A and B stand at distinct points of a circular race track of length 12

Anurocks1233

Can you please elaborate few more on the steps you did:-

\(\frac{x}{(a+b)}\) = \(16\) --> How did you reach this expression.. what is x here (I am assuming this is the first time they meet -- x distance in 16 seconds.. but why it's a+b, shouldnt it be a-b as track is circular?)
a+b because we know that this time is quicker vs the time it took to meet them first time when B ran in the reverse direction.

\(x=20\) --> How did you get this value. --> (If you considered (a+b) = 5 m/s ; then x should have been 80)
I see X=80 in my solution. there is no x=20.

\(\frac{120-x}{b-a }= 40\) [Reverse] --> what do u mean by reverse
Reverse when B reverse the direction.

where did you use the relative speed 5m/s

A and B stand at distinct points of a circular race track of length 120 meters. They run at speeds of 'a' meter per second and 'b' meter per second respectively. They meet for the first time 16 seconds after they started the race and for the second time 40 seconds from the time they started the race. If B had started in the opposite direction to the one he originally started, they would have met for the first time after 40 seconds. If B is quicker than A, find B’s speed (in meter per second).

A. 3
B. 4
C. 5
D. 8
E. 10


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