AptitudeB2A Basics to Advanced

Time Speed & Distance A descriptive book for all competitive exams

Neeraja Anand V Pragadeeswara Prabhu CN

NOTION PRESS

AptitudeB2A Basics to Advanced – Time Speed & Distance NOTION PRESS India. Singapore. Malaysia. ISBN 9781637811207 This book has been published with all reasonable efforts taken to make the material error-free after the consent of the author. No part of this book shall be used, reproduced in any manner whatsoever without written permission from the author, except in the case of brief quotations embodied in critical articles and reviews. The Author of this book is solely responsible and liable for its content including but not limited to the views, representations, descriptions, statements, information, opinions, and references [“Content”]. The Content of this book shall not constitute or be construed or deemed to reflect the opinion or expression of the Publisher or Editor. Neither the Publisher nor Editor endorse or approve the Content of this book or guarantee the reliability, accuracy or completeness of the Content published herein and do not make any representations or warranties of any kind, express or implied, including but not limited to the implied warranties of merchantability, fitness for a particular purpose. The Publisher and Editor shall not be liable whatsoever for any errors, omissions, whether such errors or omissions result from negligence, accident, or any other cause or claims for loss or damages of any kind, including without limitation, indirect or consequential loss or damage arising out of use, inability to use, or about the reliability, accuracy or sufficiency of the information contained in this book.

AptitudeB2A Basics to Advanced – Time Speed & Distance Dedicated to: All the students who came to us with doubts. We were able to explain the answers in an elaborate way to make sure everyone understands. Thank you, students. Why should you learn from this book? Overcome these issues: • “Not understanding the question”. • “Not knowing the method to solve”. • “Not knowing the type of question”. • “Taking too much time to solve the question”. The topic Time, speed and Distance and its subtopics like trains, boats, races, escalators, and elevators holds more weightage in many competitive exams. The reason for providing a comprehensive book on a certain topic is to make the learners understand over 240 different model questions from this topic in detail. The questions provided in this book are the most repeated models from the topic Time, Speed and Distance. The answers for the questions are explained in a detailed manner, keeping in mind the various doubts asked by the thousands of students when the authors delivered the session on this topic over a cumulative period of 16+ years. The goal of this book is to make sure every reader understands the methods in depth to solve the questions from this topic. The descriptive explanations not only help a learner to solve that model but also makes him/her to understand twisted questions and solve them easily. Level of difficulty of the questions range from basics to advance with most questions having additional methods (shortcuts) to solve them.

All the best -Neeraja Anand V & Pragadeeswara Prabhu CN Author Information Neeraja Anand V

Pragadeeswara Prabhu CN

Quantitative Aptitude and Logical Reasoning Trainer since 2016.

Quantitative Aptitude and Logical Reasoning Trainer since 2011.

Founder, Newbie’s Career Masters, Karnataka.

Co-Founder, Channel B.Tech, Tamil Nadu.

[email protected]

[email protected]

AptitudeB2A Basics to Advanced – Time Speed & Distance

AptitudeB2A Basics to Advanced – Time Speed & Distance Table of contents 1.

2.

3.

4.

5.

6.

7.

8.

Problems based on formulae 1.1. Level Easy 1.2. Level Moderate 1.3. Exercise 1.4. Detailed Solutions Problems based on Proportionality 2.1. Level Easy 2.2. Level Moderate 2.3. Exercise 2.4. Detailed Solutions Problems based on Unit conversion 3.1. Level Easy 3.2. Level Moderate 3.3. Exercise 3.4. Detailed Solutions Problems based on equations 4.1. Level Easy 4.2. Level Moderate 4.3. Level Hard 4.4. Exercise 4.5. Detailed Solutions Average speed 5.1. Level Easy 5.2. Level Moderate 5.3. Level Hard 5.4. Exercise 5.5. Detailed Solutions Relative Speed 6.1. Level Easy 6.2. Level Moderate 6.3. Level Hard 6.4. Exercise 6.5. Detailed Solutions Problems based on trains 7.1. Level Easy 7.2. Level Moderate 7.3. Level Hard 7.4. Exercise 7.5. Detailed solutions Problems based on boats and streams 8.1. Level Easy 8.2. Level Moderate 8.3. Level Hard 8.4. Exercise 8.5. Detailed Solutions

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1 1 1 2 3 5 5 7 8 9 12 12 12 13 14 15 15 16 16 17 19 25 25 26 26 27 28 30 30 31 33 34 36 46 47 48 51 52 53 57 57 58 60 61 63

AptitudeB2A Basics to Advanced – Time Speed & Distance 9.

10.

Problems based Elevators and Escalators 9.1. Level Easy 9.2. Level Moderate 9.3. Exercise 9.4. Detailed Solutions Problems based on races 10.1. Level Easy 10.2. Level Moderate 10.3. Level Hard 10.4. Exercise 10.5. Detailed Solution

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69 69 71 72 74 78 78 79 82 85 86

AptitudeB2A Basics to Advanced – Time Speed & Distance Explanation: Given, time = 3 hours and distance = 135 miles D We know that S = T

1. Problems based on formulae Speed is a parameter used to represent the distance covered by a certain object over unit time. It could be represented in various units like kilometres per hour (kmph), miles per hour (mph) or meters per second (mps) etc. In this complete chapter, we will represent ‘speed’ using the term S, ‘distance’ using the term D and ‘time’ using the term T. D S= T Assume that a person has travelled 200 miles in 5 hours. When you find the distance covered by him in one hour, you get the speed of that person. 200 S= = 40 mph 5 Since the distance is given in miles and the time is given in hours, the unit for speed in the above case is miles per hour or mph. The above formula can be rewritten to calculate time and distance as shown below. To find the time taken to cover a certain distance at certain speed: D T= S To find the distance covered at certain speed for certain time: D =S×T

Therefore, required speed =

100

Therefore, required time = 2 = 50 sec. Hence, the total distance covered by her (in meters) = (150 + 100) = 250 meters The total time taken to cover 250 meters = (300 + 50) = 350 seconds.

