Question

Chetan, Bhushan and Shailu start running at the same time from the same point on a circular track of 70 meters radius. Chetan and Bhushan run clockwise and Shailu runs counter clockwise. If Chetan meets Shailu every 66 seconds and Bhushan meets Shailu every 110 seconds, how frequently (in seconds) will Chetan meet Bhushan?

Answer 1

The circumference of the track is 70*2*π= 140π
Let the speed of the Chetan be c m/s, Bhushan be b m/s and Shailu be s m/s

Since Chetan and Shailu is running in opposite directions, they will meet whenever their speed made up to 140π. So you can set an equation like this,
(c+s)*66 = 140π

We can express Chetan's speed in terms of s.
c+s = 140π/66
c = 140π/66 - s   ------------ (1)

Similarly Bhushan and Shailu's speed can be put into an equation like this.(b+s)*110 = 140π  

We can express Bhushan's speed in terms of s like this.
b+s = 140π/110
b = 140π/110 - s    ------------ (2)

From the equations, we can see that Chetan is faster than Bhushan. 
When Chetan meet Bhushan is the moment when Chetan overlaps Bhushan. That is the moment when the difference of their speed make up to 140π. We can find that moment (at x sec) using the equation below.

(c - b$[( \frac{140 \pi }{66}-s )-( \frac{140 \pi }{110}-s )] x = 140 \pi$) * x = 140π

Substituting previous equations (1) and (2), we can change the above equation to this.

$[( \frac{140 \pi }{66}-s )-( \frac{140 \pi }{110}-s )] x = 140 \pi$

$( \frac{140 \pi }{66}- \frac{140 \pi }{110}) x = 140 \pi$

(6.6640- 3.9984)x = 140π
x = 165

So Chetan will meet Bhushan every 165 seconds. 

Answer 2

Answer:

The circumference of the track is 70*2*π= 140π

Let the speed of the Chetan be c m/s, Bhushan be b m/s and Shailu be s m/s

Since Chetan and Shailu is running in opposite directions, they will meet whenever their speed made up to 140π. So you can set an equation like this,

(c+s)*66 = 140π

We can express Chetan's speed in terms of s.

c+s = 140π/66

c = 140π/66 - s ------------ (1)

Similarly Bhushan and Shailu's speed can be put into an equation like this.(b+s)*110 = 140π

We can express Bhushan's speed in terms of s like this.

b+s = 140π/110

b = 140π/110 - s ------------ (2)

From the equations, we can see that Chetan is faster than Bhushan.

When Chetan meet Bhushan is the moment when Chetan overlaps Bhushan. That is the moment when the difference of their speed make up to 140π. We can find that moment (at x sec) using the equation below.

(c - b[( \frac{140 \pi }{66}-s )-( \frac{140 \pi }{110}-s )] x = 140 \pi[(

66

140π

−s)−(

110

140π

−s)]x=140π ) * x = 140π

Substituting previous equations (1) and (2), we can change the above equation to this.

[( \frac{140 \pi }{66}-s )-( \frac{140 \pi }{110}-s )] x = 140 \pi[(

66

140π

−s)−(

110

140π

−s)]x=140π

( \frac{140 \pi }{66}- \frac{140 \pi }{110}) x = 140 \pi(

66

140π

−

110

140π

)x=140π

(6.6640- 3.9984)x = 140π

x = 165

So Chetan will meet Bhushan every 165 seconds.