x-axis.They also make an angle of 0° eleven times and 180° eleven times in a span of 12 hours • Speed – Time Graphs -. then the graph shows an accelerating/decelerating graph The graph is accelerating one if speed increases with time and is a decelerating one if speed decreases with time.If a speed time graph is a straight line making a certain slope with the axes. then speed remains constant with increasing time.The hands of the clock make an angle of 90° twenty two times in a period of 12 hours -.Important formulae/properties – Time and Distance -. . Savg = 48 km/hr . 60 km/hr and 80 km/hr respectively.Examples – Average Speed • If Mike covers a distance of 300 km in three stretches of 100 km each with speeds 30 km/hr. Let us recall the formula for average speed when time is not constant 𝑛 Savg = 1 1 1 1 s1 s2 s3 : : : … :s 𝑛 Savg = 3 1 1 1 : : 30 60 80 = 3×30×60×80 80×60 : 30×80 : 60×30 = 48 Hence. then what is the average speed of Mike throughout the journey? Solution: It is given that Mike travels 100 km each with varying speeds. Thus time to travel each stretch of 100 km is different. Thus Ram and Shyam meet at 10:10 a. Shyam (running at a speed of 10 km/hr.5 km = 2. (∵ distance = speed × time) At this point. At what time do Ram and Shyam first meet each other? Solution: The distance between points A and B is 5 km. .m. Ram and Shyam are moving towards each other and hence their relative speed is (10 + 5) = 15 km/hour. for the first time. or 15 minutes after Shyam started from A). Recall that time required = distance travelled/speed Time taken to cover 2.m. Ram starts at 9 a. at 10. and returns to A at the same speed.e.m.5 km at a combined speed of 15 km/hr. Both of them cover a total of 2.m. reaches B and comes back to A at the same speed. So.m.m. from A at a speed of 5 km/hr.. Ram starts at 9 a. from A at a speed of 5 km/hr. Shyam starts at 9:45 a. 5 km apart.5 × 60/15 = 10 minutes. from A at a speed of 10 km/hr. When Ram reaches B (i.00 a.5 km away from A.Examples – Relative Speed in Linear Motion • Ram and Shyam run a race between points A and B.) is 15/60 × 10 = 2. he reaches B at 10:00 a. reaches B. (Note: Assume all trains to be of negligible length. Hence. and that between the faster train and the third train will be 75 km.Examples – Relative Speed . and 150 km/hr. Since. and were headed in the same direction.Trains • Two super-fast trains. contd. It crossed the faster train an hour and a half after it crossed the slower one. in 30 minutes they will each travel 60 km and 75 km respectively.) Solution: The speeds of the two trains are 120 km/hr. Relative Speed = Relative Distance Time ⇒ Time = Relative distance Relative speed Assume that the time needed for the third train to cross the slower train is x km/hr. Thirty minutes later. in 30 minutes. Find the third train’s speed. the relative distance between the third train and the slower train will be 60 km. Hence. another super-fast train left Beijing Station in the same direction as the previous two. . Assume that the speed of the third train is u km/hr. left the Beijing Railway Station at exactly the same time. which travel at speeds of 120 km/hr and 150 km/hr. 5 𝑢.120 ∴ 𝑢 − 150 1.120 60:1.Examples – Relative Speed .5𝑢2 − 420𝑢 + 27000 = 0 ∴ 𝑢2 − 280𝑢 + 18000 = 0 ∴ 𝑢 − 100 𝑢 − 180 = 0 ∴ 𝑢 = 100 or 𝑢 = 180 Since the third train is faster than both the other trains.5 𝑢−120 = 75 𝑢.120 = 75𝑢.5 75 60 :1. its speed must be 180 km/hr.5𝑢. .120 60 Thus.9000 1. the third train crosses the second train (faster one) 1.5 hours after it crosses the slower train. 𝑢 − 120 = 𝑥 Also.5𝑢 − 120 = 75𝑢 − 9000 ∴ 1.Trains ∴ 𝑥 = 60 𝑢. ∴ 𝑢 − 150 = ∴ 𝑢 − 150 = 75 𝑥:1. Similarly. if t2 is the time for which the boat travels upstream in the second case. . while going downstream. it can go 40 km upstream and 55 km down-stream.Examples – Relative Speed – Boat and Stream • A boat goes 30 km upstream and 44 km downstream in 10 hours. While going upstream. the boat moves against the stream and hence the resultant relative speed is (x – u) km/hour. Suppose that the time it takes to travel 30 km upstream is t1 ∴ (𝑥 – 𝑢) = 30 𝑡1 and 𝑥 + 𝑢 = 44 10 –𝑡1 Similarly. the relative speed is (x + u) km/hour. In 13 hours. The speed of stream is: