Question
The distance of the School and house of Suresh is 80 km.
One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed speed of Suresh?Solution
D=80km ATQ- 80= S×(t+1)...(1) 80=(S+4)t...(2) Equation (1) and (2) both are equal S×(t+1)=(S+4)t st+s=st+4t s=4t put the value of S in (1) 80=4t(t+1) 20=t(t+1) t=4 S=16 new speed =16+4=20km/h
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