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A boy walks to his school at a distance of 6 km with constant speed of 2.5 kmh^-1 and walks back with a constant speed of 4 kmh^-1. His average speed for round trip expressed (in kmh^-1), is

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24/13

40/13

answer is B.

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Detailed Solution

Average speed of a boy $= \frac{2 v_{1} v_{2}}{v_{1} + v_{2}} = \frac{2 \times 2.5 \times 4}{2.5 + 4}$

$= \frac{20}{6.5} = \frac{40}{13} km / h$

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