Example 5: A train started from station X at 10:00 am and reached station Y at 01:00 pm the same day. What was the speed of the train if the distance between the two stations is 165 miles? Explanation: Given, distance = 165 mph and time From 10:00 am to 01:00 pm, duration = 3 hours 𝐷 We know that 𝑆 = 𝑇

Sonya walks at a speed of 2.4 mph. How much time will she take to walk 14.4 miles? Explanation: Given, speed = 2.4 mph and distance = 14.4 miles D We know that T = S 2.4

=

144 24

= 45 mph.

Hannah walked for 5 minutes at a speed of 0.5 meters per second and then covered 100 meters by running at a speed of 2 meters per second. What is the total distance covered by her (in meters) and the total time taken by her (in seconds)? Explanation: Step 1: Find the distance covered in 5 minutes at a speed of 0.5 mps 5 Time = 5 × 60 = 300 sec and speed = mps 10 We know that 𝐷 = 𝑆 × 𝑇 5 Required distance = 300 × 10 = 150 meters Step 2: Find the time taken to cover 100 meters by running at a speed of 2 mps. D We know that T = S

Example 1:

14.4

3

Example 4:

1.1. Level Easy

Therefore, required time =

135

= 6 hours.

Required speed =

Example 2:

165 3

= 55 mph.

1.2. Level Moderate

A flight leaves New York at 04:00 am and reaches Vegas at 09:00 am. If the flight did not stop in between and was flying at a constant speed of 300 mph, how far is Vegas from New York? Explanation: Given, speed = 300 mph and time (from 04:00 am to 09:00 am) = 5 hours We know that 𝐷 = 𝑆 × 𝑇 Required distance = 300 × 5 = 1500 miles.

Example 6: Mike leaves home to office every day at 08:00 am and reaches the office 10 miles away from his home at 09:40 am. On a particular day, he started 20 minutes late from home. By how much should he increase the speed so that he will reach at the usual time? Explanation: Given, distance = 10 miles Duration from 08:00 am to 09:40 am 5 = 1 hour 40 minutes = hours

Example 3: Justin drove a car for 3 hours and reached 135 miles. What should have been his speed to achieve this distance?

3

40

40 minutes = 60 hours. Therefore 1 hour 40 minutes 40 2 5 = (1 + ) = (1 + ) = hours 60

1

3

3

AptitudeB2A Basics to Advanced – Time Speed & Distance We know that 𝑆 =

𝐷 𝑇

Therefore, usual speed =

10 5 3

a. Statement 1 alone is true. b. Statement 2 alone is true. c. Both statements are true. d. Neither statements are true. Explanation: Given, usual speed = 2 mps. Usual time = 40 minutes = 40 × 60 = 2400 seconds We know that 𝐷 = 𝑆 × 𝑇 Therefore, total distance = 2 × 2400 = 4800 m. From statement 1: Speed = (2+3) = 5 mps 𝐷 We know that 𝑇 =

= 6 mph.

Now, distance = 10 miles Time = 1 hour 40 minutes – 20 minutes 4 = 1 hour 20 minutes = 3 hours 20

20 minutes = 60 hours Therefore 1 hour 20 minutes 20 1 4 = (1 + 60) = (1 + 3) = 3 hours Therefore, required speed = 1

10 4 3

=

30 4

1

= 7 mph. 2

1

Change in speed = (7 2 − 6) = 1 2 mph

4800

1

Hence, he should increase the speed by 1 2 mph to reach at the usual time.

Example 7: Jane travelled the first quarter of the total distance to reach school at a speed of 2 mph. Her speed for the next two quarters was 6 mph and her speed for the last quarter was 4 mph. If the total distance is 12 miles, what is the time taken by her to reach school? Explanation: Given, total distance = 12 miles Each quarter of the total distance 1 = 12 × 4 = 3 miles. Total time taken to reach school = Time taken to cover the first quarter + Time taken to cover the second and the third quarter + Time taken to cover the last quarter. 𝐷 We know that 𝑇 = 𝑆 Total time taken to cover the first quarter i.e., 3 miles at a speed 2 mph 3 1 = 2 = 1 2 hours = 1 hour 30 minutes. 1

1

1.

hours = 2 × 60 = 30 minutes 2 1

Therefore 1 2 hours = 1 hour 30 minutes Total time taken to cover next two quarters 6 i.e., (3+3) = 6 miles at a speed 6 mph = 6 = 1 hour. Total time taken to cover the last quarter 3 i.e., 3 miles at a speed 4 mph = hours = 45 minutes. 3

3

𝑆

Total time = 5 = 960 seconds. Change in time = (2400 − 960) = 1440 seconds 1440 = = 24 minutes. 60 When his speed is increased by 3 mps, he would take 24 minutes less. Statement 1 is true. From statement 2: Given speed = 2 mps till first 1200 meters, and speed = (2+2) = 4 mps for the remaining 3600 meters 𝐷 We know that 𝑇 = 𝑆 Time taken to cover the first 1200 meters 1200 = 2 = 600 seconds. Time taken to cover the remaining 3600 meters 3600 = = 900 seconds. 4 Total time taken = (600+900) = 1500 seconds 1500 = 60 = 25 minutes. When his speed is increased by 2 mps after crossing 1200 meters, he would take 25 minutes for the whole journey. Statement 2 is true. Since both statements are true, option (c) must be marked as correct answer.

