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Arithmetic II

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Time & Speed
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CAT 2025 Lesson : Time & Speed - Basics of Ratios

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1. Introduction

About 22 to 33 questions are asked from Time & Speed in almost every entrance test. The basic concepts of time, speed and distance and the relationship between them are elementary. However, these are word problems which can be made difficult with wordplay.

In this lesson, we will start with the basics and then proceed to the enhanced concepts. Key to solving these questions is developing an ability to read the question clearly and visualise/understand the case at hand.

1.1 Relationship between Time, Speed and Distance

When a distance D is covered at a speed of S at time T, then

D=S×T,S=DTD = S \times T , S = \dfrac{D}{T} and T=DST = \dfrac{D}{S}

When TT is constant, D∝SD \propto S, which means D is directly proportional to S
An increase in speed results in an increase in the distance covered and vice-versa.

When SS is constant, D∝TD \propto T, which means D is directly proportional to T
An increase in time taken results in an increase in the distance covered and vice-versa.

When DD is constant, S∝1TS \propto \dfrac{1}{T} , which means S is inversely proportional to T
An increase in speed results in a decrease in the time taken and vice-versa.

Example 1

A car takes 33 hours to cover a distance, if it travels at a speed of 8080 km/hr. What should be its speed to cover the same distance in 22 hours?

Solution

Distance covered = 3×803 \times 80 = 240 km/hr

Speed required to cover the same distance in 22 hours =2402=120= \dfrac{240}{2} = 120 km/hr

Answer: 120120 km/hr

Example 2

An aeroplane covers a certain distance at a speed of 240240 km/hr in 55 hours. To cover the same distance in 1231 \dfrac{2}{3} hours, it must travel at a speed of

Solution

Distance = 240×5=1200240 \times 5 =1200 km/hr

Speed = DistanceTime \dfrac{\text{Distance}}{\text{Time}}

= 120053 \dfrac{1200}{\dfrac{5}{3}}

Required speed = 1200×351200 \times \dfrac{3}{5} km/hr = 720720 km/hr

Answer: 720720 km/hr

1.2 Ratios of Distance, Speed or Time

Distance and Speed have a direct relationship., i.e. D∝SD \propto S
∴\therefore Where time is constant, ratio of distances == ratio of speeds

Distance and Time have a direct relationship, i.e. D∝TD \propto T
∴\therefore Where speed is constant, ratio of distances == ratio of time taken

Speed and Time, however, have an inverse relationship, i.e. S∝1TS \propto \dfrac{1}{T}
∴ Where distance is constant, ratio of speed == ratio of reciprocal of time taken.

For instance, if for a given a distance, the ratio of speeds of three people is s1:s2:s3s_{1} : s_{2} : s_{3}, then the ratio of their time taken is 1s1:1s2:1s3\dfrac{1}{s_{1}} : \dfrac{1}{s_{2}} : \dfrac{1}{s_{3}}

Similarly, for a given distance, if the ratio of time taken of three people is t1:t2:t3t_{1} : t_{2} : t_{3}, then the ratio for their speeds is 1t1:1t2:1t3 \dfrac{1}{t_{1}} : \dfrac{1}{t_{2}} : \dfrac{1}{t_{3}}

Example 3

Students of Class A take a quarter of the time taken by the students of Class B to get to the playground. Assuming all students of a class move at the same speed, what is the speed of the students of Class B, if Class A students move at 22 m/s?

Solution

We know that S∝1TS \propto \dfrac{1}{T}.

Therefore, as the distance (to the playground) is constant, ratio of speeds is the reciprocal of time taken.

For Classes A and B,
Ratio of times taken = 1:4 1 : 4

∴ Ratio of speeds = 11:14 \dfrac{1}{1} : \dfrac{1}{4} = 4:14 : 1

∴ Speed of Class B students = 24 \dfrac{2}{4} = 0.50.5 m/s

Answer: 0.50.5 m/s

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