Solution: Assume that the speed of the boat in still water is x km/hour and the speed of the stream is u km/hour. The boat goes 30 km upstream and 44 km downstream in 10 hours. then contd. . we get 16𝑡1 3 2 – 5𝑡1 = 2 ⇒ 𝑡1 = 6 hours. the speed of the stream is 3 km/hour. ∴(x – u) = 5 and (x + u) = 11 On solving the two above obtained equations for u. we can get a relation between t1 and t2 𝑡1 𝑡2 = 3 4 ⇒ 𝑡2 = 55 4𝑡1 3 … (1) Comparing the R.S. Thus. of the two downstream equations.H. we get u = 3 km/hour.H. we get 44 10 –𝑡2 = 13 –𝑡 This on further simplification gives 4t2 – 5t1 = 2 Substituting the value of t2 from (1) in the equation above. of the two upstream equations.S.Examples – Relative Speed – Boat and Stream (𝑥 – 𝑢) = 40 𝑡2 and 𝑥 + 𝑢 = 13 –𝑡 55 2 Comparing the R. so that they both reach the finish line together? Solution: Since the radius of the circular track is 14 metres. its circumference = 2 × × 14 = 88 7 Let the speeds of the faster and slower runners be SF and SS respectively. Now.• Examples – Relative Speed in Circular Motion and Race Two runners are running on a circular track of radius 14 metres. the two runners run on a straight 100 metre track in the same direction. When the two runners start running simultaneously in the same direction. Relative speed. when they run in the same direction. 22 Relative speed. 𝑆𝐹 + 𝑆𝑆 = 88 10 = 8. they meet every 10 seconds. . they meet each other every 22 seconds. they meet each other every 10 seconds. then how much head-start (in metres) must the faster runner give the slower one. they meet every 22 seconds. Hence.8 ∵ relative speed = circumference 𝑥 : 𝑦 … (ii) contd. Hence. 𝑆𝐹 − 𝑆𝑆 = 88 22 =4 (∵ relative speed = circumference ) 𝑥 – 𝑦 … (i) When they run in opposite directions. Now. If the runners run at the same speed as before. When they start running simultaneously in opposite directions. 4 6.4 = 2.4 ∴ 𝒙 = 𝟏𝟎𝟎 = 𝟔𝟐.4 𝑥 100 Distance covered by the slower runner after he is given the head start Distance covered by the faster runner ∴ ∴ ∴ = 100.4 6. 𝟓 𝐦𝐞𝐭𝐫𝐞𝐬 .4∴ SS = 8.4 Now.𝑥 100 𝑥 100 2. the time for which the two runners run before reaching the finish will be the same. Let x be the head-start that the faster runner gives the slower one.8 – 6.Examples – Relative Speed in Circular Motion and Race Solving equation (i) and (ii) simultaneously. Hence. we get. 𝑆𝑆 𝑆𝐹 = 2.4 2.4 6.8/2 = 6.4 =1− =1− 𝟒 × 𝟔.𝟒 = 4 6. Once the slower runner is ahead of the faster one by x metres. SF = 12. we move on to the second part of the question. Hence.Examples – Race • A hare and a turtle decided to race each other. Let the hare’s speed be x km/min. then ran back another 2d/3 km to the oak tree. slow but steady and honest. won the race by 8 minutes. The turtle. ∴ Total distance covered by the hare = 2𝑑 3 + 2𝑑 3 + 𝑑 = 7𝑑 3 7𝑑 3 7𝑑 3𝑥 ∴ Time taken by the hare to complete the race = 𝑥 = minutes contd. the race beginning at an oak tree and finishing at a pine tree. the hare got caught cheating and was forced to hop back to the oak tree and start again. However. . then finally ran d km from the oak to the pine tree. after it had already covered 2/3rd of the total distance. the hare ran 2d/3 kilometres. The turtle’s speed was only half of that of the hare. got caught cheating. so the turtle’s speed = x/2 km/min. How long did the hare take to complete the race? Solution: Let the total distance from the oak tree to the pine tree be d kilometres. hence (Time taken by the hare to complete the race) – (Time taken by the turtle to complete the race) = 8 ∴ ∴ 7𝑑 3𝑥 𝑑 𝑥 − 2𝑑 𝑥 = 𝑑 7 𝑥 3 −2 = × =8 𝑑 𝑥 1 3 = 24 minutes 𝟕 𝒅 𝟕 ∴ 𝐓𝐢𝐦𝐞 𝐭𝐚𝐤𝐞𝐧 𝐛𝐲 𝐭𝐡𝐞 𝐡𝐚𝐫𝐞 𝐭𝐨 𝐜𝐨𝐦𝐩𝐥𝐞𝐭𝐞 𝐭𝐡𝐞 𝐫𝐚𝐜𝐞 = × = × 𝟐𝟒 𝟑 𝒙 𝟑 = 𝟓𝟔 𝐦𝐢𝐧𝐮𝐭𝐞𝐬 .Examples – Race Time taken by the turtle to complete the race = 𝑥 = 2 𝑑 2𝑑 𝑥 Since the turtle beat the hare by 8 minutes. m. contd.72 minutes ≈ 12 minutes 43 seconds.m. x minutes after 6:00 pm. Returning before 7:00 p.5x = 110° ∴ x ≈ 12. . observes that the hands of his watch form an angle of 110°.5x) – (6x) = 110° ∴ 180 – 5. The minute hand is behind the hour hand. The number of minutes that he has been away is: Solution: When the hour hand and minute hand form an angle of 110°. we have two possibilities 1. The speed of the minute hand is 6° per minute and the speed of the hour hand is 0. he notices that again the hands of his watch form an angle of 110°.5° per minute.Examples – Clocks – Angle between hour hand and minute hand • A man on his way to dinner shortly after 6:00 p. Initial distance between the hour and the minute hands at 6:00 pm is 180° ∴ (180 + 0. 