If a person goes from town A to town B by a speed of 50 km/h and comes back with a speed of 150 km/h then average speed of the person is:

100 km/h

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75 km/h

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0 km/h

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200 km/h

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Solution

The correct option is B 75 km/h
$A B = - t_{1} = \frac{d}{50}$

$B A = t_{2} = \frac{d}{150}$

$a v e r a g e s p e e d = \frac{t o t a l d i s t a n c e}{t o t a l t i m e}$

$= \frac{d + d}{\frac{d}{50} + \frac{d}{150}}$

$= \frac{2 d \times 150}{3 d + d}$

$= 75 k m h^{- 1}$

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If a person goes from town A to town B by a speed of 50 km/h and comes back with a speed of 150 km/h then average speed of the person is:

The correct option is B 75 km/hAB=−t1=d50 BA=t2=d150 averagespeed=totaldistancetotaltime =d+dd50+d150 =2d×1503d+d =75kmh−1

The correct option is B 75 km/h
$A B = - t_{1} = \frac{d}{50}$

$B A = t_{2} = \frac{d}{150}$

$a v e r a g e s p e e d = \frac{t o t a l d i s t a n c e}{t o t a l t i m e}$

$= \frac{d + d}{\frac{d}{50} + \frac{d}{150}}$

$= \frac{2 d \times 150}{3 d + d}$

$= 75 k m h^{- 1}$