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Question:

A man travels three-fifths of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and return at a speed 5c in the same time, then

Explanation:

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

$[t i m e = \frac{d i s \tan c e}{s p e e d}]$

He then travels a distance $\frac{2 x}{5}$ at a speed 2b. Hence, time taken to travel this distance = $\frac{2 x}{(10 b)} = \frac{x}{(5 b)} .$ So total time taken in going from A to B = $\frac{x}{(5 a)} + \frac{x}{(5 b)} .$ 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 = $\frac{2 x}{(5 c)}$ .

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

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

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