2.

4

[ 4 hours = 4 × 60 = 45 minutes, 3

therefore hours = 45 minutes] 4 Therefore, required total time = 1 hour 30 minutes + 1 hour + 45 minutes = 3 hours 15 minutes.

3.

Example 8: A man walking at a speed of 2 mps has reached his destination in 40 minutes. Statement 1: If his speed were increased by 3 mps, he would have taken 24 minutes less. Statement 2: If his speed were increased by 2 mps after crossing 1200 meters, he would have taken 25 minutes for the whole journey.

4.

2

1.3. Exercise How much distance can a man cover at a running speed of 3 mps for 30 seconds? a. 90 m b. 45 m c. 30 m d. 60 m e. 75 m What is the speed of a car that travels 120 miles in 6 hours? a. 30 mph b. 25 mph c. 24 mph d. 20 mph e. 15 mph How long will it take for Charlie to cover 40 miles at a speed of 2.5 mph? a. 20 hours b. 18 hours c. 16 hours d. 12 hours e. 10 hours A bird flies 240 m towards east at a speed of 20 mps and then returns to the starting point at a speed of 16 mps. What is the total duration for which the bird has flown? a. 21 seconds b. 24 seconds c. 27 seconds d. 30 seconds e. 33 seconds

AptitudeB2A Basics to Advanced – Time Speed & Distance 5.

Speed of a train from station A to B is 40 mph. Its speed from station B to C is half of the speed from A to B. Its speed from station C to D is double the speed from A to B. If the distances from A to B, B to C and C to D, are same, time taken from B to D is how many times the time taken from A to B? a. 2 times b. 2.5 times c. 3 times d. 3.5 times e. 4 times 6. Assume that the speed of Alan is ‘m’ and the speed of Bob is ‘n’. They both travelled for ‘t’ minutes each. What is the difference between the distances travelled by each if Alan is faster than Bob? a. (m-n)/t b. t/(m-n) c. mt-n d. m-nt e. t(m-n) 7. Distance travelled by a woman at the rate of 3 mps is 360 meters. Distance travelled by a man at a speed of 2.5 mps is 450 meters. If the time taken by the woman and the time taken by the man are interchanged and their distances are unaltered, what would be the difference between their altered speeds? a. 0.50 mps b. 1.00 mps c. 1.25 mps d. 1.75 mps e. 2.00 mps 8. What is the ratio between the distances covered by Alex, Jack and Ryan if their speeds are in the ratio 1 : 2 : 3 and the durations for which they have travelled are in the ratio 4 : 3 : 2 in that order? a. 4:1:1 b. 2:6:6 c. 2:3:3 d. 4:3:3 e. 4:3:1 9. David can walk at a speed of 2 mph. John can walk at a speed of 2.05 mph. If both start together, how long will it take for John to cover 1 mile more than the distance covered by David? a. 20 hours b. 18 hours c. 15 hours d. 10 hours e. 8.33 hours 10. Vince has two friends Bill and Mike. Bill lives on the 20th floor and Mike lives on the 5th floor. Vince is on the 15th floor of the same building. While using the stairs, he takes 40 seconds per floor to climb up and 20 seconds per floor to climb down. If he decides to meet a friend, I. he can meet Bill sooner than he can meet Mike. II. he can meet Mike sooner than he can meet Bill. a. Statement (I) is true b. Statement (II) is true c. Both statements are false d. Cannot be determined based on the given data. Answer keys: 1-a 2-d 6-e 7-d

3-c 8-c

4-c 9-a

1.4. Detailed Solutions 1. Option (a) Given, speed = 3 mps and time = 30 sec We know that 𝐷 = 𝑆 × 𝑇 Therefore, required distance = 3 × 30 = 90 meters. 2.

Option (d)

Given, distance = 120 miles and time = 6 hours 𝐷 We know that 𝑆 = 𝑇

Therefore, required speed = 3.

120 6

= 20 mph.

Option (c)

Given, distance = 40 miles and speed = 2.5 mph 𝐷 We know that 𝑇 = 𝑆 40

Therefore, required time = 2.5 = 4.

400 25

= 16 hours.

Option (c) 𝐷

We know that 𝑇 = 𝑆 Time taken by the bird to fly 240 meters towards east 240 at the speed 20 mps = 20 = 12 seconds. Time taken by the bird to return back 240 meters at the 240 speed 16 mps = 16 = 15 seconds. Therefore, the required total time = 12 + 15 = 27 seconds. 5.

Option (b)

Let 𝑆1 , 𝑆2 , 𝑆3 be the speed of the train from A to B, B to C and C to D respectively and 𝑇1 , 𝑇2 , 𝑇3 be the time taken by the train to cover A to B, B to C and C to D respectively. Assume that the distance from A to B, B to C, C to D as d meters each since the distance from A to B, B to C and C to D is same. Given, 𝑆1 = 40 mph then 𝑆2 = 20 mph and 𝑆3 = 80 mph We know that, 𝐷 𝑇= 𝑆 𝑑 𝑑 𝑇1 = 𝑆 = 40 hours 1

𝑑

𝑑

𝑇2 = 𝑆 = 20 hours 2

𝑑

𝑑

𝑇3 = 𝑆 = 80 hours 3

Time taken from B to D = Time taken from B to C + Time taken from C to D 𝑑 𝑑 𝑑 i.e., 𝑇2 + 𝑇3 = 20 + 80 = 16 Time taken from B to D is, how many times the time taken from A to B, is found by the equation

5-b 10-c

=

3

𝑇2 +𝑇3 𝑇1

=

𝑑 16 𝑑 40

40

= 16 = 2.5 times

AptitudeB2A Basics to Advanced – Time Speed & Distance Alternate solution:

8.

