Travelling at 3/4 of the normal speed, a person reaches his workplace 15 minutes late. How many minutes does he take usually to reach the workplace?
let s be his normal speed
t = his normal time
as we know thatÂ
D = $$s\times t$$
then
$$D = (\frac{3}{4})s\times (t+15)$$
here the distance is the same we can equate this to his regular day which is D = $$s\times t$$
$$s\times t =Â (\frac{3}{4})s\times (t+15)$$
t=45 minutes
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