Question
A driver moving at half of his normal speed takes 45
minutes more than the typical time to reach his destination. Determine the typical time taken by him.Solution
ATQ,
Let the usual speed of the driver be '2s' km/h and the distance be 'd' km.
Let the usual time taken be 't' hours.
ATQ;
(d/2s) = t
Or, d = 2ts ........ (I)
Now, if he travels at half his speed:
(d/s) = t + (45/60)
Using equation (I), we have;
(2ts/s) = t + (3/4)
Or, 2t = t + (3/4)
Or, t = (3/4)
Therefore, the usual time taken by the driver to reach his destination = (3/4) hours or 45 minutes.
If n(A) = 25, n(B) = 40 and n(A βͺ B) = 50, then n(A β© B) equals
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