Let 𝑆1 , 𝑆2 , 𝑆3 be the speed of the train from A to B, B to C and C to D respectively and 𝑇1 , 𝑇2 , 𝑇3 be the time taken by the train to cover A to B, B to C and C to D respectively. Given, 𝑆1 = 40 mph then 𝑆2 = 20 mph and 𝑆3 = 80 mph We know that speed and time are inversely proportional. Therefore, the ratio of time taken 1 1 1 = 𝑇1 : 𝑇2 : 𝑇3 = 40 : 20 : 80 [multiply each number with LCM of 40, 20 and 80 i.e., 80] We get, 𝑇1 : 𝑇2 : 𝑇3 = 2: 4: 1 𝑇1 = 2𝑥 hours 𝑇2 = 4𝑥 hours 𝑇3 = 1𝑥 hours Time taken from B to D = Time taken from B to C + Time taken from C to D i.e. 𝑇2 + 𝑇3 = 4𝑥 + 1𝑥 = 5𝑥 Time taken from B to D is, how many times the time taken from A to B, is found by the equation 𝑇 +𝑇 5𝑥 = 2𝑇 3 = 2𝑥 = 2.5 times.

Let 𝐷𝐴 , 𝐷𝐽 , 𝐷𝑅 be the distances covered by Alex, Jack, and Ryan respectively, 𝑆𝐴 , 𝑆𝐽 , 𝑆𝑅 be the speeds of Alex, Jack, and Ryan respectively and 𝑇𝐴 , 𝑇𝐽 , 𝑇𝑅 be the time taken by Alex, Jack and Ryan respectively. Given, 𝑆𝐴 : 𝑆𝐽 : 𝑆𝑅 = 1: 2: 3, Then 𝑆𝐴 = 1𝑥, 𝑆𝐽 = 2𝑥 and 𝑆𝑅 = 3𝑥 𝑇𝐴 : 𝑇𝐽 : 𝑇𝑅 = 4: 3: 2, Then 𝑇𝐴 = 4𝑦, 𝑇𝐽 = 3𝑦 and 𝑇𝑅 = 2𝑦 We know that, 𝐷 =𝑆×𝑇 𝐷𝐴 = 𝑆𝐴 × 𝑇𝐴 = 1𝑥 × 4𝑦 = 4𝑥𝑦 𝐷𝐽 = 𝑆𝐽 × 𝑇𝐽 = 2𝑥 × 3𝑦 = 6𝑥𝑦 𝐷𝑅 = 𝑆𝑅 × 𝑇𝑅 = 3𝑥 × 2𝑦 = 6𝑥𝑦 𝐷𝐴 : 𝐷𝐽 : 𝐷𝑅 = 4𝑥𝑦: 6𝑥𝑦: 6𝑥𝑦 = 4: 6: 6 = 2: 3: 3 Therefore, the ratio between the distances covered by Alex, Jack and Ryan is 2:3:3. 9.

Option (e)

Let 𝑆𝐴 , 𝑆𝐵 be the speed of Alan and the speed of Bob respectively. Given, 𝑆𝐴 = 𝑚 and 𝑆𝐵 = 𝑛 We know that, 𝐷 =𝑆×𝑇 The distance travelled by Alan in t minutes = 𝑚 × 𝑡 The distance travelled by Bob in t minutes = 𝑛 × 𝑡 Given, Alan is faster than Bob which means Alan covers more distance than Bob. Therefore, the required difference = (𝑚 × 𝑡) − (𝑛 × 𝑡) = 𝑡(𝑚 − 𝑛) 7.

Option (d)

Let 𝑇𝑊 be the time taken by the woman to cover 360 meters at the speed 3 mps and 𝑇𝑚 be the time taken by the man to cover 450 meters at the speed 2.5 mps. 𝐷 We know that 𝑇 = 𝑇𝑊 =

360 3 450

Option (a)

We know that 𝐷 = 𝑆 × 𝑇 The distance covered by David after the 1st hour = 2 × 1 = 2 miles. The distance covered by John after the same 1st hour = 2.05 × 1 = 2.05 miles. Now, the difference in their distances covered after the 1st hour = 2.05 − 2 = 0.05 miles which means John has covered 0.05 miles more than David after the 1st hour. Now, the distance covered by David after 2nd hour = 2 × 2 = 4 miles. The distance covered by John after the same 2nd hour = 2.05 × 2 = 4.10 miles. Now, the difference in their distances covered after the 2nd hour = 4.10 − 4 = 0.10 miles which means John has covered 0.10 miles more than David after the 2nd hour. After every hour, John covers 0.05 miles more than David. The time taken for John to cover 1 mile more than the distance covered by David 1 100 = = = 20 hours.

1

6.

Option (c)

0.05

5

10. Option (c)

𝑆

= 120 seconds

Number of floors between Vince and Bill (from 15th floor to 20th floor) = 5 floors. Given, Vince takes 40 seconds per floor to climb up, which means the time taken by Vince to climb up 5 floors = 5 × 40 = 200 seconds. Now, number of floors between Vince and Mike (from 15th floor to 5th floor) = 10 floors. Given, Vince takes 20 seconds per floor to climb down, then the time taken by Vince to climb down 10 floors = 10 × 20 = 200 seconds. Therefore, he takes same time to meet Bill and mike, so the answer is both statements are false.

