Tap A and B together can fill a tank in 6 hours while B and C together can fill the same tank in 9 hours. If A fill the tank for 4 hours and C fill the tank for 6hours then the remaining tank is filled by B in 5 hours. Then, find in how many hours tap C alone can fill the tank?

Options

A. A)17.5 hours
B. B)21 hours
C. C)24 hours
D. D)22.5 hours

Answer

To solve the problem step by step, we will first determine the efficiencies of taps A, B, and C based on the information provided.

### Step 1: Determine the efficiencies of A, B, and C
- Tap A and B together can fill the tank in 6 hours. Therefore, their combined efficiency is:
\[ \text{Efficiency of A + B} = \frac{1}{6} \text{ tank/hour} \]
- Tap B and C together can fill the tank in 9 hours. Therefore, their combined efficiency is:
\[ \text{Efficiency of B + C} = \frac{1}{9} \text{ tank/hour} \]

### Step 2: Express efficiencies in terms of units
Let the efficiency of A be \( a \), B be \( b \), and C be \( c \). We can express the above equations as:
1. \( a + b = \frac{1}{6} \) (1)
2. \( b + c = \frac{1}{9} \) (2)

### Step 3: Calculate the total work done by A and C
- Tap A works for 4 hours, so the work done by A is:
\[ \text{Work by A} = 4a \]
- Tap C works for 6 hours, so the work done by C is:
\[ \text{Work by C} = 6c \]

### Step 4: Calculate the remaining work done by B
- After A and C have worked, the remaining work is filled by B in 5 hours. Therefore, the work done by B is:
\[ \text{Work by B} = 5b \]

### Step 5: Set up the equation for total work
The total work done to fill the tank is 1 (the whole tank):
\[ 4a + 6c + 5b = 1 \quad (3) \]

### Step 6: Substitute equations (1) and (2) into (3)
From equation (1), we can express \( b \) as:
\[ b = \frac{1}{6} - a \]
From equation (2), we can express \( c \) as:
\[ c = \frac{1}{9} - b \]
Substituting \( b \) into the equation for \( c \):
\[ c = \frac{1}{9} - \left(\frac{1}{6} - a\right) = \frac{1}{9} - \frac{1}{6} + a \]
Finding a common denominator (18):
\[ c = \frac{2}{18} - \frac{3}{18} + a = a - \frac{1}{18} \]

### Step 7: Substitute \( b \) and \( c \) back into equation (3)
Now substitute \( b \) and \( c \) into equation (3):
\[ 4a + 6\left(a - \frac{1}{18}\right) + 5\left(\frac{1}{6} - a\right) = 1 \]
Expanding this:
\[ 4a + 6a - \frac{6}{18} + \frac{5}{6} - 5a = 1 \]
Combine like terms:
\[ (4a + 6a - 5a) + \left(-\frac{1}{3} + \frac{5}{6}\right) = 1 \]
This simplifies to:
\[ 5a + \left(-\frac{1}{3} + \frac{5}{6}\right) = 1 \]
Finding a common denominator for the fractions:
\[ -\frac{2}{6} + \frac{5}{6} = \frac{3}{6} = \frac{1}{2} \]
So:
\[ 5a + \frac{1}{2} = 1 \]
Subtracting \(\frac{1}{2}\) from both sides:
\[ 5a = \frac{1}{2} \]
Thus:
\[ a = \frac{1}{10} \]

### Step 8: Find \( b \) and \( c \)
Now substitute \( a \) back into the equations for \( b \) and \( c \):
\[ b = \frac{1}{6} - \frac{1}{10} = \frac{5}{30} - \frac{3}{30} = \frac{2}{30} = \frac{1}{15} \]
\[ c = \frac{1}{9} - b = \frac{1}{9} - \frac{1}{15} \]
Finding a common denominator (45):
\[ c = \frac{5}{45} - \frac{3}{45} = \frac{2}{45} \]

### Step 9: Calculate the time taken by C to fill the tank alone
The time taken by C to fill the tank alone is given by:
\[ \text{Time} = \frac{\text{Work}}{\text{Efficiency}} = \frac{1}{c} = \frac{1}{\frac{2}{45}} = \frac{45}{2} = 22.5 \text{ hours} \]

### Final Answer
Tap C alone can fill the tank in **22.5 hours**.