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

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Solution

The correct option is D
A man walks from his home to market with a speed of 5km/h. Distance =2.5 km and time $= \frac{d}{v} = \frac{2.5}{5} = \frac{1}{2} h r .$

and he returns back with speed of 7.5 km/h in rest of time of 10minutes.

Distance =7.5 $\times \frac{10}{60}$ =1.25 km

so,Average speed

$= \frac{(2.5 + 1.25) k m}{(\frac{40}{60}) h r} = \frac{45}{8} k m / h r .$

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