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

Answer

To solve the problem step by step, we will analyze the contents of both bottles A and B, set up equations based on the information given, and find the required quantities.

### Step 1: Understand the composition of the mixtures in both bottles.

- **Bottle A**: The amount of water is double that of acid.
Let's denote the amount of acid in bottle A as \( A_a \) and the amount of water as \( W_a \).
Therefore, if we let the amount of acid be \( x \), then the amount of water will be \( 2x \).
Thus, the ratio of acid to water in bottle A is \( 1:2 \).

- **Bottle B**: The amount of acid is three times that of water.
Let the amount of acid in bottle B be \( A_b \) and the amount of water be \( W_b \).
If we let the amount of water be \( y \), then the amount of acid will be \( 3y \).
Thus, the ratio of acid to water in bottle B is \( 3:1 \).

### Step 2: Set up the equations for the mixtures.

Let \( X \) be the amount taken from bottle A and \( Y \) be the amount taken from bottle B. We want to prepare a total of 5 liters of diluted sulphuric acid containing equal amounts of acid and water.

Since we want equal amounts of acid and water in the final mixture, we can express this requirement mathematically:

1. The total volume of the mixture is:
\[ X + Y = 5 \quad \text{(Equation 1)} \]

2. The acid content from both bottles:
- From bottle A: The acid content is \( \frac{1}{3}X \) (since the ratio is \( 1:2 \)).
- From bottle B: The acid content is \( \frac{3}{4}Y \) (since the ratio is \( 3:1 \)).

Therefore, the total acid content is:
\[ \frac{1}{3}X + \frac{3}{4}Y \quad \text{(Equation 2)} \]

3. The water content from both bottles:
- From bottle A: The water content is \( \frac{2}{3}X \).
- From bottle B: The water content is \( \frac{1}{4}Y \).

Therefore, the total water content is:
\[ \frac{2}{3}X + \frac{1}{4}Y \quad \text{(Equation 3)} \]

### Step 3: Set up the equation for equal acid and water.

Since we want equal amounts of acid and water in the final mixture, we can set the total acid content equal to the total water content:
\[ \frac{1}{3}X + \frac{3}{4}Y = \frac{2}{3}X + \frac{1}{4}Y \]

### Step 4: Solve the equations.

Rearranging the equation gives:
\[ \frac{1}{3}X - \frac{2}{3}X + \frac{3}{4}Y - \frac{1}{4}Y = 0 \]
\[ -\frac{1}{3}X + \frac{2}{4}Y = 0 \]
\[ -\frac{1}{3}X + \frac{1}{2}Y = 0 \]

Multiplying through by 6 to eliminate the fractions:
\[ -2X + 3Y = 0 \quad \Rightarrow \quad 2X = 3Y \quad \Rightarrow \quad \frac{X}{Y} = \frac{3}{2} \]

### Step 5: Substitute back into the total volume equation.

From Equation 1:
\[ X + Y = 5 \]
Let \( X = 3k \) and \( Y = 2k \) (from the ratio \( \frac{3}{2} \)):
\[ 3k + 2k = 5 \quad \Rightarrow \quad 5k = 5 \quad \Rightarrow \quad k = 1 \]

Thus:
\[ X = 3(1) = 3 \text{ liters from bottle A} \]
\[ Y = 2(1) = 2 \text{ liters from bottle B} \]

### Final Answer:
We should take **3 liters from bottle A** and **2 liters from bottle B** to prepare 5 liters of diluted sulphuric acid containing equal amounts of acid and water.

---