𝑇𝑚 = 2.5 = 180 seconds Now, 𝑆𝑊 and 𝑆𝑚 be the speed of the woman and the speed of the man after interchanging the time. After interchanging the duration for completing the distances, time allotted for the woman is 180 seconds and the time allotted for the man is 120 seconds. 𝐷 We know that 𝑆 = 𝑇 360

𝑆𝑊 = 180 = 2 mps 450

𝑆𝑚 = 120 = 3.75 mps Difference in altered speeds = 3.75 − 2 = 1.75 mps. 4

AptitudeB2A Basics to Advanced – Time Speed & Distance Explanation: Let 𝑆1 be the former speed of the car and 𝐷1 be the former distance covered by the car. Let 𝑆2 be the latter speed of the car and 𝐷2 be the latter distance covered by the car for the same duration. Given, 𝑆1 = 𝑆 mph, 𝐷1 = 40 miles, 150 3 𝑆2 = 100 𝑆 = 2 𝑆 mph [If x% of S is increased from S, the result is 𝑥 𝑆 + (100 × 𝑆)] We know, Distance covered is directly proportional to speed. 𝐷1 𝑆1 = 𝐷2 𝑆2 40 𝑆 = 3 𝐷2 2𝑆 3 𝐷2 = 40 × 2 = 60 miles. Therefore, the car covers 60 miles at 50% increased speed for the same duration.

2. Problems based on Proportionality There are two kinds of proportionalities. Direct proportion and inverse proportion. Direct proportion between two variables ‘a’ and ‘b’ suggests that when the value of ‘a’ increases, the value of ‘b’ also increases and vice versa. Inverse proportion between two variables ‘a’ and ‘b’ suggests that when the value of ‘a’ increases, the value of ‘b’ decreases and vice versa. Distance covered is directly proportional to speed. For instance, for a constant period ‘T’, if the speed of an object is doubled, the distance covered by the object will also be doubled. Let 𝑆1 be the speed of an object and 𝐷1 be the distance covered by the object for a certain duration. Let 𝑆2 be the speed of an object and 𝐷2 be the distance covered by the object for the same duration. 𝐷1 𝑆1 = 𝐷2 𝑆2 Distance covered is directly proportional to time. For instance, for a constant speed ‘S’, if the duration doubled, the distance covered by the object will also be doubled. Let 𝑇1 be the duration in which an object covered a distance 𝐷1 at speed S. Let 𝑇2 be the duration in which an object covered a distance 𝐷2 at the same speed S. 𝐷1 𝑇1 = 𝐷2 𝑇2 Since distance is directly proportional to speed and time, we can represent this proportionality as mentioned below: 𝐷1 𝑆1 × 𝑇1 = 𝐷2 𝑆2 × 𝑇2 Time taken to cover a certain distance is inversely proportional to the speed of the object. If the speed of the object is increased, the time taken to reach the destination decreases. Let 𝑇1 be the time taken to cover a distance at a speed of 𝑆1 and 𝑇2 be the time taken to cover the same distance at the speed of 𝑆2 . The relations between these parameters can be expressed as: 1 𝑇1 𝑇1 𝑆2 𝑆1 = → = 1 𝑇2 𝑇2 𝑆1 𝑆2

Example 10: At a speed of 50 mph, Alex can reach the destination in 4 hours. At what speed should he travel to reach the destination in 2 hours? Explanation: Let 𝑇1 be the former time taken by Alex to cover a distance at a former speed of 𝑆1 and 𝑇2 be the latter time taken to cover the same distance at the latter speed of 𝑆2 . Given, 𝑆1 = 50 mph, 𝑇1 = 4 hours, 𝑇2 = 2 hours We know, Time taken to cover a certain distance is inversely proportional to the speed of the object. 𝑇1 𝑆2 = 𝑇2 𝑆1 4 𝑆2 = 2 50 𝑆2 = 100 mph. Therefore, the Alex must travel at the speed of 100 mph to reach the destination in 2 hours.

Example 11: Speeds at which James and Ryan jog are the same. If James jog for 20 minutes every morning and Ryan jogs for 25 minutes every morning, what is the ratio between the distances jogged by James and Ryan in that order? Explanation: Let 𝑇1 be the duration in which James has covered a distance 𝐷1 at speed S. Let 𝑇2 be the duration in which Ryan has covered a distance 𝐷2 at the same speed S. Given, 𝑇1 = 20 minutes and 𝑇2 = 25 minutes We know that the distance covered is directly proportional to time. 𝐷1 𝑇1 = 𝐷2 𝑇2

2.1. Level Easy

Example 9: A car can cover 40 miles travelling at a speed of S mph for a certain duration. If the speed of the car is increased by 50%, what will be the distance covered by the car for the same duration?

5

AptitudeB2A Basics to Advanced – Time Speed & Distance 𝐷1 20 4 = = 𝐷2 25 5 𝐷1 : 𝐷2 = 4: 5 The ratio between the distances jogged by James and Ryan is 4:5. Alternate solution: Let 𝑇1 be the duration in which James has covered a distance 𝐷1 at speed S. Let 𝑇2 be the duration in which Ryan has covered a distance 𝐷2 at the same speed S. Given, 𝑇1 = 20 minutes and 𝑇2 = 25 minutes We know that the distance covered is directly proportional to time. 𝐷1 : 𝐷2 = 𝑇1 : 𝑇2 = 20: 25 = 4: 5 Therefore, the ratio between the distances jogged by James and Ryan is 4:5.

