A cistern can be filled by two pipes in 20 and 30 minutes respectively. Both pipes being opened, when the first pipe must be turned off so that the cistern may be filled in 10 minutes more.

Options

A. after 10 minutes
B. after 12 minutes
C. after 20 minutes
D. after 8 minutes

Answer

To solve the problem step by step, let's break it down:

### Step 1: Determine the rates of the pipes
- Pipe A fills the cistern in 20 minutes. Therefore, its rate is:
\[ \text{Rate of Pipe A} = \frac{1 \text{ cistern}}{20 \text{ minutes}} = \frac{1}{20} \text{ cisterns per minute} \]

- Pipe B fills the cistern in 30 minutes. Therefore, its rate is:
\[ \text{Rate of Pipe B} = \frac{1 \text{ cistern}}{30 \text{ minutes}} = \frac{1}{30} \text{ cisterns per minute} \]

### Step 2: Combine the rates of both pipes
When both pipes are opened, their combined rate is:
\[ \text{Combined Rate} = \frac{1}{20} + \frac{1}{30} \]

To add these fractions, find a common denominator (which is 60):
\[ \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60} \]
Thus,
\[ \text{Combined Rate} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \text{ cisterns per minute} \]

### Step 3: Set up the equation for filling the cistern
Let \( x \) be the number of minutes both pipes are open before Pipe A is turned off. In those \( x \) minutes, the amount of the cistern filled is:
\[ \text{Amount filled by both pipes} = \frac{1}{12} \times x \]

After Pipe A is turned off, Pipe B continues to fill the cistern for an additional 10 minutes. The amount filled by Pipe B in 10 minutes is:
\[ \text{Amount filled by Pipe B} = \frac{1}{30} \times 10 = \frac{10}{30} = \frac{1}{3} \text{ cisterns} \]

### Step 4: Write the total equation
The total amount filled must equal 1 cistern:
\[ \frac{x}{12} + \frac{1}{3} = 1 \]

### Step 5: Solve for \( x \)
First, convert \( \frac{1}{3} \) to have a common denominator of 12:
\[ \frac{1}{3} = \frac{4}{12} \]
Now substitute back into the equation:
\[ \frac{x}{12} + \frac{4}{12} = 1 \]
Multiply through by 12 to eliminate the denominator:
\[ x + 4 = 12 \]
Now, solve for \( x \):
\[ x = 12 - 4 = 8 \]

### Conclusion
Pipe A should be kept open for **8 minutes** before being turned off.

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