A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank ?

Options

A. 20 hrs
B. 25 hrs
C. 35 hrs
D. Cannot be determined

Answer

To solve the problem, we need to determine how long pipe A alone will take to fill the tank. Let's break it down step by step.

### Step 1: Define the rates of work for each pipe
Let the time taken by pipe A to fill the tank alone be \( x \) hours.

- The rate of work of pipe A is \( \frac{1}{x} \) (i.e., it fills \( \frac{1}{x} \) of the tank in one hour).
- Since pipe B is twice as fast as A, the time taken by pipe B to fill the tank alone will be \( \frac{x}{2} \) hours. Thus, the rate of work of pipe B is \( \frac{2}{x} \).
- Pipe C is twice as fast as pipe B, so the time taken by pipe C to fill the tank alone will be \( \frac{x}{4} \) hours. Therefore, the rate of work of pipe C is \( \frac{4}{x} \).

### Step 2: Combine the rates of work
According to the problem, when all three pipes A, B, and C work together, they fill the tank in 5 hours. Therefore, their combined rate of work is:

\[ \frac{1}{5} \text{ (tank per hour)} \]

Now, we can express the combined rate of work of pipes A, B, and C:

\[ \frac{1}{x} + \frac{2}{x} + \frac{4}{x} = \frac{1}{5} \]

### Step 3: Simplify the equation
Combining the rates on the left side gives:

\[ \frac{1 + 2 + 4}{x} = \frac{7}{x} \]

So, we have:

\[ \frac{7}{x} = \frac{1}{5} \]

### Step 4: Solve for \( x \)
To find \( x \), we can cross-multiply:

\[ 7 \cdot 5 = 1 \cdot x \]

This simplifies to:

\[ x = 35 \]

### Conclusion
Therefore, pipe A alone will take **35 hours** to fill the tank.

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