Question

A man walks on a straight road from his home to a market $2.5km$ away with a speed of $5km/h$. 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 $0to40min$ is equal to

• A. $5km/h$
• B. $\frac{25}{4}km/h$
• C. $\frac{30}{4}km/h$
• D. $\frac{45}{8}km/h$

Answer 1

The correct option is D $\frac{45}{8}km/h$

A man walks from his home to market with a speed of 5 km/h.

Distance = 2.5 km and time = $\frac{d}{v}=\frac{2.5}{5}=\frac{1}{2}hr.$

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

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

So, Average Speed

$=\frac{(2.5+1.25)km}{(40+60)hr}=\frac{45}{8}km/hr.$