X began his journey from point A to point B. Two hours later, Y set off from point B towards A. By the time X had covered one-fifth of the total distance, Y had also managed to cover the same distance. If Y's speed is four times that of X's speed, what is the difference in the time (in hours) taken by X and Y to reach their respective destinations?

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Let 1/5 of the distance covered by X = s*(2+t), where s is the speed of X and (2+t) is the time taken by X to cover 1/5 of the total distance. Let t be the time taken by Y to cover the same 1/5 of the distance. Since Y's speed is four times that of X's speed, we have 1/5 of the distance covered by Y = 4s*t. Thus, we can set up the equation: s(2+t) = 4st. Dividing through by s and solving for t gives us t =…

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What this question tests

What it tests

Your application of distance = rate × time to trains, boats, and commuters, including relative speed.

Common trap

Mixing units (miles per hour vs. minutes) or averaging speeds — average speed is total distance over total time.