The question asks: A pipe can fill a tank in 21 minutes, while another pipe can completely empty the filled tank in 24 minutes. If both pipes are opened simultaneously into an empty tank, how long will it take (in minutes) to fill one-fourth of the tank?

1. Problem Description

The question asks: A pipe can fill a tank in 21 minutes, while another pipe can completely empty the filled tank in 24 minutes. If both pipes are opened simultaneously into an empty tank, how long will it take (in minutes) to fill one-fourth of the tank?

2. Solution Steps

Let the volume of the tank be $V$.

The rate at which the filling pipe fills the tank is $R_1 = \frac{V}{21}$ (volume per minute).

The rate at which the emptying pipe empties the tank is $R_2 = \frac{V}{24}$ (volume per minute).

When both pipes are open, the net rate of filling is $R = R_1 - R_2 = \frac{V}{21} - \frac{V}{24}$.

$R = V (\frac{1}{21} - \frac{1}{24}) = V (\frac{24 - 21}{21 \times 24}) = V (\frac{3}{21 \times 24}) = V (\frac{1}{7 \times 24}) = \frac{V}{168}$.

So, the net rate of filling is $\frac{V}{168}$ (volume per minute).

We want to find the time it takes to fill one-fourth of the tank, which is $\frac{V}{4}$.

Let $t$ be the time in minutes.

Then, $\frac{V}{168} \times t = \frac{V}{4}$.

$t = \frac{V}{4} \times \frac{168}{V} = \frac{168}{4} = 42$.

So, it will take 42 minutes to fill one-fourth of the tank.

3. Final Answer

42