Two pipes X and Y can fill up a tank in 32 minutes and 16 minutes respectively. If both the pipes are opened simultaneously, after how much time should Y be closed, so that the tank is full in 20 minutes?

Options

A. 6 minutes
B. 10 minutes
C. 4 minutes
D. 8 minutes

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Determine the rates of work for pipes X and Y
- Pipe X can fill the tank in 32 minutes. Therefore, the rate of work of pipe X is:
\[ \text{Rate of X} = \frac{1}{32} \text{ tank per minute} \]
- Pipe Y can fill the tank in 16 minutes. Therefore, the rate of work of pipe Y is:
\[ \text{Rate of Y} = \frac{1}{16} \text{ tank per minute} \]

### Step 2: Calculate the combined rate of work when both pipes are open
- When both pipes are open, their combined rate of work is:
\[ \text{Combined Rate} = \text{Rate of X} + \text{Rate of Y} = \frac{1}{32} + \frac{1}{16} \]
- To add these fractions, we find a common denominator (which is 32):
\[ \text{Combined Rate} = \frac{1}{32} + \frac{2}{32} = \frac{3}{32} \text{ tank per minute} \]

### Step 3: Set up the equation for the total work done
- Let Y be closed after \( t \) minutes. Therefore, for the first \( t \) minutes, both pipes work together, and for the remaining \( 20 - t \) minutes, only pipe X works.
- The amount of work done by both pipes in \( t \) minutes is:
\[ \text{Work by X and Y} = \text{Combined Rate} \times t = \frac{3}{32} t \]
- The amount of work done by pipe X alone in \( 20 - t \) minutes is:
\[ \text{Work by X alone} = \text{Rate of X} \times (20 - t) = \frac{1}{32} (20 - t) \]

### Step 4: Write the total work equation
- The total work done must equal 1 tank:
\[ \frac{3}{32} t + \frac{1}{32} (20 - t) = 1 \]

### Step 5: Simplify the equation
- Multiply through by 32 to eliminate the denominator:
\[ 3t + (20 - t) = 32 \]
- Simplifying gives:
\[ 3t + 20 - t = 32 \]
\[ 2t + 20 = 32 \]
\[ 2t = 12 \]
\[ t = 6 \]

### Conclusion
- Therefore, pipe Y should be closed after **6 minutes**.