Two pipes A and B together can fill a tank in 5 days. If A fill twice as efficiently as he actually did and B fill 1/3 as efficiently as he actually did then, they can fill the tank in 3 days. In how much time B alone can fill it?

Options

A. 25 days
B. 18 days
C. 15 days
D. 20 days

Answer

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

### Step 1: Define Variables
Let:
- \( x \) = the efficiency of pipe A (amount of tank filled by A in one day)
- \( y \) = the efficiency of pipe B (amount of tank filled by B in one day)

### Step 2: Set Up the First Equation
According to the problem, together pipes A and B can fill the tank in 5 days. Therefore, their combined efficiency can be expressed as:
\[ x + y = \frac{1}{5} \quad \text{(since they fill the tank in 5 days)} \]

### Step 3: Set Up the Second Equation
If pipe A filled twice as efficiently and pipe B filled one-third as efficiently, their new efficiencies would be:
- Efficiency of A: \( 2x \)
- Efficiency of B: \( \frac{1}{3}y \)

Together, they can fill the tank in 3 days, giving us:
\[ 2x + \frac{1}{3}y = \frac{1}{3} \quad \text{(since they fill the tank in 3 days)} \]

### Step 4: Eliminate Fractions
To eliminate the fractions in the second equation, multiply the entire equation by 3:
\[ 3(2x) + 3\left(\frac{1}{3}y\right) = 3\left(\frac{1}{3}\right) \]
This simplifies to:
\[ 6x + y = 1 \]

### Step 5: Solve the System of Equations
Now we have a system of two equations:
1. \( x + y = \frac{1}{5} \)
2. \( 6x + y = 1 \)

We can solve these equations simultaneously. From the first equation, we can express \( y \) in terms of \( x \):
\[ y = \frac{1}{5} - x \]

Substituting this expression for \( y \) into the second equation:
\[ 6x + \left(\frac{1}{5} - x\right) = 1 \]
This simplifies to:
\[ 6x - x + \frac{1}{5} = 1 \]
\[ 5x + \frac{1}{5} = 1 \]
Subtract \( \frac{1}{5} \) from both sides:
\[ 5x = 1 - \frac{1}{5} \]
\[ 5x = \frac{5}{5} - \frac{1}{5} = \frac{4}{5} \]
Dividing both sides by 5:
\[ x = \frac{4}{25} \]

### Step 6: Find \( y \)
Now substitute \( x \) back into the equation for \( y \):
\[ y = \frac{1}{5} - \frac{4}{25} \]
Convert \( \frac{1}{5} \) to a fraction with a denominator of 25:
\[ y = \frac{5}{25} - \frac{4}{25} = \frac{1}{25} \]

### Step 7: Calculate Time for B Alone
The efficiency of pipe B is \( y = \frac{1}{25} \). Therefore, the time taken by pipe B alone to fill the tank is the reciprocal of its efficiency:
\[ \text{Time for B alone} = \frac{1}{y} = \frac{1}{\frac{1}{25}} = 25 \text{ days} \]

### Final Answer
Pipe B alone can fill the tank in **25 days**.