Time Speed Distance problem 2 by Dhaval Furia

Algebra Level pending

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

10 : 27 10:27 am 10 : 18 10:18 am 10 : 25 10:25 am 10 : 45 10:45 am

Dhaval Furia by Dhaval Furia , 26, India

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1 solution

Speed of A is 60 100 − 60 \dfrac{60}{100-60} or 1.5 1.5 times that of B .

Time taken by A to cover ( 100 − 60 100 ) (\frac{100-60}{100}) ' t h ^{th} or

( 2 5 ) (\frac{2}{5}) ' t h ^{th} of the path is 12 12 sec. So the time required by B to cover the same path length is

3 2 × 12 = 18 \dfrac {3}{2}\times 12=18 sec. Hence the time required to cover 60 % 60\% or ( 3 5 ) (\frac{3}{5}) ' t h ^{th} of the path is

18 × 3 5 × 5 2 = 27 18\times \dfrac{3}{5}\times \dfrac{5}{2}=\boxed {27} sec.

So B returns to P at 10 : 27 \boxed {10 : 27} am

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