Two taps X and Y together can fill a tank in 36 hours, Y and Z in 48 hours and X and Z in 72 hours. If all the three taps are opened, then in how much time (in hours) the tank will be completely filled?

Options

A. 16
B. 32
C. 22
D. 36

Answer

To solve the problem, we need to find the individual work rates of taps X, Y, and Z based on the information given about their combined work rates.

### Step-by-Step Solution:

1. **Understand the Work Rates**:
- Let the work done by tap X in one hour be \( x \).
- Let the work done by tap Y in one hour be \( y \).
- Let the work done by tap Z in one hour be \( z \).

2. **Set Up the Equations**:
- From the problem, we know:
- X and Y together can fill the tank in 36 hours:
\[ x + y = \frac{1}{36} \]
- Y and Z together can fill the tank in 48 hours:
\[ y + z = \frac{1}{48} \]
- X and Z together can fill the tank in 72 hours:
\[ x + z = \frac{1}{72} \]

3. **Add the Equations**:
- We can add all three equations:
\[ (x + y) + (y + z) + (x + z) = \frac{1}{36} + \frac{1}{48} + \frac{1}{72} \]
- This simplifies to:
\[ 2x + 2y + 2z = \frac{1}{36} + \frac{1}{48} + \frac{1}{72} \]

4. **Calculate the Right Side**:
- To add the fractions on the right side, we need a common denominator. The least common multiple (LCM) of 36, 48, and 72 is 144.
- Convert each fraction:
\[ \frac{1}{36} = \frac{4}{144}, \quad \frac{1}{48} = \frac{3}{144}, \quad \frac{1}{72} = \frac{2}{144} \]
- Now, add them:
\[ \frac{4}{144} + \frac{3}{144} + \frac{2}{144} = \frac{9}{144} \]

5. **Simplify the Equation**:
- Substitute back into the equation:
\[ 2x + 2y + 2z = \frac{9}{144} \]
- Divide everything by 2:
\[ x + y + z = \frac{9}{288} \]

6. **Find the Time Taken by All Three Taps**:
- The combined work rate of all three taps is:
\[ x + y + z = \frac{1}{32} \]
- Therefore, the time taken to fill the tank when all three taps are opened is the reciprocal of their combined work rate:
\[ \text{Time} = \frac{1}{\frac{1}{32}} = 32 \text{ hours} \]

### Final Answer:
The tank will be completely filled in **32 hours**.