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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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