Question

What is the average speed if a man travels $\frac{1}{2}$ the distance at speed of 40 kmph, $\frac{1}{3}$ the distance at 50 kmph and rest of the distance at 60 kmph?

Answer 1

Let the distance be d km.
Time taken for $\frac{d}{2}=\frac{\left(\frac{d}{2}\right)}{40}$
Time taken for $\frac{d}{3}=\frac{\left(\frac{d}{3}\right)}{50}$
Time taken for $\frac{d}{6}$ (which is the balance distance) $=\frac{\left(\frac{d}{6}\right)}{60}$
Therefore total time taken $=\frac{d}{80}+\frac{d}{150}+\frac{d}{360}=\frac{d(45+24+10)}{3600}=d\left(\frac{79}{3600}\right)hrs$.
Hence, average speed $=\frac{distancetravelled}{timetaken}$
$=\frac{d}{79d}\times 3600=\frac{3600}{79}=45.57kmph$