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Q.

A man walks on a straight road from his home to a market 2.5 km. away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. The average speed of man over the interval of time 0 to 40 min, is equal to

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a

5 km/h

b

$\frac{25}{4}$ km/h

c

$\frac{30}{4}$ km/h

d

$\frac{45}{8}$ km/h

answer is D.

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Detailed Solution

Time for onward journey,

$t_{1} = \frac{2.5}{5} Hr = 30 \min$

Distance traveled in 10 min in return journey

$d_{2} = 7.5 \times \frac{1}{6} = 1.25 k m$

$\Rightarrow V_{a v g} = \frac{(2.5 + 1.25) k m}{(\frac{2}{3}) H r} = \frac{45}{8} k m h^{- 1}$

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A man walks on a straight road from his home to a market 2.5 km. away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. The average speed of man over the interval of time 0 to 40 min, is equal to