Two pipes A and B can fill a tank in 15 hours and 20 hours respectively while a third pipe C can empty the full tank in 25 hours. All the three pipes are opened in the begining. After 10 hours, C is closed. In how much time, will the tank be full?

Options

A. 12 hrs
B. 13 hrs
C. 16 hrs
D. 18 hrs

Answer

To solve the problem step by step, we need to determine how much of the tank is filled by pipes A and B, and how much is emptied by pipe C. Let's break it down:

### Step 1: Determine the rates of filling and emptying
- Pipe A can fill the tank in 15 hours. Therefore, the rate of A = \( \frac{1}{15} \) of the tank per hour.
- Pipe B can fill the tank in 20 hours. Therefore, the rate of B = \( \frac{1}{20} \) of the tank per hour.
- Pipe C can empty the tank in 25 hours. Therefore, the rate of C = \( \frac{1}{25} \) of the tank per hour.

### Step 2: Calculate the combined rate of A, B, and C
When all three pipes are opened, the combined rate of filling the tank is:
\[ \text{Combined rate} = \text{Rate of A} + \text{Rate of B} - \text{Rate of C} \]
Substituting the rates:
\[ \text{Combined rate} = \frac{1}{15} + \frac{1}{20} - \frac{1}{25} \]

### Step 3: Find a common denominator
The least common multiple (LCM) of 15, 20, and 25 is 300. Now we convert each rate to have a denominator of 300:
- Rate of A: \( \frac{1}{15} = \frac{20}{300} \)
- Rate of B: \( \frac{1}{20} = \frac{15}{300} \)
- Rate of C: \( \frac{1}{25} = \frac{12}{300} \)

Now substituting these values:
\[ \text{Combined rate} = \frac{20}{300} + \frac{15}{300} - \frac{12}{300} = \frac{23}{300} \]

### Step 4: Calculate the amount filled in 10 hours
In 10 hours, the amount of the tank filled by the three pipes is:
\[ \text{Amount filled in 10 hours} = 10 \times \frac{23}{300} = \frac{230}{300} = \frac{23}{30} \]

### Step 5: Determine the remaining part of the tank
The total capacity of the tank is 1 (full tank). The remaining part after 10 hours is:
\[ \text{Remaining part} = 1 - \frac{23}{30} = \frac{30}{30} - \frac{23}{30} = \frac{7}{30} \]

### Step 6: Calculate the time taken to fill the remaining part with A and B
After 10 hours, pipe C is closed, and only pipes A and B are filling the tank. Their combined rate is:
\[ \text{Combined rate of A and B} = \frac{1}{15} + \frac{1}{20} \]
Using the same common denominator of 60:
- Rate of A: \( \frac{4}{60} \)
- Rate of B: \( \frac{3}{60} \)

Thus,
\[ \text{Combined rate of A and B} = \frac{4}{60} + \frac{3}{60} = \frac{7}{60} \]

### Step 7: Calculate the time to fill the remaining part
Let \( t \) be the time taken to fill the remaining \( \frac{7}{30} \):
\[ t \times \frac{7}{60} = \frac{7}{30} \]
Solving for \( t \):
\[ t = \frac{7}{30} \times \frac{60}{7} = 2 \text{ hours} \]

### Step 8: Total time taken to fill the tank
The total time taken to fill the tank is:
\[ 10 \text{ hours} + 2 \text{ hours} = 12 \text{ hours} \]

### Final Answer
The tank will be full in **12 hours**.

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