Pipes P and Q can fill a tank in 10 and 12 hours respectively and C can empty it in 6 hours. If all the three are opened at 7 a.m., at what time will one-fourth of the tank be filled ?

Options

A. 10 a.m.
B. 10 p.m.
C. 11 p.m.
D. 11 a.m.

Answer

To solve the problem step by step, we will calculate the filling and emptying rates of the pipes and then determine how long it will take to fill one-fourth of the tank.

### Step 1: Determine the filling rates of pipes P and Q
- Pipe P can fill the tank in 10 hours. Therefore, its rate of filling is:
\[ \text{Rate of P} = \frac{1 \text{ tank}}{10 \text{ hours}} = \frac{1}{10} \text{ tanks/hour} \]

- Pipe Q can fill the tank in 12 hours. Therefore, its rate of filling is:
\[ \text{Rate of Q} = \frac{1 \text{ tank}}{12 \text{ hours}} = \frac{1}{12} \text{ tanks/hour} \]

### Step 2: Determine the emptying rate of pipe C
- Pipe C can empty the tank in 6 hours. Therefore, its rate of emptying is:
\[ \text{Rate of C} = \frac{1 \text{ tank}}{6 \text{ hours}} = \frac{1}{6} \text{ tanks/hour} \]
Since it is emptying, we will consider this rate as negative:
\[ \text{Rate of C} = -\frac{1}{6} \text{ tanks/hour} \]

### Step 3: Calculate the combined rate of filling when all pipes are opened
- The combined rate of filling when P, Q, and C are opened together is:
\[ \text{Combined Rate} = \text{Rate of P} + \text{Rate of Q} + \text{Rate of C} \]
\[ = \frac{1}{10} + \frac{1}{12} - \frac{1}{6} \]

### Step 4: Find a common denominator and simplify
- The least common multiple of 10, 12, and 6 is 60. We convert each rate:
\[ \frac{1}{10} = \frac{6}{60}, \quad \frac{1}{12} = \frac{5}{60}, \quad -\frac{1}{6} = -\frac{10}{60} \]

- Now, substituting these values into the combined rate:
\[ \text{Combined Rate} = \frac{6}{60} + \frac{5}{60} - \frac{10}{60} = \frac{1}{60} \text{ tanks/hour} \]

### Step 5: Calculate the time to fill one-fourth of the tank
- To fill one-fourth of the tank, we need to fill:
\[ \frac{1}{4} \text{ tank} \]

- Using the combined rate:
\[ \text{Time} = \frac{\text{Amount of tank}}{\text{Combined Rate}} = \frac{\frac{1}{4}}{\frac{1}{60}} = \frac{60}{4} = 15 \text{ hours} \]

### Step 6: Determine the time when the tank will be one-fourth full
- Since all pipes are opened at 7 a.m., we add 15 hours to this time:
\[ 7 \text{ a.m.} + 15 \text{ hours} = 10 \text{ p.m.} \]

### Final Answer
The tank will be one-fourth full at **10 p.m.**.

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