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If a boy walks from his home to the school at 4km/hr he reaches the school 10mins earlier than d scheduled time. however if he walks at 3 km/hr, he reaches

10 mins late. the distance from the school to the house is :


Let the distance be  d KM

so when speed is 4 KM/hr    Time  T1= d/4 hr  (10 min earler than schd time)
when speed is 3 KM/hr         Time  T2=d/3 hr    (10 min late than schd time)
difference  between T1 and T2 is 20 min = 20/60 hr
                                                            = 1/3 hr
d/3-d/4= 1/3
(4d-3d)/12= 1/3
d= 4 KM
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