A tank can be filled by two pipes in 20 minutes and 30 minutes respectively. When the tank was empty, the two pipes were opened. After some time, the first pipe was stopped and the tank was filled in 18 minutes. After how much time of the start was the first pipe stopped?

Options

A. 5 minutes
B. 8 minutes
C. 10 minutes
D. 12 minutes

Answer

To solve the problem step by step, we will first determine the rates at which the two pipes fill the tank, then calculate how long the first pipe was open before it was stopped.

### Step 1: Determine the filling rates of the pipes
- Let the first pipe (A) fill the tank in 20 minutes. Therefore, the rate of pipe A is:
\[ \text{Rate of A} = \frac{1}{20} \text{ tank/minute} \]
- Let the second pipe (B) fill the tank in 30 minutes. Therefore, the rate of pipe B is:
\[ \text{Rate of B} = \frac{1}{30} \text{ tank/minute} \]

### Step 2: Find the combined rate of both pipes
- The combined rate when both pipes are open is:
\[ \text{Combined Rate} = \text{Rate of A} + \text{Rate of B} = \frac{1}{20} + \frac{1}{30} \]
- To add these fractions, we find a common denominator, which is 60:
\[ \text{Combined Rate} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \text{ tank/minute} \]

### Step 3: Calculate the total time taken to fill the tank
- The total time taken to fill the tank is given as 18 minutes. This means that the tank was filled completely in 18 minutes.

### Step 4: Determine how long each pipe was open
- Let \( t \) be the time (in minutes) that pipe A was open before it was stopped. Therefore, pipe B was open for the entire 18 minutes.
- The amount of tank filled by pipe B in 18 minutes is:
\[ \text{Amount filled by B} = \text{Rate of B} \times \text{Time} = \frac{1}{30} \times 18 = \frac{18}{30} = \frac{3}{5} \text{ of the tank} \]
- The remaining amount to fill the tank is:
\[ \text{Remaining amount} = 1 - \frac{3}{5} = \frac{2}{5} \text{ of the tank} \]

### Step 5: Calculate how much pipe A filled
- The amount filled by pipe A in \( t \) minutes is:
\[ \text{Amount filled by A} = \text{Rate of A} \times t = \frac{1}{20} \times t \]
- The total amount filled by both pipes must equal 1 tank:
\[ \frac{1}{20} t + \frac{3}{5} = 1 \]

### Step 6: Solve for \( t \)
- Convert \( \frac{3}{5} \) to a fraction with a denominator of 20:
\[ \frac{3}{5} = \frac{12}{20} \]
- Substitute this into the equation:
\[ \frac{1}{20} t + \frac{12}{20} = 1 \]
- Multiply through by 20 to eliminate the fraction:
\[ t + 12 = 20 \]
- Solve for \( t \):
\[ t = 20 - 12 = 8 \text{ minutes} \]

### Final Answer
The first pipe was stopped after **8 minutes**.

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