Understanding the Mixture Problem

This problem involves a mixture of acid and water where the concentration of acid changes when water is added. The key to solving this type of mixture problem is to understand that the amount of acid in the mixture remains constant, even though the total volume of the mixture and the percentage of acid change.

Setting up the Initial Mixture

Let the original quantity of the mixture be \(M\) litres. We are given that the mixture contains 20 percent acid. This means:

• Quantity of acid in the original mixture = 20% of \(M\) = \(0.20 \times M\) litres.
• Quantity of water in the original mixture = (100 - 20)% of \(M\) = 80% of \(M\) = \(0.80 \times M\) litres.

The total original quantity is the sum of acid and water: \(0.20M + 0.80M = M\) litres.

Analyzing the Change

When 10 litres of water are added to the mixture, the total quantity of the mixture increases, but the quantity of acid remains exactly the same. Only water is added.

• Quantity of acid in the new mixture = Quantity of acid in the original mixture = \(0.20 \times M\) litres.
• Quantity of water in the new mixture = Original water quantity + Added water = \(0.80 \times M + 10\) litres.
• Total quantity of the new mixture = Original mixture quantity + Added water = \(M + 10\) litres.

Alternatively, the total quantity of the new mixture is the sum of the new acid and water quantities: \(0.20M + (0.80M + 10) = M + 10\) litres. Both ways give the same total quantity for the new mixture.

Formulating the Equation

We are told that in the new mixture, the percentage of acid becomes 15 percent. The percentage of acid in any mixture is calculated as:

Percentage of acid \( = \left( \frac{\text{Quantity of Acid}}{\text{Total Quantity of Mixture}} \right) \times 100 \)

Using the quantities for the new mixture, we can write the equation:

\(15 = \left( \frac{0.20 \times M}{M + 10} \right) \times 100\)

Now, we need to solve this equation for \(M\).

Solving for the Original Quantity

Let's solve the equation step-by-step:

\(15 = \frac{0.20M}{M + 10} \times 100\)

Divide both sides by 100:

\(\frac{15}{100} = \frac{0.20M}{M + 10}\)

\(0.15 = \frac{0.20M}{M + 10}\)

Multiply both sides by \((M + 10)\) to remove the denominator:

\(0.15 \times (M + 10) = 0.20M\)

Distribute 0.15 on the left side:

\(0.15M + 0.15 \times 10 = 0.20M\)

\(0.15M + 1.5 = 0.20M\)

Subtract 0.15M from both sides to isolate the term with M:

\(1.5 = 0.20M - 0.15M\)

\(1.5 = 0.05M\)

Now, divide both sides by 0.05 to find the value of M:

\(M = \frac{1.5}{0.05}\)

To make the division easier, multiply the numerator and denominator by 100:

\(M = \frac{1.5 \times 100}{0.05 \times 100}\)

\(M = \frac{150}{5}\)

\(M = 30\)

So, the original quantity of the mixture was 30 litres.

Verification

Let's check if this answer is correct:

• Original mixture quantity = 30 litres
• Original acid quantity = 20% of 30 litres = \(0.20 \times 30 = 6\) litres
• Original water quantity = 30 - 6 = 24 litres

Add 10 litres of water:

• New total mixture quantity = 30 + 10 = 40 litres
• New acid quantity = 6 litres (remains unchanged)
• New water quantity = 24 + 10 = 34 litres

Calculate the percentage of acid in the new mixture:

Percentage of acid \( = \left( \frac{\text{Quantity of Acid}}{\text{Total Quantity of Mixture}} \right) \times 100 = \left( \frac{6}{40} \right) \times 100\)

\( = \frac{600}{40} = 15\) percent

The calculated percentage matches the given information (15 percent), so our answer is correct.

Summary of Calculations

Equation based on constant acid quantity:

\(0.20M = 0.15(M + 10)\)

Solving gives \(M = 30\) litres.

Revision Table: Key Concepts for Mixture Problems

Additional Information on Mixture Problems

Mixture problems are common in quantitative aptitude tests. They usually involve combining different substances or changing the concentration of a substance in a mixture. Here are some related concepts and types:

• Combining Mixtures: Problems where two or more different mixtures (with different concentrations) are combined. The total quantity and the total quantity of each component are added up to find the composition of the new mixture.
• Removing and Replacing: Problems where a part of the mixture is removed and replaced by another substance (often water or one of the original components). This changes the concentration in a more complex way, often requiring iterative calculations or formulas.
• Dilution: Adding a solvent (like water) to a solution, which decreases the concentration of the solute (like acid or salt). Our current problem is an example of dilution.
• Concentration: Refers to the amount of a substance in a defined space. In mixtures, it's often expressed as a percentage or ratio of one component to the total mixture.

Always identify what is changing and what is staying constant. This is the first crucial step in solving mixture problems. Setting up a clear equation based on the constant quantity or the final concentration is key.