Two points A and B are diametrically opposite points on a 6 km long circular road. A cyclist started from A and made two rounds. He made the first round with a certain uniform speed and then decreased his speed by 3 km/h. The interval between his two passages through point B is 50 min. Find the speed with which the cyclist made the first round.

6 km/h

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9 km/h

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12 km/h

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16 km/h

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Solution

The correct option is B 9 km/h

Let the speed with which the cyclist made the first round be x km/h
Then, the speed for second round = (x - 3) km/h
Time taken to reach $^{'} B^{'} = \frac{3}{x}$
Again, the time taken to reach 'B' in the second round $= \frac{3}{x} + \frac{3}{x - 3}$
As given $\frac{3}{x} + \frac{3}{x - 3} = \frac{50}{60}$
On solving we get, x = 9 km/h

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The correct option is B 9 km/h Let the speed with which the cyclist made the first round be x km/h Then, the speed for second round = (x - 3) km/h Time taken to reach ′B′=3x Again, the time taken to reach 'B' in the second round =3x+3x−3 As given 3x+3x−3=5060 On solving we get, x = 9 km/h