let's assume A runs 10m/s that means B runs 6m/s so we know it takes A 200 sec to finish the course and in 200 seconds B runs 1200m in my mind that means B has 8 laps remaining or they met 8 times.

Given,
The length of circular track = 100 mtrs.
Length of the race = 2000 mtrs.
Total rounds taken by the athletes to complete the race = 2000/100 = 20 rounds

Now,ratio of speed of the two athletes = 5:3
We calculate time, using the relative speed formula,

Relative Speed = Relative Distance / Time


Relative Speed = 2 (5-3)
Relative Distance = 100 mtrs.(distance of the circular track)

So, Time = 100/2 = 50 sec

Now, the distance covered by faster athlete in 50 sec = 50*5
After running 250 mtrs. both athletes will cross eachother.

Thus, In a race of 2000 mtrs after every 250 mtrs they will cross eachother, which means, they will cross eachother 2000/250 = 8 times.

The correct answer is option E.

Given the length of the race A runs 20 times the circular track to finish. Lets assume that A runs at a rate of 100m/min, thus B runs at a rate of 60m/min. Starting at the same point the difference is 40m/min. Therefore, it takes A 2.5 laps to overtake B. 20 laps in total divided by 2.5 gives us 4.

IMO: A

Given: A and B run a race, in the same direction on a circular track of 100 meters.

Asked: If the race was of 2000 meters and the ratio of speeds of A and B was 5 : 3, find the number of times the two meet?

Let the speeds of A and B be 5x & 3x m/s respectively

Time after they first met = 100/2x = 50/xseconds

Race is completed in = 2000/5x = 400/x seconds

Number of time they met = (400/x)/(50/x) = 8 times

IMO E
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total laps in the race 2000/100 ; 20
since both are running in same direction so relative speed 5-3 ; 2
time taken per lap ; 100/2 ; 50 sec
distance A travels in 50 sec ; 50*5 ;250 mtrs
total times they meet 2000/250; 8
option E