Mixture Problem: Finding Solution Amounts
What This Problem Teaches
- Setting up mixture problems using concentration and volume relationships
- Understanding that pure substance amounts are conserved in mixing
- Building equations where the unknown appears in multiple terms
- Connecting percentages to actual quantities in real scenarios
- Recognizing why simple percentage averaging doesn't work with unequal volumes
Visualizing the Problem
Let's organize what we know in a table to see the relationship between volumes, concentrations, and pure alcohol:
| Solution | Volume (L) | Concentration | Pure Alcohol (L) |
|---|---|---|---|
| 15% solution | x | 15% = 0.15 | 0.15x |
| 30% solution | 3 | 30% = 0.30 | 0.30(3) = 0.9 |
| Final mixture | 3 + x | 20% = 0.20 | 0.20(3 + x) |
The key insight: pure alcohol from both original solutions equals pure alcohol in the final mixture.
Solution: Method 1 — The Pure Alcohol Balance
Step 1 — Set up the conservation equation
Pure alcohol is conserved when mixing solutions. The total pure alcohol from both original solutions must equal the pure alcohol in the final mixture:
0.15x + 0.9 = 0.20(3 + x)
Step 2 — Expand the right side
Distribute 0.20 across the parentheses:
0.15x + 0.9 = 0.6 + 0.20x
Step 3 — Collect like terms
Move all terms with x to one side and constants to the other:
-0.05x = -0.3
Step 4 — Solve for x
Divide both sides by -0.05:
Solution: Method 2 — The Weighted Average Approach
Step 1 — Think about the target concentration
The 30% solution is 10 percentage points above our target of 20%. The 15% solution is 5 percentage points below our target. To balance these deviations, we need the right ratio of volumes.
Step 2 — Set up the weighted average equation
For a weighted average, each solution contributes proportionally to its volume:
x(0.15) + 3(0.30) = (x + 3)(0.20)
Step 3 — Simplify and solve
This gives us the same equation as Method 1, just derived differently:
0.9 - 0.6 = 0.20x - 0.15x
0.3 = 0.05x
x = 6
Both methods confirm we need 6 liters of the 15% solution.
Verification
Let's check that our answer produces exactly 20% alcohol concentration:
Check the volumes and pure alcohol amounts:
- 15% solution: 6 L containing
6 × 0.15 = 0.9 Lpure alcohol - 30% solution: 3 L containing
3 × 0.30 = 0.9 Lpure alcohol - Final mixture:
6 + 3 = 9 Lcontaining0.9 + 0.9 = 1.8 Lpure alcohol
Check the final concentration:
= 1.8 L ÷ 9 L = 0.2 = 20% ✓
Perfect! The mixture has exactly the required 20% concentration.
Common Pitfalls
✗ Mistake 1: Averaging the percentages
Students often think: "15% + 30% = 45%, so 45% ÷ 2 = 22.5%." This would only be correct if you used equal volumes. Since we use 6 L of 15% and only 3 L of 30%, the weaker solution dominates, pulling the final percentage below 22.5%.
✗ Mistake 2: Forgetting to add volumes in the final mixture
Writing the final mixture as having 3 L instead of (3 + x) L. The total volume increases when you add solutions together — this isn't a replacement, it's an addition.
✗ Mistake 3: Using percentages instead of decimals inconsistently
Mixing notation like 15% × x = 15x instead of 0.15x. When multiplying, percentages must be converted to decimals: 15% becomes 0.15.
The Underlying Pattern
All mixture problems follow the same fundamental principle: the amount of pure substance is conserved. Whether it's alcohol in solutions, salt in water, or gold in alloys, the pure substance doesn't disappear — it just gets redistributed.
If you mix volume
V₁ of concentration C₁ with volume V₂ of concentration C₂:C₁V₁ + C₂V₂ = C_final(V₁ + V₂)In our problem: 0.15x + 0.30(3) = 0.20(x + 3), which fits this pattern perfectly.
Where This Shows Up in Real Life
- Pharmacy: Diluting concentrated medications to prescribed strengths
- Food industry: Blending oils or vinegars to achieve specific acidity levels
- Manufacturing: Creating metal alloys with precise composition percentages
- Environmental science: Calculating pollution concentrations when contaminated water mixes with clean water
Four "What-If?" Problems
Pure alcohol from both solutions equals pure alcohol in final mixture: 0.15x + 0.30(3) = 0.25(x + 3)
0.15x + 0.9 = 0.25x + 0.75
0.9 - 0.75 = 0.25x - 0.15x0.15 = 0.10xx = 1.5
Final mixture: 4.5 L containing 0.15(1.5) + 0.9 = 1.125 L pure alcohol
Concentration: 1.125 ÷ 4.5 = 0.25 = 25% ✓
Answer: 1.5 liters of 15% solution
0.15x + 0.30(5) = 0.20(x + 5)
0.15x + 1.5 = 0.20x + 1.0
1.5 - 1.0 = 0.20x - 0.15x0.5 = 0.05xx = 10
Final mixture: 15 L containing 0.15(10) + 1.5 = 3 L pure alcohol
Concentration: 3 ÷ 15 = 0.20 = 20% ✓
Answer: 10 liters of 15% solution
0.15x + 0.40(3) = 0.20(x + 3)
0.15x + 1.2 = 0.20x + 0.6
1.2 - 0.6 = 0.20x - 0.15x0.6 = 0.05xx = 12
Final mixture: 15 L containing 0.15(12) + 1.2 = 3 L pure alcohol
Concentration: 3 ÷ 15 = 0.20 = 20% ✓
Answer: 12 liters of 15% solution
Total volume is 10 L, and 4 L is 30% solution, so: x = 10 - 4 = 6 L of 15% solution
Check: 0.15(6) + 0.30(4) = 0.22(10)
Left side: 0.9 + 1.2 = 2.1 L pure alcohol
Right side: 2.2 L pure alcohol
Wait! 2.1 ≠ 2.2. Let me solve this properly using the equation:0.15x + 0.30(4) = 0.22(10)0.15x + 1.2 = 2.20.15x = 1.0x = 6.67 L
Answer: 6.67 liters of 15% solution
Frequently Asked Questions
2026-07-01