Mixing Solutions: Finding the Ratio of Volumes

This problem involves mixing two different solutions, each containing water and acid in specific ratios. We are given the ratio of water to acid in solution A and solution B, and the resulting ratio when x liters of A are mixed with y liters of B. Our goal is to determine the ratio x : y.

Understanding the Composition of Solutions

Let's first break down the composition of Solution A and Solution B based on the given ratios:

• Solution A: The ratio of water to acid is 4 : 5. This means that for every 4 parts of water, there are 5 parts of acid. The total number of parts is $4 + 5 = 9$.
• The fraction of water in Solution A is $\frac{4}{9}$.
• The fraction of acid in Solution A is $\frac{5}{9}$.

• Solution B: The ratio of water to acid is 1 : 2. This means that for every 1 part of water, there are 2 parts of acid. The total number of parts is $1 + 2 = 3$.
• The fraction of water in Solution B is $\frac{1}{3}$.
• The fraction of acid in Solution B is $\frac{2}{3}$.

Calculating Water and Acid in the Mixture

Now, let's consider what happens when x liters of Solution A are mixed with y liters of Solution B.

• From x liters of Solution A:
  • Amount of water = $x \times \text{Fraction of water in A} = x \times \frac{4}{9} = \frac{4x}{9}$ liters.
  • Amount of acid = $x \times \text{Fraction of acid in A} = x \times \frac{5}{9} = \frac{5x}{9}$ liters.
• From y liters of Solution B:
  • Amount of water = $y \times \text{Fraction of water in B} = y \times \frac{1}{3} = \frac{y}{3}$ liters.
  • Amount of acid = $y \times \text{Fraction of acid in B} = y \times \frac{2}{3} = \frac{2y}{3}$ liters.

In the final mixture, the total amount of water and total amount of acid will be the sum of the amounts from Solution A and Solution B.

• Total water in the mixture = $\frac{4x}{9} + \frac{y}{3}$ liters.
• Total acid in the mixture = $\frac{5x}{9} + \frac{2y}{3}$ liters.

Setting Up and Solving the Ratio Equation

The problem states that the ratio of water and acid in the mixture becomes 8 : 13. We can write this as an equation:

$$ \frac{\text{Total Water}}{\text{Total Acid}} = \frac{8}{13} $$

Substitute the expressions for total water and total acid:

$$ \frac{\frac{4x}{9} + \frac{y}{3}}{\frac{5x}{9} + \frac{2y}{3}} = \frac{8}{13} $$

To simplify the fractions within the main fraction, multiply the numerator and the denominator of the left side by the least common multiple of the denominators (9 and 3), which is 9:

$$ \frac{9 \left( \frac{4x}{9} + \frac{y}{3} \right)}{9 \left( \frac{5x}{9} + \frac{2y}{3} \right)} = \frac{8}{13} $$

$$ \frac{4x + 3y}{5x + 6y} = \frac{8}{13} $$

Now, we can cross-multiply to solve for the ratio x : y:

$$ 13(4x + 3y) = 8(5x + 6y) $$

Distribute the numbers on both sides:

$$ 52x + 39y = 40x + 48y $$

Collect terms with x on one side and terms with y on the other side:

$$ 52x - 40x = 48y - 39y $$

$$ 12x = 9y $$

To find the ratio x : y, divide both sides by 12y:

$$ \frac{12x}{12y} = \frac{9y}{12y} $$

$$ \frac{x}{y} = \frac{9}{12} $$

Simplify the fraction $\frac{9}{12}$ by dividing both numerator and denominator by their greatest common divisor, which is 3:

$$ \frac{x}{y} = \frac{9 \div 3}{12 \div 3} = \frac{3}{4} $$

So, the ratio x : y is 3 : 4.

Conclusion on Mixing Ratio

When x liters of solution A are mixed with y liters of solution B, and the resulting water to acid ratio is 8:13, the ratio of the volumes mixed, x : y, is found to be 3 : 4.

Revision Table: Key Concepts in Ratio Problems

Additional Information on Mixing Ratios and Proportions

Mixing problems are common in quantitative aptitude and chemistry. They often involve combining substances with different concentrations or ratios and finding the resulting concentration or ratio. Here are some related points:

• Concentration: Ratios can represent concentrations. For example, the fraction of acid in a solution is its concentration by volume (or mass, depending on context).
• Weighted Average: The resulting concentration or ratio in a mixture can be thought of as a weighted average of the concentrations or ratios of the components, weighted by the volumes (or masses) mixed.
• Algebraic Method vs. Alligation Method: While the algebraic method shown here is versatile, some mixing problems can also be solved using the Alligation method, which is a shortcut based on weighted averages, particularly useful when mixing two components to get a desired ratio.
• Units: Ensure consistency in units when solving mixing problems. In this case, all volumes are in liters, so it's straightforward.

Understanding how to work with ratios, fractions, and setting up algebraic equations is fundamental to solving these types of problems efficiently.