Question
A runner, going at five-sixths of her normal speed, takes
10 minutes longer than usual to complete a race. What is her usual time?Solution
ATQ,
Let the usual speed of the runner be '6s' km/h and the distance be 'd' km.
Let the usual time taken be 't' hours.
ATQ;
(d/6s) = t
Or, d = 6ts ........ (I)
Now, running at five-sixths of her speed:
(d/(5s)) = t + (10/60)
Using equation (I), we have;
(6ts/5s) = t + (1/6)
Or, 6t = 5t + (1/6)
Or, t = (1/6)
Thus, the usual time taken by the runner to complete the race = (1/6) hours or 10 minutes.
In how many days can ‘Q’ alone complete the work?
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