Mixture Math Problems: Practice Questions with Step-by-Step Solutions
Mixture problems use percentages and concentrations to describe what happens when two or more solutions, alloys, or liquids are combined. Work through each problem by setting up the equations yourself first, then open the solution to check your reasoning.
Core Idea: Balance the Active Ingredient
Every mixture problem reduces to two equations:
- Total amount: (amount of part 1) + (amount of part 2) = (final amount)
- Active ingredient: (part 1 × its %) + (part 2 × its %) = (final amount × final %)
Use substitution to solve the system. The same idea applies to alloys (mass of pure metal), solutions (volume of solute), and dilutions (adding pure water = 0% solute).
Practice Problems and Solutions
Problem 1: Diluting an alcohol solution
Question: How many liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution?
Solution:
Let \(x\) be the liters of 20% solution to add. The final volume is:
\[ x + 40 = y \]The alcohol balance gives:
\[ 0.20x + 0.50(40) = 0.30y \]Substitute \(y = x + 40\):
\[ 0.20x + 20 = 0.30(x + 40) \] \[ 0.20x + 20 = 0.30x + 12 \] \[ 8 = 0.10x \implies x = 80 \]80 liters of 20% alcohol solution should be added.
Check: \(0.20(80) + 0.50(40) = 16 + 20 = 36\); \(0.30(120) = 36\). ✔
Problem 2: Mixing two solutions to reach a target
Question: John wants to make 100 ml of a 5% alcohol solution by mixing a 2% solution with a 7% solution. What quantities of each does he need?
Solution:
Let \(x\) and \(y\) be the volumes of the 2% and 7% solutions in ml:
\[ x + y = 100 \] \[ 0.02x + 0.07y = 0.05(100) = 5 \]From the first equation, \(y = 100 - x\). Substitute:
\[ 0.02x + 0.07(100 - x) = 5 \] \[ 0.02x + 7 - 0.07x = 5 \] \[ -0.05x = -2 \implies x = 40 \] \[ y = 100 - 40 = 60 \]Mix 40 ml of the 2% solution with 60 ml of the 7% solution.
Problem 3: Sterling silver alloy mixture
Question: Sterling Silver is 92.5% pure silver. How many grams of Sterling Silver must be mixed with a 90% silver alloy to obtain 500 g of a 91% silver alloy?
Solution:
Let \(x\) be the mass of Sterling Silver (92.5%) and \(y\) the mass of the 90% alloy:
\[ x + y = 500 \] \[ 0.925x + 0.90y = 0.91(500) = 455 \]Substitute \(y = 500 - x\):
\[ 0.925x + 0.90(500 - x) = 455 \] \[ 0.925x + 450 - 0.90x = 455 \] \[ 0.025x = 5 \implies x = 200 \]200 grams of Sterling Silver is needed.
Problem 4: Adding pure water
Question: How many kilograms of pure water must be added to 100 kg of a 30% saline solution to obtain a 10% saline solution?
Solution:
Let \(x\) be the kilograms of water added, and \(y\) the final weight:
\[ x + 100 = y \]Pure water has 0% salt. The salt content in the original solution is:
\[ 0.30(100) = 30 \ \text{kg} \]The final solution must have this same salt at 10%:
\[ 30 = 0.10y \]Substitute \(y = x + 100\):
\[ 30 = 0.10(x + 100) \] \[ 300 = x + 100 \implies x = 200 \]200 kilograms of pure water must be added.
Check: Final weight is 300 kg; \(30/300 = 10\%\). ✔
Problem 5: Calculating final percentage
Question: A 50 ml after-shave lotion at 30% alcohol is mixed with 30 ml of pure water. What is the percentage of alcohol in the new solution?
Solution:
Total volume:
\[ 50 + 30 = 80 \ \text{ml} \]Alcohol from the lotion (water contributes 0):
\[ 0.30 \times 50 = 15 \ \text{ml} \]Let \(x\) be the final percentage. Then:
\[ x \cdot 80 = 15 \] \[ x = \frac{15}{80} = 0.1875 = 18.75\% \]The new solution is 18.75% alcohol.
Problem 6: Variable addition and ratios
Question: You add \(x\) ml of a 25% alcohol solution to 200 ml of a 10% alcohol solution.
- Find the amount of alcohol in the final solution in terms of \(x\).
- Find the ratio of alcohol to total volume in terms of \(x\).
- What happens if \(x\) is very large?
- Find \(x\) so the final solution is 15% alcohol.
Solution:
a) Alcohol from the 200 ml at 10%:
\[ 200 \times 0.10 = 20 \ \text{ml} \]Alcohol from \(x\) ml at 25%:
\[ 0.25x \ \text{ml} \]Total alcohol: \(20 + 0.25x\).
b) Total volume is \(x + 200\), so the ratio is:
\[ r = \frac{20 + 0.25x}{x + 200} \]c) As \(x\) becomes very large, \(r \to 0.25 = 25\%\). The added 25% solution dominates, so the mixture approaches 25% alcohol.
d) Set \(r = 0.15\):
\[ \frac{20 + 0.25x}{x + 200} = 0.15 \] \[ 20 + 0.25x = 0.15(x + 200) \] \[ 20 + 0.25x = 0.15x + 30 \] \[ 0.10x = 10 \implies x = 100 \]You must add 100 ml to reach 15%.
Check: \((20 + 25)/(300) = 45/300 = 0.15\). ✔
Key Takeaways
- Balance the active ingredient: (volume × concentration) before = (volume × concentration) after.
- Pure water contributes 0% solute; a pure solute contributes 100%.
- Use substitution to reduce a two-variable system to a single equation in one variable.
- Always verify by recomputing the final percentage from your answer.
- When one solution is added in very large amounts, the final concentration approaches the concentration of that added solution.