A car travels the first half of the distance between two places with a speed of $60 k m p$ . The speed of the car for the rest of the distance so that its average speed becomes $90 k m p h$ .

60 k m p h

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90 k m p h

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120 k m p h

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180 k m p h

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Solution

The correct option is D $180 k m p h$
Let total distance be $2 x$ and uniform speed of next half be $y$
Time for first half $= \frac{x}{60}$
Time for second half $= \frac{x}{y}$
Average speed $= \frac{2 x}{\frac{x}{60} + \frac{x}{y}}$
$90 = \frac{2 x}{\frac{x y + 60 x}{60 y}}$
$3 x y + 180 x = 4 x y$
$x y = 180 x$
$y = 180 k m p h$
Therefore, Uniform speed for next half is 180 kmph.

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A car travels the first half of the distance between two places with a speed of 60kmp. The speed of the car for the rest of the distance so that its average speed becomes 90kmph.

The correct option is D 180 kmphLet total distance be 2x and uniform speed of next half be yTime for first half =x60Time for second half =xyAverage speed =2xx60+xy90=2xxy+60x60y3xy+180x=4xyxy=180xy=180 kmphTherefore, Uniform speed for next half is 180 kmph.

A car travels the first half of the distance between two places with a speed of $60 k m p$ . The speed of the car for the rest of the distance so that its average speed becomes $90 k m p h$ .