Example 13:

Example 12:

𝑆2 = (100 +

When traveling at the usual speed, a train reaches its destination in 12 hours. If the speed of the train is 2 increased by 16 3 %, what will be the time taken for the train to reach the destination? Explanation: Let 𝑆1 be the usual speed of the train and 𝑇1 be the usual time to reach its destination. Let 𝑆2 be the increased speed of the train and 𝑇2 be the time taken to reach the destination at the increased speed. 2 50 [ Note: 16 3 % = 3 % , If x% is increased from A, the 𝑥

resultant is (100 + 100) × 𝐴 ] Given, 𝑆1 = 𝑆1 mph, 𝑇1 = 12 hours,

The ratio between the speeds at which Sam goes to office and return home are in the ratio 5:6. What is the ratio between time taken by him to reach office from home and time taken by him to reach home from office? Explanation: Let 𝑇1 be the time taken by Sam to reach the office at a speed of 𝑆1 and 𝑇2 be the time taken by Sam to return home at the speed of 𝑆2 . Given, 𝑆1 = 5𝑥 and 𝑆2 = 6𝑥 We know that the time taken to cover a certain distance is inversely proportional to the speed of the object. 𝑇1 𝑆2 = 𝑇2 𝑆1 𝑇1 6𝑥 6 = = 𝑇2 5𝑥 5 𝑇1 : 𝑇2 = 6: 5 The ratio between time taken by Sam to reach office from home and time taken by him to reach home from office is 6:5.

50 3

100

350

7

) 𝑆1 = 300 𝑆1 = 6 𝑆1 mph

We know that the time taken to cover a certain distance is inversely proportional to the speed of the object. 𝑇1 𝑆2 = 𝑇2 𝑆1 7 12 6 𝑆1 = 𝑇2 𝑆1 6 𝑇2 = 12 × 7 2

𝑇2 = 10 hours. 7

2

Therefore, the train takes 10 7 hours to reach its destination at the increased speed.

Example 14: There are three different trains A, B and C that run between stations X and Y. Average speed of train A is 30 mph, train B is 45 mph, train C is 60 mph. What is the ratio between the time taken by the trains A, B and C to run between X and Y? Explanation: Let 𝑆𝐴 , 𝑆𝐵 and 𝑆𝐶 be the speed of A, B and C respectively, 𝑇𝐴 , 𝑇𝐵 and 𝑇𝐶 be the time taken by A, B and C respectively and D be the distance between X and Y. Given, 𝑆𝐴 = 30 mph, 𝑆𝐵 = 45 mph and 𝑆𝐶 = 60 mph We know, 𝐷 𝑇= 𝑆 𝐷 𝐷 𝑇𝐴 = = 𝑆𝐴 30 𝐷 𝐷 𝑇𝐵 = = 𝑆𝐵 45 𝐷 𝐷 𝑇𝐶 = = 𝑆𝐶 60 𝐷 𝐷 𝐷 1 1 1 Required ratio = 𝑇𝐴 : 𝑇𝐵 : 𝑇𝐶 = 30 : 45 : 60 = 30 : 45 : 60 To simplify the ratio, multiply each term with the LCM of the denominators. LCM of 30, 45 and 60 is 180. 1 1 1 ( : : ) × 180 = 6: 4: 3 30 45 60

Alternate solution: Let 𝑇1 be the time taken by Sam to reach the office at a speed of 𝑆1 and 𝑇2 be the time taken by Sam to return home at the speed of 𝑆2 . Assume, 𝑆1 = 5 mps and 𝑆2 = 6 mps We know that the time taken to cover a certain distance is inversely proportional to the speed of the object. 1 1 1 1 𝑇1 : 𝑇2 = : = : 𝑆1 𝑆2 5 6 To simplify the ratio, multiply each term with the LCM of the denominators. LCM of 5 and 6 is 30 1 1 ( : ) × 30 = 6: 5 5 6 𝑇1 : 𝑇2 = 6: 5 Therefore, the ratio between time taken by him to reach office from home and time taken by him to reach home from office is 6:5.

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AptitudeB2A Basics to Advanced – Time Speed & Distance Alternate solution:

Explanation: We know that, 𝐷1 𝑆1 × 𝑇1 = 𝐷2 𝑆2 × 𝑇2 𝐷1 = 400, 𝐷2 = ? , 𝑆1 = 2, 𝑆2 = 4, 𝑇1 = 𝑇1 , 𝑇2 = 3𝑇1 400 2 × 𝑇1 = 𝐷2 4 × 3𝑇1 400×4×3 𝐷2 = = 2400 m 2 Therefore, he can cover 1200 m.

Let 𝑆𝐴 , 𝑆𝐵 and 𝑆𝐶 be the speed of A, B and C respectively and 𝑇𝐴 , 𝑇𝐵 and 𝑇𝐶 be the time taken by A, B and C respectively. Given, 𝑆𝐴 = 30 mph, 𝑆𝐵 = 45 mph and 𝑆𝐶 = 60 mph Time taken to cover a certain distance is inversely proportional to the speed of the object. 1 1 1 1 1 1 𝑇𝐴 : 𝑇𝐵 : 𝑇𝐶 = : : = : : = 6: 4: 3 𝑆𝐴 𝑆𝐵 𝑆𝐶 30 45 60 Therefore, the ratio between the time taken by the trains A, B and C to run between X and Y is 6:4:3.

