A swimming pool has 3 drain pipes. The first two pipes A and B, operating simultaneously, can empty the pool in half the time that C, the 3rd pipe, alone takes to empty it. Pipe A, working alone, takes half the time taken by pipe B. Together they take 6 hours 40 minutes to empty the pool. Time taken by pipe A to empty the pool, (in hours) is

Options

A. 15
B. 10
C. 30
D. 7

Answer

To solve the problem, we need to find the time taken by pipe A to empty the swimming pool. Let's break down the information given step by step.

### Step 1: Define Variables
Let:
- The time taken by pipe C to empty the pool = \( x \) hours.
- The time taken by pipe B to empty the pool = \( 2y \) hours (since A takes half the time of B).
- The time taken by pipe A to empty the pool = \( y \) hours.

### Step 2: Relationship Between Pipes
From the problem, we know:
1. Pipes A and B together can empty the pool in half the time that pipe C takes alone, which means:
\[ \frac{1}{y} + \frac{1}{2y} = \frac{2}{x} \]

### Step 3: Convert Time Together
The time taken by pipes A and B together is given as 6 hours 40 minutes, which can be converted to hours:
\[ 6 \text{ hours } 40 \text{ minutes} = 6 + \frac{40}{60} = 6 + \frac{2}{3} = \frac{20}{3} \text{ hours} \]
Thus, we have:
\[ \frac{1}{y} + \frac{1}{2y} = \frac{3}{2y} = \frac{3}{\frac{20}{3}} = \frac{9}{20} \]

### Step 4: Solve for y
From the equation \(\frac{3}{2y} = \frac{9}{20}\):
\[ 3 \cdot 20 = 9 \cdot 2y \]
\[ 60 = 18y \]
\[ y = \frac{60}{18} = \frac{10}{3} \text{ hours} \]

### Step 5: Find Time for Pipe A
Since \( y \) is the time taken by pipe A to empty the pool, we have:
\[ \text{Time taken by pipe A} = \frac{10}{3} \text{ hours} \approx 3.33 \text{ hours} \]

### Final Answer
The time taken by pipe A to empty the pool is \( \frac{10}{3} \) hours or approximately 3 hours and 20 minutes.

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