A cyclist travelling at three-fourths of his usual speed

Solution

ATQ,

Let the usual speed of the cyclist be '4s' km/h and the distance travelled be 'd' km.

Let the usual time taken by the cyclist to reach his destination be 't' hours.

ATQ;

(d/4s) = t

Or, d = 4ts ........ (I)

Now, if he travels at three-fourths of his speed:

(d/(3s)) = t + (12/60)

Using equation (I), we have;

(4ts/3s) = t + (1/5)

Or, 4t = 3t + (1/5)

Or, t = (1/5)

Thus, the usual time taken by the cyclist to reach his destination = (1/5) hours or 12 minutes.