Question 33

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

Solution

$$\frac{\frac{3x}{5}}{3a} + \frac{\frac{2x}{5}}{2b} = \frac{2x}{5c}$$

(Where x is distance between A and B; $$\frac{\frac{3x}{5}}{3a}$$ = time taken to cover the distance with speed 3a ; $$\frac{\frac{2x}{5}}{2b}$$ = time taken to cover the distance with speed 2b; $$\frac{2x}{5c}$$ = time taken to cover the distance x from B to A then return.)

$$\frac{\frac{3x}{5}}{3a} + \frac{\frac{2x}{5}}{2b} = \frac{2x}{5c}$$

Or $$\frac{1}{a} + \frac{1}{b} = \frac{2}{c}$$


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