Question
A traveler at four-fifths of his normal speed takes 20
minutes longer than his usual time to reach his destination. What is his usual travel time?Solution
ATQ,
Let the usual speed of the traveler be '5s' km/h and the distance be 'd' km.
Let the usual time taken be 't' hours.
ATQ;
(d/5s) = t
Or, d = 5ts ........ (I)
Traveling at four-fifths of his normal speed:
(d/(4s)) = t + (20/60)
Using equation (I), we have;
(5ts/4s) = t + (1/3)
Or, 5t = 4t + (1/3)
Or, t = (1/3)
Thus, the usual time taken by the traveler to reach his destination = (1/3) hours or 20 minutes.
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