A boy walks to his school at a distance of 6 km with constant speed of 2.5 km/hour and walks back with a constant speed of 4 km/hr. His average speed for round trip expressed in km/hour, is

$\frac{24}{13}$

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$\frac{40}{13}$

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$3$

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$\frac{1}{2}$

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Solution

The correct option is B
$\frac{40}{13}$

Distance average speed = $\frac{2 v_{1} v_{2}}{v_{1} + v_{2}} = \frac{2 \times 2.5 \times 4}{2.5 + 4}$
= $\frac{200}{65} = \frac{40}{13} k m / h r$

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A boy walks to his school at a distance of 6 km with constant speed of 2.5 km/hour and walks back with a constant speed of 4 km/hr. His average speed for round trip expressed in km/hour, is

Q. A boy walks to his school at a distance of

6 k m

with a speed of

2.5 k m / h

and walks back with a constant speed by

4 k m / h

. His average speed for trip expressed in

k m / h

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A boy walks to his school at a distance of 6 km with constant speed of 2.5 km/hour and walks back with a constant speed of 4 km/hr. His average speed for round trip expressed in km/hour, is

The correct option is B 4013 Distance average speed = 2v1v2v1+v2=2×2.5×42.5+4 =20065=4013km/hr

The correct option is B
$\frac{40}{13}$

Distance average speed = $\frac{2 v_{1} v_{2}}{v_{1} + v_{2}} = \frac{2 \times 2.5 \times 4}{2.5 + 4}$
= $\frac{200}{65} = \frac{40}{13} k m / h r$