Problem Setup

The goal is to find the volumes from Bottle A and Bottle B needed to create 5 liters of a diluted acid mixture where the acid and water amounts are equal.

Bottle A Composition

In Bottle A, the amount of water is double the amount of acid. This means for every 1 part acid, there are 2 parts water.

• Ratio of Acid to Water = 1:2
• Total parts = 1 + 2 = 3
• Concentration of Acid in Bottle A = $ \frac{1}{3} $

Bottle B Composition

In Bottle B, the amount of acid is 3 times the amount of water. This means for every 1 part water, there are 3 parts acid.

• Ratio of Water to Acid = 1:3
• Total parts = 1 + 3 = 4
• Concentration of Acid in Bottle B = $ \frac{3}{4} $

Final Mixture Specifications

• Total required volume = 5 liters
• The final mixture must contain an equal amount of acid and water.
• Concentration of Acid in the Final Mixture = $ \frac{1}{2} $

Formulating the Equations

Let $x$ be the volume in liters taken from Bottle A.

Let $y$ be the volume in liters taken from Bottle B.

1. Total Volume Constraint: The sum of volumes from both bottles must equal the final volume. $ x + y = 5 \quad (1) $
2. Total Acid Constraint: The total acid contributed by the volumes $x$ and $y$ must equal the amount of acid in the final 5-liter mixture. $ (\text{Volume from A} \times \text{Acid Concentration in A}) + (\text{Volume from B} \times \text{Acid Concentration in B}) = (\text{Final Volume} \times \text{Final Acid Concentration}) $ $ \left( x \times \frac{1}{3} \right) + \left( y \times \frac{3}{4} \right) = \left( 5 \times \frac{1}{2} \right) $ $ \frac{x}{3} + \frac{3y}{4} = \frac{5}{2} \quad (2) $

Solving the System of Equations

1. Isolate one variable: From equation (1), we can express $x$ as: $ x = 5 - y $
2. Substitution: Substitute this expression for $x$ into equation (2): $ \frac{5 - y}{3} + \frac{3y}{4} = \frac{5}{2} $
3. Clear denominators: Multiply the entire equation by the least common multiple of 3, 4, and 2, which is 12: $ 12 \times \left( \frac{5 - y}{3} \right) + 12 \times \left( \frac{3y}{4} \right) = 12 \times \left( \frac{5}{2} \right) $ $ 4(5 - y) + 9y = 30 $
4. Simplify and solve for $y$: $ 20 - 4y + 9y = 30 $ $ 20 + 5y = 30 $ $ 5y = 30 - 20 $ $ 5y = 10 $ $ y = \frac{10}{5} = 2 $ So, $y = 2$ liters.
5. Solve for $x$: Substitute the value of $y$ back into equation (1): $ x + 2 = 5 $ $ x = 5 - 2 = 3 $ So, $x = 3$ liters.

Result

To prepare 5 liters of diluted acid with equal amounts of acid and water, 3 liters should be taken from Bottle A and 2 liters from Bottle B.