A borewell pump fills an overhead tank in 8 hours when the tank is sound. Because of a crack in the base the same pump now needs 10 hours to fill it. How long would the crack alone take to drain a full tank?

The pump fills 1/8 of the tank an hour; with the crack the net gain is only 1/10 an hour. The crack therefore drains 1/8 − 1/10 = 5/40 − 4/40 = 1/40 of the tank an hour, so on its own it would empty a full tank in 40 hours. 2 hours takes the extra two hours of filling as the crack's own time, which confuses a delay with a rate. 18 hours adds the two times. 80 hours multiplies 8 by 10 but forgets to divide by the 2-hour difference.