Question 38

A man went to his office on cycle at the rate of 10 km/hr and reached late by 6 minutes. When he increased the speed by 2 km/hr, he reached 6 minutes before time. What is the distance between his office and his departure point ?

Solution

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Let the distance be d. 

D/10 = T+ 1/10

D/12 = T - 1/10

 

So, D = 12 kilometers. 

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