Time, Speed & Distance

 Back to Questions

$P$ and $Q$ start running in opposite directions (towards each other) on a circular track starting at diametrically opposite points. They first meet after $P$ has run for 75m and then they next meet after $Q$ has run 100 m after their first meeting. Assume that both of them are running a constant speed.

The length of the track (in metre) is:









 Hide Ans

Option(C) is correct

Both $P$ and $Q$ are traveling at a constant speed, so the ratio of the distance covered by both $P$ and $Q$ will be same.

Now, assume that the length of track be $x$.

So, for the time of 1st meet,

$P$ travels $= 75 \text{ m}$ and $Q$ travels $=x/2-75 \text{ m}$

Thus, ratio $\dfrac{P}{Q}=\dfrac{75}{x/2-75}$ ---------(1)

Now, for the time of second meet.

$Q$ travels $= 100 \text{ m}$ and $P$ travels $=x-100 \text{ m}$

Thus, ratio $\dfrac{P}{Q}=\dfrac{x-100}{100}$ --------(2)

From (1) and (2),


Upon solving, we get,

$x=\textbf{250 m}$

Edit: Based on the comments from Ramesh and Paritosh Kumar, final answer has been changed from option(D) to option(C). The solution has been modified too.

(6) Comment(s)


Why have we done x/2 for Q when we are finding time for first meet ? Not able to get it.

Paritosh Kumar

250 has to be the answer, explanation as below :

Assume track length to be T.

For first meet A travels 75, B travels T/2 - 75.

For second meet B travels 100, A travels T-100.

Solving 75/(T/2-75)=(T-100)/100

substituting options or by solving, 250 is the solution.


Thank you Paritosh for letting me know. Changed the correct option choice and modified the solution, taking inputs from your comment.


Ratio of the distance covered by both $P$ and $Q$ will be same

Suppose length of track is $2x$

$\dfrac{75}{(x-75)}= \dfrac{(2x-100)}{100}$


Length of track will be 250


yeah only speeds are constant, they have mentioned.


Nothing being said abt their speeds are equal !!!!!!!