Question

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 immediately turns and walks back home with a speed of 7.5 km/h. The average speed of the man over the interval of time $0$ to $40min$ 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 $\frac{45}{8}km/h$

Average speed $=\frac{totaldistance}{totaltime}$
Time taken to go to the market $=\frac{distance}{speed}=\frac{2.5}{5}hour=\frac{1}{2}hours=30minutes$
Time taken to return back $=\frac{distance}{speed}=\frac{2.5}{7.5}hour=1/3hours=20minutes$
Distance travelled in 10 minutes while returning back$=\frac{10}{60}\times 7.5=1.25km$ ( Speed is constant at $7.5$ km/hr )
Hence Distance traveled in $40$ minutes($30$ mins towards the market and $10$ mins towards home) $=2.5+1.25=3.75km$
So average speed in $0$ to $40$ minutes $=\frac{3.75km}{\frac{40}{60}hr}$
$=\frac{45}{8}km/hr$