Pipes A and B can fill a tank in 16 hours and 24 hours, respectively, whereas pipe C can empty the full tank in 40 hours. All three pipes are opened together, but pipe A is closed after 10 hours. After how many hours will the remaining tank be filled?
पाइप A और Bक्रमशः 16 घंटे और 24 घंटे में एक टैंक भर सकते हैं, जबकि पाइप C, 40 घंटे में पूरा टैंक खाली कर सकता है। सभी तीन पाइप एक साथ खोले जाते हैं, लेकिन 10 घंटे के बाद पाइप A बंद कर दिया जाता है। कितने घंटे बाद शेष टैंक भर जाएगा?

Options

A. `15.5`
B. `12.5`
C. ` 20`
D. ` 10`

Answer

To solve the problem step by step, we need to determine how much of the tank is filled by pipes A, B, and C, and then calculate the remaining time to fill the tank after pipe A is closed.

### Step 1: Determine the filling and emptying rates of the pipes.
- **Pipe A** fills the tank in 16 hours, so its rate is:
\[ \text{Rate of A} = \frac{1}{16} \text{ tank/hour} \]

- **Pipe B** fills the tank in 24 hours, so its rate is:
\[ \text{Rate of B} = \frac{1}{24} \text{ tank/hour} \]

- **Pipe C** empties the tank in 40 hours, so its rate is:
\[ \text{Rate of C} = -\frac{1}{40} \text{ tank/hour} \]

### Step 2: Calculate the combined rate of all three pipes when they are opened together.
The combined rate when all three pipes are open is:
\[ \text{Combined Rate} = \text{Rate of A} + \text{Rate of B} + \text{Rate of C} \]
Substituting the rates:
\[ \text{Combined Rate} = \frac{1}{16} + \frac{1}{24} - \frac{1}{40} \]

To add these fractions, we need a common denominator. The least common multiple (LCM) of 16, 24, and 40 is 240.

Converting each rate to have a denominator of 240:
- Rate of A:
\[ \frac{1}{16} = \frac{15}{240} \]
- Rate of B:
\[ \frac{1}{24} = \frac{10}{240} \]
- Rate of C:
\[ -\frac{1}{40} = -\frac{6}{240} \]

Now, adding these together:
\[ \text{Combined Rate} = \frac{15}{240} + \frac{10}{240} - \frac{6}{240} = \frac{19}{240} \text{ tank/hour} \]

### Step 3: Calculate the amount of tank filled in the first 10 hours.
In 10 hours, the amount of tank filled by all three pipes is:
\[ \text{Amount filled in 10 hours} = \text{Combined Rate} \times 10 = \frac{19}{240} \times 10 = \frac{190}{240} = \frac{19}{24} \text{ of the tank} \]

### Step 4: Determine the remaining amount of the tank to be filled.
The remaining amount of the tank is:
\[ \text{Remaining tank} = 1 - \frac{19}{24} = \frac{5}{24} \]

### Step 5: Calculate the combined rate of pipes B and C after pipe A is closed.
After 10 hours, only pipes B and C are working. The combined rate of pipes B and C is:
\[ \text{Rate of B} + \text{Rate of C} = \frac{1}{24} - \frac{1}{40} \]

Finding a common denominator (LCM of 24 and 40 is 120):
- Rate of B:
\[ \frac{1}{24} = \frac{5}{120} \]
- Rate of C:
\[ -\frac{1}{40} = -\frac{3}{120} \]

Now, adding these together:
\[ \text{Combined Rate of B and C} = \frac{5}{120} - \frac{3}{120} = \frac{2}{120} = \frac{1}{60} \text{ tank/hour} \]

### Step 6: Calculate the time required to fill the remaining tank.
The time required to fill the remaining \(\frac{5}{24}\) of the tank is:
\[ \text{Time} = \frac{\text{Remaining tank}}{\text{Combined Rate of B and C}} = \frac{\frac{5}{24}}{\frac{1}{60}} = \frac{5}{24} \times 60 = \frac{5 \times 60}{24} = \frac{300}{24} = 12.5 \text{ hours} \]

### Final Answer:
The remaining tank will be filled after **12.5 hours**.

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