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 $0$ to $40min$. is

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

Answer 1

Answer: (D)

A man walks from his home to market with a speed of $5km/h$.
Distance $=2.5km$ and
time $=\frac{d}{v}=\frac{2.5}{5}=\frac{1}{2}hr$
and he returns back with speed of $7.5km/h$ in rest of time of $10$ minutes.
Distance $=7.5\times \frac{10}{60}=1.25km$
Average speed $=\frac{\text{ Total distance }}{\text{ Total time }}$
$=\frac{(2.5+1.25)km}{(40/60)hr}=\frac{45}{8}km/hr$