Question

Brigadier Rastogi travels from A to B at 40 km/hr on bike, from B to C at 10 km/hr on cycle. The distance AB equals BC. Then he travels from C to A via B at 24 km/hr by autorickshaw. Find his average speed.

• A. 12.2 km/hr
• B. 12.4 km/hr
• C. 19.2 km/hr
• D. 19.4 km/hr

Hint:

Average speed over multiple legs is total distance divided by total time; never average the given speeds directly unless distances (or times) are equal as required.

Answer 1

The Correct Option is C

Let \(AB=BC=d\). Then \(CA\) via \(B\) is \(CB+BA=2d\). Total distance \(=d+d+2d=4d\).
Total time \(=\dfrac{d}{40}+\dfrac{d}{10}+\dfrac{2d}{24} = d\!\left(\dfrac{1}{40}+\dfrac{1}{10}+\dfrac{1}{12}\right) = d\cdot \dfrac{25}{120}= \dfrac{5d}{24}.\)
Average speed \(=\dfrac{\text{total distance}}{\text{total time}} =\dfrac{4d}{(5d/24)}=\dfrac{96}{5}=19.2\ \text{km/hr}.\)