CAT 1996 — QA Question 25

Let the total distance be x. So the man travels a distance $\frac{3x}{5}$ at a speed 3a. Therefore, total time taken to travel this distance = $\frac{3x}{\left(15a\right)}=\frac{x}{\left(5a\right)}$

$\left[time=\frac{dis\tan ce}{speed}\right]$

He then travels a distance $\frac{2x}{5}$ at a speed 2b. Hence, time taken to travel this distance = $\frac{2x}{\left(10b\right)}=\frac{x}{\left(5b\right)}.$ So total time taken in going from A to B = $\frac{x}{\left(5a\right)}+\frac{x}{\left(5b\right)}.$ Now he travels from B to A and comes back. So total distance travelled = 2x at an average speed 5c.
Hence, time taken to return = $\frac{2x}{\left(5c\right)}$.

Since the time taken in both the cases remains the same, we can write $\frac{x}{5a}+\frac{x}{5b}=\frac{2x}{5c}$

Therefore, $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}.$