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# A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B?

## CAT Question

By Ananya #Saloni · 4 weeks ago ·CAT
Prateek Saxena answered 4 weeks ago
2 hours is answer
in first case distance is same, so D = S1T = S2(T-2). From this we get S1 = S2(T-2)/T. Also when they are travelling towards each other. D = (S1 + S2)(80/60). So, S1 = 4S2/(3T-4). Now equate it with S1 = S2(T-2)/T. we will get T = 4 or T = 2/3. But Tâ‰ 2/3 because time taken by fast motorcyclist is T -2 which will be negative if we take T = 2/3. therefore T = 4, and reqd time = T - 2 = 2hours

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