in Quantitative Aptitude edited by
1 vote
1 vote

Two ants $\text{A}$ and $\text{B}$ start from a point $\text{P}$ on a circle at the same time, with $\text{A}$ moving clock-wise and $\text{B}$ moving anti-clockwise. They meet for the first time at $10:00$ am when $\text{A}$ has covered $60\%$ of the track. If $\text{A}$ returns to $\text{P}$ at $10:12$ am, then $\text{B}$ returns to $\text{P}$ at

  1. $10:25$am
  2. $10:18$am
  3. $10:27$am
  4. $10:45$am
in Quantitative Aptitude edited by
13.4k points

1 Answer

1 vote
1 vote

Ans is an option (C)

Let the diameter of the track be  $d$ mtr. Then the length of the track is  $\pi d$ mtr. Let the speeds of ant$_{1}$ and ant$_{2}$ be  $v_{1},v_{2}$ mtr/min  respectively.

Then, let in time  $t$ minutes,  ant$_{1}$ covers $60$% of the track i.e. distance of  $0.6\pi d$ mtr.

Then in the same time  $t$ minutes, ant$_{2}$  covers $40$%  of the track i.e. distance of  $0.4\pi d$ mtr.

For ant$_{1}$ we have following equations:   $t=\frac{0.6\pi d}{v_{1}} \longrightarrow(1)$ and  $t+12=\frac{\pi d}{v_{1}}\longrightarrow(1)$

Dividing $(2)$ by $(1),$ we get  $t=18$ minutes

For ant$_{2}$ we have following equations:   $t=\frac{0.4\pi d}{v_{2}}\longrightarrow(3)$ and  $t+y=\frac{\pi d}{v_{2}}\longrightarrow(4) \; (t+y$ is the complete time taken by ant$_{2}$ to complete the track) .

Dividing $(4)$ by $(3),$ we get $y=27$ minutes

After $t$ minutes, it is  $10:00$ A.M when the two ants meet for the first time. From there, it takes $27$ minutes for ant$_{2}$ to reach point P.

So,  ant$_{2}$  reaches point P at  $10:27\;\text{A.M.}$

438 points

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true