## Fill in the Blanks

### 1. Introduction

About $2$ to $3$ 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

$D = S \times T , S = \dfrac{D}{T}$ and $T = \dfrac{D}{S}$

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

When $S$ is constant, $D \propto T$, 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 $D$ is constant, $S \propto \dfrac{1}{T}$ , 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 $3$ hours to cover a distance, if it travels at a speed of $80$ km/hr. What should be its speed to cover the same distance in $2$ hours?

### Solution

Distance covered = $3 \times 80$ = 240 km/hr

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

Answer: $120$ km/hr

Where more than two variables are involved, like in the following example, the variation and product constancy methods might require more time. Please use the standard worker-day method in these cases.

### Example 2

An aeroplane covers a certain distance at a speed of $240$ km/hr in $5$ hours. To cover the same distance in $1 \dfrac{2}{3}$ hours, it must travel at a speed of

### Solution

Distance = $240 \times 5 =1200$ km/hr

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

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

Required speed = $1200 \times \dfrac{3}{5}$ km/hr = $720$ km/hr

Answer: $720$ km/hr

#### 1.2 Ratios of Distance, Speed or Time

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

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

Speed and Time, however, have an inverse relationship, i.e. $S \propto \dfrac{1}{T}$
∴ 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 $s_{1} : s_{2} : s_{3}$, then the ratio of their time taken is $\dfrac{1}{s_{1}} : \dfrac{1}{s_{2}} : \dfrac{1}{s_{3}}$

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

### 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 $2$ m/s?

### Solution

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

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 = $1 : 4$

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

∴ Speed of Class B students = $\dfrac{2}{4}$ = $0.5$ m/s

Answer: $0.5$ m/s