Solving the Pipe Filling and Emptying Problem

This problem involves calculating the time taken for a pipe to empty a tank, given the filling times of two other pipes and the combined time when all three are operational.

Understanding Pipe Rates

We can represent the rate at which each pipe works as a fraction of the tank it can fill or empty in one hour.

• Pipe A fills the tank in 16 hours. Its filling rate is $\frac{1}{16}$ of the tank per hour.
• Pipe B fills the tank in 24 hours. Its filling rate is $\frac{1}{24}$ of the tank per hour.
• Pipe C empties the tank in $x$ hours. Its emptying rate is $\frac{1}{x}$ of the tank per hour.

Combined Operation Analysis

When all three pipes are opened together, the net rate is the sum of filling rates minus the emptying rate. The problem states that the tank is full in $20\frac{4}{7}$ hours.

First, let's convert the combined time into an improper fraction:

$20\frac{4}{7} \text{ hours} = \frac{(20 \times 7) + 4}{7} \text{ hours} = \frac{140 + 4}{7} \text{ hours} = \frac{144}{7} \text{ hours}$

The combined rate of filling the tank is the reciprocal of the time taken:

Combined Rate = $\frac{1}{\frac{144}{7}} = \frac{7}{144}$ of the tank per hour.

Setting Up the Equation

The combined rate is calculated as:

(Rate of Pipe A) + (Rate of Pipe B) - (Rate of Pipe C) = Combined Rate

Substituting the values we have:

$$ \frac{1}{16} + \frac{1}{24} - \frac{1}{x} = \frac{7}{144} $$

Calculating the Value of $x$

Let's solve the equation step-by-step:

1. Combine the filling rates of pipes A and B: Find a common denominator for 16 and 24, which is 48. $$ \frac{1}{16} + \frac{1}{24} = \frac{3}{48} + \frac{2}{48} = \frac{5}{48} $$
2. Substitute this back into the main equation: $$ \frac{5}{48} - \frac{1}{x} = \frac{7}{144} $$
3. Isolate the term with $x$: Rearrange the equation to solve for $\frac{1}{x}$. $$ \frac{1}{x} = \frac{5}{48} - \frac{7}{144} $$
4. Find a common denominator to subtract the fractions: The least common multiple (LCM) of 48 and 144 is 144. Convert $\frac{5}{48}$ to an equivalent fraction with a denominator of 144. $$ \frac{5}{48} = \frac{5 \times 3}{48 \times 3} = \frac{15}{144} $$
5. Perform the subtraction: $$ \frac{1}{x} = \frac{15}{144} - \frac{7}{144} $$ $$ \frac{1}{x} = \frac{15 - 7}{144} $$ $$ \frac{1}{x} = \frac{8}{144} $$
6. Simplify the fraction: Both 8 and 144 are divisible by 8. $$ \frac{8}{144} = \frac{1}{18} $$
7. Solve for $x$: Since $\frac{1}{x} = \frac{1}{18}$, we can conclude that: $$ x = 18 $$

Therefore, pipe C alone can empty the full tank in 18 hours.