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Two cyclists X and Y start at the same time from place A and go towards place B at a speed of 6 km/h and 8 km/h,respectively. Despite stopping for 15 minutes during the journey, Y reaches 10 minutes earlier than X. The distance between the places A and B is:
Time = distance/speed
Time taken by x to reach destination = Time taken by y to reach destination + 10 min
Let the Distance be d.
$$\frac{d}{6} = \frac{d}{8}+ \frac{15}{60}+ \frac{10}{60}$$
$$\frac{d}{6} = \frac{d}{8}+ \frac{1}{4}+ \frac{1}{6}$$
$$\frac{d}{6} = \frac{3d + 6 + 4}{24}$$
d= $$\frac{3d + 10}{4}$$
4d = 3d + 10
d = 10 km
$$\therefore$$ The distance between the places A and B is 10 km.
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