Example 17: Megan and Ayesha took part in a race. Until Ayesha completes the first half of the distance, the ratio between the speeds of Megan and Ayesha was 11:12. By what percentage should Megan increase her speed so that both will finish the race at the same time? (Ayesha’s speed is constant) Explanation: LCM of 11 and 12 is 132 Let 264 meters, the second multiple of 132 be the total distance and speed of Megan and speed of Ayesha be 11 mps and 12 mps, respectively. Then the first half and the second half of the distance is 132 meters each. We know, 𝐷 𝑇 = 𝑆 and 𝐷 = 𝑆 × 𝑇 The time taken by Ayesha to complete the first half 132 i.e., 132 meters = 12 = 11 seconds In that same 11 seconds Megan only covers = 11 × 11 = 121 meters. The time taken by Ayesha to complete the race from the 132 first half i.e., remaining 132 meters = 12 = 11 seconds In that same 11 seconds Megan should cover remaining = (264 − 121) = 143 meters to finish the race at the same time. 𝐷 We know 𝑆 = 𝑇

2.2. Level Moderate

Example 15: Two cars A and B travel from X to Y and return from Y to X. Speed of car A is S throughout the journey. When traveling from X to Y speed of car A was 150% of speed of car B. When traveling from Y to X, speed of car B is double the speed of car A. What is the ratio between time taken by car A and time taken by car B to complete the journey? Explanation: From X to Y: Speed of car A = 𝑆 150 Speed of car A = 100 × Speed of car B 𝑆=

150 100

× Speed of car B 100

2

Speed of car B = 150 × 𝑆 = 3 × 𝑆 From Y to X: Speed of car A = 𝑆 Speed of car B = 2𝑆 Assume that the distance between X and Y is D. 𝐷 Time taken by car A from X to Y = 𝑆 𝐷

3𝐷

𝑆 3 𝐷

2𝑆

Time taken by car B from X to Y = 2 = Time taken by car A from Y to X =

𝑆 𝐷

Time taken by car B from Y to X = 2𝑆 Total time taken by car A to finish the journey 𝐷 𝐷 2𝐷 = + = 𝑆 𝑆 𝑆 Total time taken by car B to finish the journey 3𝐷 𝐷 2𝐷 = + = 2𝑆 2𝑆 𝑆 Ratio between time taken by cars A and B 2𝐷 2𝐷 = : = 1: 1 𝑆 𝑆 Therefore, the ratio between time taken by car A and time taken by car B to complete the journey is 1:1.

143

The required speed = 11 = 13 mps. Percentage change of speed from 11 mps to 13 mps 2 = × 100 = 18.18% 11 Therefore, the Megan must increase her speed by 18.18% after the first half, so that both will finish the race at the same time.

Example 18: There are three trains A, B and C running between stations X and Y. In the time taken by train A to cover 120 miles, train B covers 96 miles. In the time taken by train B to cover 150 miles, train C covers 100 miles. If train A takes 24 hours to reach station Y from station X, when will train C reach station Y if it starts from station X at 03:00 am on a Monday?

Example 16: Brian covers 400 meters at a speed of 2 mps in certain time. If he doubles the speed and triples the time, how much distance can he cover?

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AptitudeB2A Basics to Advanced – Time Speed & Distance Explanation: Let 𝑆𝐴 , 𝑆𝐵 , 𝑆𝐶 be the speed of three trains A, B and C respectively and 𝑇𝐴 and 𝑇𝐶 be the total time taken by train A and train C respectively to reach station Y from station X. When the time is constant, distance covered is directly proportional to speed. 𝐷1 𝑆1 = 𝐷2 𝑆2 Given, time taken by train A to cover 120 miles = time taken by train B covers 96 miles. 120 𝑆𝐴 = 96 𝑆𝐵 𝑆𝐴 5 = − − − 𝐸𝑞(1) 𝑆𝐵 4 Given, the time taken by train B to cover 150 miles = the time taken by train C covers 100 miles. 150 𝑆𝐵 = 100 𝑆𝐶 𝑆𝐵 3 = − − − 𝐸𝑞(2) 𝑆𝐶 2 We know, Time taken to cover a certain distance is inversely proportional to the speed of the object. 𝑇𝐴 𝑆𝐶 = 𝑇𝐶 𝑆𝐴 Given, 𝑇𝐴 = 24 hours 5 From equation (1) → 𝑆𝐴 = 4 𝑆𝐵

13. Zach covers 220 m in 16 seconds. How much distance can he cover in 28 seconds, if he maintains the speed? a. 315 m b. 335 m c. 345 m d. 365 m e. 385 m 14. Distances covered by A and B are in the ratio 4:5 and their corresponding times are in the ratio 6:15. What is the ratio between their speeds? a. 2:1 b. 3:1 c. 3:2 d. 4:1 e. 4:3 15. The usual speed of Jim is ‘S’ and the usual time taken by him to reach his office is ‘T’. On a certain day, he travels at four-fifth of his usual speed. By what percentage will the time taken to reach his office increases? a. 10% b. 20% c. 25% d. 33.33% e. 50% 16. By the time A covers 300 meters, B covers 240 meters. By the time B covers 500 meters, C covers 300 meters. If A takes 48 minutes to cover a certain distance, how long will C take to cover the same distance? a. 100 minutes b. 120 minutes c. 125 minutes d. 150 minutes e. 240 minutes 17. The ratio between time taken by A, B and C to cover a certain distance is 8 : 12 : 15. If the speed of A is 1 increased by 33 3 %, speed of B is increased by 50% and speed of C is increased by 25%, what will be the ratio between the time taken by A, B and C to cover the same distance? a. 3:4:5 b. 5:4:3 c. 3:4:6 d. 6:4:3 e. 4:3:5 18. On day 1, a man reaches his office in 20 minutes. On day 2, he reaches in 18 minutes. On day 3, he reaches in 24 minutes. If his speed on day 3 was 60 mph, what were his speeds on day 1 and day 2 in that order? a. 72 & 75 mph b. 72 & 80 mph c. 80 & 64 mph d. 45 & 50 mph e. 50 & 45 mph 19. Speeds of three trains A, B and C starting from station M are in the ratio 6:8:9. Train A travels between stations M and P, train B travels between stations M and O and train C travels between stations M and N. If the ratio between the distances between stations M and P, M and O and M and N is 10:15:24, what is the ratio between the time taken by trains A, B and C to reach their destinations? a. 40:64:45 b. 45:40:64 c. 64:45:40 d. 40:45:64 e. 64:40:45