5x) = 110° ∴ 5.72 minutes ≈ 52 minutes 43 seconds ∴ The man leaves at 06:12:43 pm and returns at 06:52:43 pm ∴ He is away for 40 minutes. ∴ 6x – (180 + 0. .Examples – Clocks – Angle between hour hand and minute hand 2.5x – 180 = 110° ∴ x ≈ 52. The minute hand is ahead of the hour hand. Time on Sangeeta’s watch Actual time 62 minutes --------------> 60 minutes 620 minutes --------------> 600 minutes contd.m. According to Sangeeta’s watch.40 a.40 a.m. IST. After purchasing they found that when 60 minutes elapses on a correct clock (IST).m.m. that is.. . her watch has shown time elapsing by 10 hours. it was now 10 p. Sangeeta’s watch shows time elapsing by 62 minutes. 20 minutes (= 620 minutes). It is given that when the actual time elapses by 60 minutes. Sangeeta’s wristwatch registers 62 minutes whereas Swati’s wristwatch registers 56 minutes.. Later in the day Sangeeta’s wristwatch reads 10 p.Examples – Clocks – Gaining and Losing Time • Sangeeta and Swati bought two wristwatches from Jamshedpur Electronics at 11. then the time on Swati’s wristwatch is: Solution: It was 11. when Sangeeta and Swati bought the watches. Hence. 00 p. Swati’s watch shows time elapsing by 56 minutes.Examples – Clocks – Gaining and Losing Time It is given that when the actual time elapses by 60 minutes.40 a.m. 20 mins) = 9. the time on her watch is 11. . the time on Swati’s watch elapsed by (600 × 56)/60 = 560 minutes Hence. Time on Swati’s watch Actual time 56 minutes --------------> 60 minutes 560 minutes --------------> 600 minutes Hence. + (9 hrs.m. Hence. 18 p. The minute hand has to gain 150° − 90° = 60° over the hour hand to make a right angle with it.54 p.m. The minute and hour hands will also form a right angle when the minute hand gains 150 + 90 = 240 degrees over the hour hand. Time = 240 52 1 = 480 min 11 = 43 7 11 min This means that the hands of the watch will at right angles for the second time at 5:43:38.Examples – Clocks – Relative Speed • At what time between 5 o’clock and 6 o’clock will the hands of a watch be at right angles? Solution: At 5 o’clock. the angle between the minute hand and the hour hand is 150 degrees. Time = 60 1 52 = 120 min 11 = 10 10 11 min This means that the hands of the watch will at right angles for the first time at 5:10:54.m. . Shyam gets to the top of the escalator after having taken 25 steps. ∴ Shyam takes 6/3 = 2 seconds for a step and Yom takes 6/2 = 3 seconds for a step ∴ Shyam took 25 × 2 = 50 seconds to go up ∴ Height of the stairway = (25 + 50x) steps Yom took 20 × 3 = 60 seconds to go up Similarly. they would have to take (20 + 60 × 1/2) = 50 steps . While Yom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. how many steps would they have to take to walk up? Solution: Assume that Shyam takes 3 steps and Yom takes 2 steps in 6 seconds and let the escalator moves up by x steps per second. If the escalator were turned off. Shyam takes three steps for every two of Yom's steps. 20 + 60x = 25 + 50x ∴ x = 1/2 ∴ If the escalator was turned off.Examples . in Yom's case. the height of the stairway = (20 + 60x) steps As both of them reached the top of the escalator together. The escalator moves at a constant speed.Escalators • Shyam and Yom walk up an escalator. The red line depicts a graph of constant speed. the red line depicts a graph wherein the speed is increasing at a constant rate i.Speed – Time Graphs Assume that speed is defined on the y-axis and time is defined on the x-axis. The deceleration is uniform. In this case.e. the red line depicts a graph wherein the speed is decreasing at a constant rate i.e. . The acceleration is uniform. In this case. .Speed – Time Graphs The red curve in this case is a parabolic curve which shows that the increase in speed is non uniform but the acceleration is uniform. The red curve in this case is a parabolic curve which shows that the decrease in speed is non uniform but the deceleration is uniform.