Pipes A, B and C together can fill a tank in 4 hours. Pipes A and B together can fill the tank in 9 hours. If pipes A, B and C are opened simultaneously, and after 3 hours, pipes A and B are closed, then how many hours will C alone require to fill the tank?
(a)1.5
(b)1.8
(c)2
(d)2.25

Options

A. 1.5
B. 1.8
C. 2
D. 2.25

Answer

To solve the problem step by step, we will first determine the rates at which each pipe fills the tank and then calculate how long pipe C will take to fill the tank alone after pipes A and B have been closed.

### Step 1: Determine the filling rates of the pipes
- **Pipes A, B, and C together fill the tank in 4 hours.**
- Therefore, their combined rate is \( \frac{1}{4} \) of the tank per hour.

- **Pipes A and B together fill the tank in 9 hours.**
- Therefore, their combined rate is \( \frac{1}{9} \) of the tank per hour.

### Step 2: Find the rate of pipe C
Let the rate of pipe C be \( r_C \). We can express the rates mathematically:
\[ r_A + r_B + r_C = \frac{1}{4} \quad \text{(1)} \]
\[ r_A + r_B = \frac{1}{9} \quad \text{(2)} \]

From equation (2), we can substitute \( r_A + r_B \) in equation (1):
\[ \frac{1}{9} + r_C = \frac{1}{4} \]

### Step 3: Solve for \( r_C \)
To isolate \( r_C \), we subtract \( \frac{1}{9} \) from \( \frac{1}{4} \):
\[ r_C = \frac{1}{4} - \frac{1}{9} \]

To perform this subtraction, we need a common denominator. The least common multiple of 4 and 9 is 36:
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
Thus,
\[ r_C = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

### Step 4: Calculate the work done by A, B, and C in 3 hours
In 3 hours, the amount of work done by pipes A, B, and C together is:
\[ \text{Work done} = 3 \times \left( \frac{1}{4} \right) = \frac{3}{4} \]

### Step 5: Determine the remaining work
The total work to fill the tank is 1 (the whole tank), so the remaining work after 3 hours is:
\[ \text{Remaining work} = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 6: Calculate the time required for pipe C to fill the remaining work
Since pipe C fills at a rate of \( \frac{5}{36} \) of the tank per hour, we can find the time \( t \) required for C to fill the remaining \( \frac{1}{4} \):
\[ \frac{5}{36} \times t = \frac{1}{4} \]
To solve for \( t \):
\[ t = \frac{1}{4} \div \frac{5}{36} = \frac{1}{4} \times \frac{36}{5} = \frac{36}{20} = \frac{9}{5} = 1.8 \text{ hours} \]

### Final Answer
Thus, pipe C alone will require **1.8 hours** to fill the tank.