Question

Two taps a and b fill a swimming pool together in 3 hours. Alone, it takes tap a 4 hours less than b to fill the same pool. How many hours does b tap to fill the pool separately?

Answer 1

Let the larger tap, A, take t hours to fill the tank, alone.

The smaller tap, B, takes (t+3) hours to fill the tank, alone.

A fills (1/t)th of the tank in 1 hour.

B fills [1/(t+3)]th of the tank in 1 hour.

So A and B take (1/t) + [1/(t+3)] = (t+3+t)/[t(t+3)] th part of the tank in one hour which is half the tank. Thus

(t+3+t)/[t(t+3)] = 1/2 or

2(2t+3) = t^2+3t, or

4t+6 = t^2+3t, or

t^2-t+6 = 0

(t-3)(t+2) = 0

t = 3 hours. [-2 hours has no meaning and so rejected]’

So the larger tap takes 3 hours to fill the tank, alone, while the smaller tap takes 6 hours to fill the tank, alone.

Check: (1/3)+(1/6) = (2/6)+(1/6) = (3/6) = 1/2 of the tank in one hour. Correct.