Two pipes P and Q can fill a tank in 8 and 24 hours respectively. There is a leak which can empty the full tank in 12 hours. If leak is exactly at `2//3^(rd)` of the height from the base, then in how many hours, will the empty tank will be filled, if all pipes are opened simultaneously?

Options

A. 8 hr
B. 6 hr
C. 10 hr
D. 12 har

Answer

To solve the problem step by step, we need to determine how long it will take to fill the tank when both pipes P and Q are opened simultaneously along with the leak.

### Step-by-Step Solution:

**Step 1: Determine the work done by each pipe and the leak.**

- Pipe P can fill the tank in 8 hours.
- Pipe Q can fill the tank in 24 hours.
- The leak can empty the tank in 12 hours.

**Step 2: Calculate the efficiencies of each pipe and the leak.**

- Efficiency of Pipe P = 1 tank / 8 hours = 1/8 tanks per hour.
- Efficiency of Pipe Q = 1 tank / 24 hours = 1/24 tanks per hour.
- Efficiency of the leak = -1 tank / 12 hours = -1/12 tanks per hour (negative because it is emptying the tank).

**Step 3: Find the total efficiency when all pipes are opened simultaneously.**

Total efficiency = Efficiency of P + Efficiency of Q + Efficiency of leak
\[ = \frac{1}{8} + \frac{1}{24} - \frac{1}{12} \]

To add these fractions, we need a common denominator. The least common multiple (LCM) of 8, 24, and 12 is 24.

- Convert each fraction:
- \(\frac{1}{8} = \frac{3}{24}\)
- \(\frac{1}{24} = \frac{1}{24}\)
- \(-\frac{1}{12} = -\frac{2}{24}\)

Now, adding these:
\[ \frac{3}{24} + \frac{1}{24} - \frac{2}{24} = \frac{3 + 1 - 2}{24} = \frac{2}{24} = \frac{1}{12} \]

So, the total efficiency when all pipes are open is \( \frac{1}{12} \) tanks per hour.

**Step 4: Determine the total work to be done.**

The total work to fill the tank is 1 tank.

**Step 5: Calculate the time taken to fill the tank with the total efficiency.**

Time = Total Work / Total Efficiency
\[ = 1 \text{ tank} / \frac{1}{12} \text{ tanks per hour} = 12 \text{ hours} \]

**Step 6: Adjust for the leak being at 2/3 of the height.**

Since the leak is at 2/3 of the height, for the first 2/3 of the tank, the leak does not affect the filling.

- Work done in the first 2/3 of the tank:
\[ \text{Work} = \frac{2}{3} \text{ tank} \]

Time taken to fill 2/3 of the tank with P and Q only:
\[ \text{Efficiency of P + Q} = \frac{1}{8} + \frac{1}{24} = \frac{3}{24} + \frac{1}{24} = \frac{4}{24} = \frac{1}{6} \text{ tanks per hour} \]
\[ \text{Time} = \frac{2/3}{1/6} = 4 \text{ hours} \]

**Step 7: Calculate the remaining work and time with the leak.**

Remaining work = 1 - 2/3 = 1/3 tank.

Now, we will consider the leak:
\[ \text{Total efficiency with leak} = \frac{1}{12} \text{ tanks per hour} \]
\[ \text{Time to fill remaining 1/3 tank} = \frac{1/3}{1/12} = 4 \text{ hours} \]

**Step 8: Total time taken.**

Total time = Time to fill first 2/3 + Time to fill remaining 1/3
\[ = 4 \text{ hours} + 4 \text{ hours} = 8 \text{ hours} \]

### Final Answer:
The empty tank will be filled in **8 hours** when all pipes are opened simultaneously.