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ATQ,
Let the usual speed of the athlete be '4s' km/h and the distance be 'd' km.
Let the usual time taken be 't' hours.
ATQ;
(d/4s) = t
Or, d = 4ts ........ (I)
When she runs at three-quarters of her speed:
(d/(3s)) = t + (25/60)
Using equation (I), we have;
(4ts/3s) = t + (5/12)
Or, 4t = 3t + (5/12)
Or, t = (5/12)
Therefore, the usual time taken by the athlete to finish the marathon = (5/12) hours or 25 minutes.
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