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Arithmetic II

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Time & Speed

Time And Speed

MODULES

Basics of Ratios
Average Speed
Product Constancy
Relative Speed
Trains
Walking To & Fro
Circular Track
Boats, Races & Clock
Escalators
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Time & Speed 1
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Time & Speed 2
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Time & Speed 3
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Time & Speed 4
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PRACTICE

Time & Speed : Level 1
Time & Speed : Level 2
Time & Speed : Level 3
ALL MODULES

CAT 2025 Lesson : Time & Speed - Basics of Ratios

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1. Introduction

About
222 to 333 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}D=S×T,S=TD​ and T=DST = \dfrac{D}{S}T=SD​

When
TTT is constant, D∝SD \propto SD∝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
SSS is constant, D∝TD \propto TD∝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
DDD is constant, S∝1TS \propto \dfrac{1}{T}S∝T1​ , 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 333 hours to cover a distance, if it travels at a speed of 808080 km/hr. What should be its speed to cover the same distance in 222 hours?

Solution

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

Speed required to cover the same distance in
222 hours =2402=120= \dfrac{240}{2} = 120=2240​=120 km/hr

Answer:
120120120 km/hr

Example 2

An aeroplane covers a certain distance at a speed of 240240240 km/hr in 555 hours. To cover the same distance in 1231 \dfrac{2}{3}132​ hours, it must travel at a speed of

Solution

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

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

=
120053 \dfrac{1200}{\dfrac{5}{3}}35​1200​

Required speed =
1200×351200 \times \dfrac{3}{5}1200×53​ km/hr = 720720720 km/hr

Answer:
720720720 km/hr

1.2 Ratios of Distance, Speed or Time

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

Distance and Time have a direct relationship, i.e.
D∝TD \propto TD∝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}S∝T1​
∴ 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}s1​:s2​:s3​, then the ratio of their time taken is 1s1:1s2:1s3\dfrac{1}{s_{1}} : \dfrac{1}{s_{2}} : \dfrac{1}{s_{3}}s1​1​:s2​1​:s3​1​

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

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 222 m/s?

Solution

We know that
S∝1TS \propto \dfrac{1}{T}S∝T1​.

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 1:4

∴ Ratio of speeds =
11:14 \dfrac{1}{1} : \dfrac{1}{4} 11​:41​ = 4:14 : 14:1

∴ Speed of Class B students =
24 \dfrac{2}{4}42​ = 0.50.50.5 m/s

Answer:
0.50.50.5 m/s

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