Using two pumps

Hi, Ramel!

This is a working together problem.

The rate of the 8 hp pump is:
$$\frac{1}{8}$$

The rate of the 3 hp pump is: 1/12
$$\frac{1}{12}$$

The time together will be 4 hours and the time that the 3 hp pump works alone would be $x$ hrs. We want to fill only 1 tank. The formula we now need then is:

$$\left (\frac{1}{8} \times 4 \right ) + \left (\frac{1}{12} \times 4 \right ) + \left (\frac{1}{12} \times x \right ) = 1$$

we get...

$$\frac{4}{8} + \frac{4}{12} + \frac{x}{12} = 1$$

Adding the fractions, we get:

$$\frac{5}{6} + \frac{x}{12} = 1$$

Subtract $\frac{5}{6}$ to get:

$$\frac{x}{12} = \frac{1}{6}$$

The value of $x$ then is $2$.

It means that the 3 hp pump must at least work for additional $\color{blue}{2hours}$ to fill the tank.

Alternate solutions are encouraged.

Thanks!