Question

A person X is running around a circular track completing one round every 40 seconds. Another person Y running in the opposite direction meets every 15 seconds. The time, expressed in seconds, taken by Y to complete one round is
A. 12.5
B. 24
C. 25
D. 55

Answer 1

Hint: First, assume the time taken by person Y to complete one round of circular track. Then, find the distance covered by X and Y in one second. Then, the sum of the distance covered by X and Y in one second is equal to $\dfrac{c}{{15}}$. After that do the calculation to get the desired result.

Complete step-by-step solution:
It is given that X completes one round of circular track in 40 seconds and meets another person Y every 15 seconds who is running in the opposite direction.
Let, the distance of the circular track be $c$ and time taken by person Y to complete one round of circular track be $t$ sec.
Now, the distance covered by person X in 1 sec is given by,
$ \Rightarrow \dfrac{c}{{40}}$
Now, the distance covered by person X in 1 sec is given by,
$ \Rightarrow \dfrac{c}{t}$
Now the distance covered by both X and Y in 1 sec will be equal to the sum of distance covered by X in 1 sec and distance covered by Y in 1 sec.
$ \Rightarrow \dfrac{c}{{15}} = \dfrac{c}{{40}} + \dfrac{c}{t}$
Cancel out the common factors from both sides of the equation,
$ \Rightarrow \dfrac{1}{{15}} = \dfrac{1}{{40}} + \dfrac{1}{t}$
Move the constant part to another side,
$ \Rightarrow \dfrac{1}{{15}} - \dfrac{1}{{40}} = \dfrac{1}{t}$
Take LCM on the left side,
$ \Rightarrow \dfrac{{8 - 3}}{{120}} = \dfrac{1}{t}$
Subtract the value in the numerator,
$ \Rightarrow \dfrac{5}{{120}} = \dfrac{1}{t}$
Cancel out the common factors from the left side,
$ \Rightarrow \dfrac{1}{{24}} = \dfrac{1}{t}$
Cross multiply the terms,
$\therefore t = 24$
Thus, the time taken by person Y to complete the circular track is 24 sec.

Hence, option (B) is the correct answer.

Note: The concept of relativity is very popular in solving such types of problems. In this concept, we obtain a pseudo frame by making one person stationary and giving its actual speed to the other person but in exactly the opposite direction. This concept can also be applied to car problems in which one car moves away from the other at a different speed.