An automobile driver travels from plane to a hill station 120 Km distant at an average speed of 30 Km per hour. He then makes the return trip at an average speed of 25 Km per hour. He covers another 120 Km distance on plane at an average speed of 50 KM per hour. His average speed over the distance of 360 KM will be

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$\frac{3}{\frac{1}{30} + \frac{1}{25} + \frac{1}{50}} km / hr .$

$\frac{30 + 25 + 50}{3} km / hr .$

105 km/hr.

${(30.25.50)}^{1 / 3} km / hr .$

answer is C.

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Detailed Solution

Avg Speed of first 120 km is 30 km/h

Avg Speed of second 120 km is 25 km/h

Avg Speed third 120 km is 50 km/h

Total time taken is $= \frac{120}{30}$

Avg Speed $= \frac{Total dist}{Total time taken}$

$= \frac{360}{\frac{120}{30} + \frac{120}{25} + \frac{120}{50}} = \frac{3}{\frac{1}{30} + \frac{1}{25} + \frac{1}{50}}$

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