John cycles at 10 km/h to reach his destination at 2 PM. If he travels at 15 km/h, he will reach at 12 noon itself. At what speed should he travel to reach the place at 1 PM?
Time difference= 2PM -12 noon =2 hours ATQ- 10t = 15(t-2) 10t = 15t-30 30 =5t t = 6 hours Distance= d = 10t = 10 ×6 =60 km Time to reach by 1 PM= t-1 6-1= 5 hours Speed = 60 km /5 hours= 12 km/h
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