Two persons X and Y leave place P for place Q at 7:00 a.m. and 7:10 a.m. respectively along the same path. X walks at a speed of 4.8 km/h and Y walks at a speed of 6 km/h. How many kilometres from place P will X meet Y?

Explanation

In 10 minutes (1/6 hour), X travels 4.8 \times \frac{1}{6} = 0.8 km. The relative speed of Y with respect to X is 6 - 4.8 = 1.2 km/h. Time taken for Y to catch X is \frac{0.8}{1.2} = \frac{2}{3} hours. Distance from P = 6 \times \frac{2}{3} = 4 km.

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