Two bottles A and B contain diluted acid. A, the amount of water is double the amount of acid while in bottle B, the amount of acid is 3 times that of water. How much mixture (in litres) should be taken from each bottle A and B respectively in order to prepare 5 litres diluted acid containing equal amount of acid and water?

Options

A. 1, 4
B. 4, 1
C. 2, 3
D. 3, 2

Answer

To solve the problem, we need to determine how much mixture should be taken from each bottle A and B to prepare a total of 5 liters of diluted acid containing equal amounts of acid and water.

### Step 1: Understand the Composition of Each Bottle
- **Bottle A**: The amount of water is double the amount of acid.
- Let the amount of acid in bottle A be \( x \) liters.
- Then, the amount of water in bottle A is \( 2x \) liters.
- Total mixture in bottle A = \( x + 2x = 3x \) liters.

- **Bottle B**: The amount of acid is three times that of water.
- Let the amount of water in bottle B be \( y \) liters.
- Then, the amount of acid in bottle B is \( 3y \) liters.
- Total mixture in bottle B = \( 3y + y = 4y \) liters.

### Step 2: Set Up the Equations for the Desired Mixture
We want to prepare 5 liters of a mixture containing equal amounts of acid and water. Therefore, we need:
- 2.5 liters of acid
- 2.5 liters of water

Let \( a \) be the amount taken from bottle A and \( b \) be the amount taken from bottle B.

### Step 3: Express the Amount of Acid and Water from Each Bottle
- From bottle A:
- Acid from A = \( \frac{x}{3x} \cdot a = \frac{a}{3} \)
- Water from A = \( \frac{2x}{3x} \cdot a = \frac{2a}{3} \)

- From bottle B:
- Acid from B = \( \frac{3y}{4y} \cdot b = \frac{3b}{4} \)
- Water from B = \( \frac{y}{4y} \cdot b = \frac{b}{4} \)

### Step 4: Set Up the Total Acid and Water Equations
We need the total acid to equal 2.5 liters and the total water to equal 2.5 liters:
1. Total Acid:
\[ \frac{a}{3} + \frac{3b}{4} = 2.5 \]
2. Total Water:
\[ \frac{2a}{3} + \frac{b}{4} = 2.5 \]

### Step 5: Solve the Equations
To solve these equations, we can eliminate \( a \) and \( b \) by multiplying through by the least common multiple of the denominators.

1. For the acid equation, multiply by 12:
\[ 4a + 9b = 30 \quad \text{(Equation 1)} \]

2. For the water equation, multiply by 12:
\[ 8a + 3b = 30 \quad \text{(Equation 2)} \]

Now we have a system of linear equations:
- \( 4a + 9b = 30 \) (1)
- \( 8a + 3b = 30 \) (2)

### Step 6: Solve the System of Equations
From Equation (1), we can express \( a \) in terms of \( b \):
\[ 4a = 30 - 9b \implies a = \frac{30 - 9b}{4} \]

Substituting \( a \) into Equation (2):
\[ 8\left(\frac{30 - 9b}{4}\right) + 3b = 30 \]
\[ 2(30 - 9b) + 3b = 30 \]
\[ 60 - 18b + 3b = 30 \]
\[ 60 - 15b = 30 \]
\[ -15b = 30 - 60 \]
\[ -15b = -30 \implies b = 2 \]

Now substitute \( b = 2 \) back into Equation (1) to find \( a \):
\[ 4a + 9(2) = 30 \]
\[ 4a + 18 = 30 \]
\[ 4a = 12 \implies a = 3 \]

### Final Answer
Thus, the amounts to be taken from each bottle are:
- From Bottle A: **3 liters**
- From Bottle B: **2 liters**