A person delayed by 15 minutes if he goes at 8 km/h but he goes 18 km/h then he is delayed by just 3 minutes. find the distance between office and home
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distance = rate*time
Let t = time in hours to arrive 10 mins early.
Difference in time is 15 mins or 1/4 hours more to arrive 5 mins late
d = 5(t)
d = 4(t + 1/4)
5t = 4(t + 1/4)
5t = 4t + 1
t = 1
d = 5t = 5(1) = 5 km
Let t = time in hours to arrive 10 mins early.
Difference in time is 15 mins or 1/4 hours more to arrive 5 mins late
d = 5(t)
d = 4(t + 1/4)
5t = 4(t + 1/4)
5t = 4t + 1
t = 1
d = 5t = 5(1) = 5 km
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