A bus travelling the first one third distance at a speed of $10 k m / h$ , the next one third at $20 k m / h$ and the last one third at $60 k m / h$ . The average speed of the bus is

9 k m / h

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16 k m / h

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18 k m / h

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48 k m / h

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Solution

The correct option is C $18 k m / h$
$T o t a l d i s t a n c e = x A s s = v t t_{1} = \frac{\frac{x}{3}}{10} = \frac{x}{30} t_{2} = \frac{\frac{x}{3}}{20} = \frac{x}{60} t_{3} = \frac{\frac{x}{3}}{60} = \frac{x}{180} A v e r a g e v e l o c i t y = \frac{x}{t o t a l t i m e t a k e n} t o t a l t i m e t a k e n = \frac{x + 3 x + 6 x}{180} = \frac{10 x}{180} T h u s A v e r a g e v e l o c i t y = \frac{x \times 180}{10 x} = 18 k m / h$

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