Solving Pipe and Cistern Time Calculation

This problem involves calculating the time taken by an emptying pipe based on the combined filling and emptying rates of multiple pipes connected to a cistern.

Understanding Pipe Rates

We first determine the rate at which each pipe fills or empties the cistern. The rate is the fraction of the cistern that can be filled or emptied per minute.

• Let the time taken by the first pipe to fill the cistern be $T_1 = 99$ minutes. The rate of the first pipe ($R_1$) is $\frac{1}{99}$ of the cistern per minute.
• Let the time taken by the second pipe to fill the cistern be $T_2 = 90$ minutes. The rate of the second pipe ($R_2$) is $\frac{1}{90}$ of the cistern per minute.
• Let the time taken by the pipe at the bottom to empty the cistern be $T_e$ minutes. The rate of this emptying pipe ($R_e$) is $\frac{1}{T_e}$ of the cistern per minute.
• When all three pipes are opened simultaneously, the cistern gets filled in $T_{net} = 50$ minutes. The net rate ($R_{net}$) is $\frac{1}{50}$ of the cistern per minute.

Calculating the Net Rate

The net rate when all pipes are open is the sum of the filling rates minus the emptying rate:

$$R_{net} = R_1 + R_2 - R_e$$

We know the values for $R_1$, $R_2$, and $R_{net}$, so we can plug them into the equation:

$$\frac{1}{50} = \frac{1}{99} + \frac{1}{90} - \frac{1}{T_e}$$

Finding the Emptying Pipe's Rate

To find the rate of the emptying pipe ($R_e = \frac{1}{T_e}$), we rearrange the equation:

$$\frac{1}{T_e} = \frac{1}{99} + \frac{1}{90} - \frac{1}{50}$$

Performing the Calculation

To solve this, we need to find a common denominator for 99, 90, and 50. The least common multiple (LCM) of 99, 90, and 50 is 4950.

• Convert $\frac{1}{99}$ to a fraction with denominator 4950: $\frac{1}{99} = \frac{1 \times 50}{99 \times 50} = \frac{50}{4950}$
• Convert $\frac{1}{90}$ to a fraction with denominator 4950: $\frac{1}{90} = \frac{1 \times 55}{90 \times 55} = \frac{55}{4950}$
• Convert $\frac{1}{50}$ to a fraction with denominator 4950: $\frac{1}{50} = \frac{1 \times 99}{50 \times 99} = \frac{99}{4950}$

Now substitute these back into the equation for $\frac{1}{T_e}$:

$$\frac{1}{T_e} = \frac{50}{4950} + \frac{55}{4950} - \frac{99}{4950}$$

$$\frac{1}{T_e} = \frac{50 + 55 - 99}{4950}$$

$$\frac{1}{T_e} = \frac{105 - 99}{4950}$$

$$\frac{1}{T_e} = \frac{6}{4950}$$

Determining the Time Taken to Empty

Simplify the fraction $\frac{6}{4950}$:

$$\frac{6}{4950} = \frac{1}{825}$$

So, the rate of the emptying pipe is $\frac{1}{825}$ of the cistern per minute.

Therefore, the time taken by the pipe at the bottom to empty the completely filled cistern is:

$$T_e = 825 \text{ minutes}$$