2

From equation (2) → 𝑆𝐶 = 3 𝑆𝐵 2 24 3 𝑆𝐵 = 𝑇𝐶 5 𝑆 4 𝐵 5 3 𝑇𝐶 = 24 × 4 × 2 = 45 hours. Total time = 3:00 am on Monday + 45 hours = 12:00 am Wednesday. [3:00 am on Monday + 24 hours = 3:00 am on Tuesday, 3:00 am on Tuesday + 21 hours = 12:00 am Wednesday] Therefore, the train C reaches station Y at 12:00 am on Wednesday of same week, if it starts from station X at 03:00 am on a Monday. 2.3. Exercise 11. At 45 mph, Brad can reach his destination in 4 hours. At what speed must he travel to reach the same destination in 5 hours? a. 25 mph b. 28 mph c. 33 mph d. 36 mph e. 40 mph 12. Speeds of Vincent and Vega are in the ratio 3:5. When Vega covers 240 meters, what distance will Vincent cover? a. 120 m b. 144 m c. 160 m d. 192 m e. 400 m 8

AptitudeB2A Basics to Advanced – Time Speed & Distance 220 16 = 𝐷2 28 28 𝐷2 = 220 × = 385 meters.

20. Based on the above question, if train A takes 6 hours to reach station P from station M at a speed of 30 mph, what is the distance between stations M and O? a. 210 miles b. 240 miles c. 250 miles d. 260 miles e. 270 miles Answer keys: 11-d 12-b 16-a 17-c

13-e 18-b

14-a 19-d

16

14. Option a Let 𝑇𝐴 be the duration in which A has covered a distance 𝐷𝐴 at speed 𝑆𝐴 . Let 𝑇𝐵 be the duration in which B has covered a distance 𝐷𝐵 at the speed 𝑆𝐵 . Given, 𝑇𝐴 = 6𝑥, 𝑇𝐵 = 15𝑥, 𝐷𝐴 = 4𝑦 and 𝐷𝐵 = 5𝑦 We know, 𝐷 𝑆= 𝑇 𝐷𝐴 4𝑦 𝑆𝐴 = = 𝑇𝐴 6𝑥 𝐷𝐵 5𝑦 𝑆𝐵 = = 𝑇𝐵 15𝑥 4𝑦 5𝑦 2 1 Required ratio between their speeds = 6𝑥 : 15𝑥 = 3 : 3 To simplify the ratio, multiply each term with the LCM of the denominators. LCM of 3 and 3 is 3 2 1 ( : ) × 3 = 2: 1 3 3 Therefore, the ratio between their speeds is 2:1.

15-c 20-e

2.4. Detailed Solutions 11. Option d Let 𝑇1 be the former time taken by Brad to cover a distance at a former speed of 𝑆1 and 𝑇2 be the latter time taken to cover the same distance at the latter speed of 𝑆2 . Given, 𝑆1 = 45 mph, 𝑇1 = 4 hours, 𝑇2 = 5 hours We know that the time taken to cover a certain distance is inversely proportional to the speed of the object. 𝑇1 𝑆2 = 𝑇2 𝑆1 4 𝑆2 = 5 45 4 𝑆2 = 45 × 5 = 36 mph. Therefore, the Brad must travel at the speed 36 mph to reach the destination in 5 hours.

15. Option c Let 𝑇1 be the usual time taken to cover a distance at a usual speed of 𝑆1 and 𝑇2 be the time taken to cover the same distance at the speed of 𝑆2 . 4 Given, 𝑆1 = 𝑆, 𝑇1 = 𝑇 and 𝑆2 = 5 𝑆 We know, Time taken to cover a certain distance is inversely proportional to the speed of the object. 4 𝑇1 𝑆2 𝑇 5𝑆 5 = → = → 𝑇2 = 𝑇 𝑇2 𝑆1 𝑇2 𝑆 4 5 Percentage change in time from T to 4 𝑇

12. Option b Let 𝑆1 and 𝑆2 be the speed of the Vincent and Vega respectively and 𝐷1 and 𝐷2 be the distance covered by Vincent and Vega respectively for the same duration. Given, 𝑆1 = 3𝑥 mps, 𝑆2 = 5𝑥 mps and 𝐷2 = 240 meters We know that the distance covered is directly proportional to speed. 𝐷1 𝑆1 = 𝐷2 𝑆2 𝐷1 3𝑥 = 240 5𝑥 3 𝐷1 = 240 × = 144 meters.

5

𝑇−𝑇

5

−1

1

= 4 𝑇 × 100 = 4 1 = 4 × 100 = 25%. Therefore, the percentage of time taken to reach his office increases by 25% 16. Option a

5

13. Option e

Let 𝑆𝐴 , 𝑆𝐵 , 𝑆𝐶 be the speed of A, B and C respectively and 𝑇𝐴 and 𝑇𝐶 be the total time taken by A and C respectively to reach the distance. When the time is constant, distance covered is directly proportional to speed. 𝐷1 𝑆1 = 𝐷2 𝑆2 Given, the time taken by A to cover 300 meters = The time taken by B covers 240 meters. 300 𝑆𝐴 = 240 𝑆𝐵 𝑆𝐴 5 = − − − (1) 𝑆𝐵 4

Let 𝑇1 be the former duration in which Zach has covered a distance 𝐷1 at speed S. Let 𝑇2 be the latter duration in which he has covered a distance 𝐷2 at the same speed S. Given, 𝑇1 = 16 seconds 𝐷1 = 220 meters 𝑇2 = 28 seconds We know, Distance covered is directly proportional to time. 𝐷1 𝑇1 = 𝐷2 𝑇2

9