A tank can be filled by two taps X and Y in 5 hrs and 10 hrs respectively while another tap Z empties the tank in 20 hrs. In how many hours can the tank be filled if all 3 taps kept open ?
A. 5
B. 4
C. 7
D. 8

Options

A. C
B. B
C. D
D. A

Answer

To solve the problem of how long it will take to fill the tank when all three taps (X, Y, and Z) are open, we can follow these steps:

### Step 1: Determine the filling rates of taps X and Y, and the emptying rate of tap Z.

- Tap X fills the tank in 5 hours. Therefore, the rate of tap X is:
\[ \text{Rate of X} = \frac{1 \text{ tank}}{5 \text{ hours}} = \frac{1}{5} \text{ tanks per hour} \]

- Tap Y fills the tank in 10 hours. Therefore, the rate of tap Y is:
\[ \text{Rate of Y} = \frac{1 \text{ tank}}{10 \text{ hours}} = \frac{1}{10} \text{ tanks per hour} \]

- Tap Z empties the tank in 20 hours. Therefore, the rate of tap Z is:
\[ \text{Rate of Z} = -\frac{1 \text{ tank}}{20 \text{ hours}} = -\frac{1}{20} \text{ tanks per hour} \]

### Step 2: Combine the rates of all three taps.

To find the net rate of filling the tank when all three taps are open, we add the rates of X and Y and subtract the rate of Z:
\[ \text{Net Rate} = \text{Rate of X} + \text{Rate of Y} + \text{Rate of Z} \]
\[ \text{Net Rate} = \frac{1}{5} + \frac{1}{10} - \frac{1}{20} \]

### Step 3: Find a common denominator and calculate the net rate.

The least common multiple (LCM) of 5, 10, and 20 is 20. We can express each rate with a denominator of 20:
\[ \text{Rate of X} = \frac{4}{20}, \quad \text{Rate of Y} = \frac{2}{20}, \quad \text{Rate of Z} = -\frac{1}{20} \]
Now, substituting these values into the net rate equation:
\[ \text{Net Rate} = \frac{4}{20} + \frac{2}{20} - \frac{1}{20} = \frac{4 + 2 - 1}{20} = \frac{5}{20} = \frac{1}{4} \text{ tanks per hour} \]

### Step 4: Calculate the time to fill the tank.

If the net rate of filling the tank is \(\frac{1}{4}\) tanks per hour, then the time to fill 1 tank is the reciprocal of the net rate:
\[ \text{Time} = \frac{1 \text{ tank}}{\frac{1}{4} \text{ tanks per hour}} = 4 \text{ hours} \]

### Conclusion

Thus, the tank can be filled in **4 hours** when all three taps are kept open.