It takes 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter is used for 9 hours, only half of the pool is filled. How long would each pipe take to fill the swimming pool ?

Let the time taken by the first pipe be x hours and the time taken by the second pipe be y hours.

In 1 hour,

The first pipe can fill $\dfrac{1}{x}$ th of the pool

The second pipe can fill $\dfrac{1}{y}$ th part of pool.

Given,

It takes 12 hours to fill the pool.

So, in 1 hour both of them will fill $\dfrac{1}{12}$ th part of the pool.

$\therefore \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12}$ ………..(1)

Given,

If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter is used for 9 hours, only half of the pool is filled.

$\therefore \dfrac{4}{x} + \dfrac{9}{y} = \dfrac{1}{2}$ ………….(2)

Multiplying equation (1) by 4, we get :

$\Rightarrow \dfrac{4}{x} + \dfrac{4}{y} = \dfrac{4}{12} \\[1em] \Rightarrow \dfrac{4}{x} + \dfrac{4}{y} = \dfrac{1}{3} …….(3)$

Subtracting equation (3) from (2), we get :

$\Rightarrow \dfrac{4}{x} + \dfrac{9}{y} - \Big(\dfrac{4}{x} + \dfrac{4}{y}\Big)= \dfrac{1}{2} - \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{4}{x} - \dfrac{4}{x} + \dfrac{9}{y} - \dfrac{4}{y} = \dfrac{3 - 2}{6} \\[1em] \Rightarrow \dfrac{5}{y} = \dfrac{1}{6} \\[1em] \Rightarrow y = 5 \times 6 \\[1em] \Rightarrow y = 30.$

Substituting value of y in equation (1), we get :

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12} \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{1}{30} = \dfrac{1}{12} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{12} - \dfrac{1}{30} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 - 2}{60} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{3}{60} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{20} \\[1em] \Rightarrow x = 20.$

Hence, pipe will larger diameter will fill pool in 20 hours and with smaller diameter in 30 hours.