Ethanol Fuel Mixture: Find the Right Blend
What This Problem Teaches
Picture This
| Fuel Type | Ethanol % | Amount (gallons) | Pure Ethanol (gallons) |
|---|---|---|---|
| E85 (high ethanol) | 85% | x | 0.85x |
| E25 (low ethanol) | 25% | 20 - x | 0.25(20 - x) |
| Final mixture | 50% | 20 | 10 |
The key insight: pure ethanol from both sources must equal the pure ethanol in the final mixture.
Solution: Method 1 — The Component Balance
We track the amount of pure ethanol from each source. This is the standard approach for mixture problems because it focuses on what we're actually trying to control: the concentration of the active ingredient.
Step 1 — Define the variable
Let x = gallons of E85 fuel (85% ethanol) needed.
Since the tank holds 20 gallons total: 20 - x = gallons of E25 fuel (25% ethanol) needed.
Step 2 — Write the ethanol equation
The pure ethanol from both sources must equal the pure ethanol in our target mixture:
0.85x + 0.25(20 - x) = 0.50 × 20
Step 3 — Solve the equation
First, simplify the right side:
Distribute on the left:
Combine like terms:
0.60x = 5
x = 5/0.60 = 25/3 = 8⅓
Step 4 — Find both amounts
E85 needed: x = 8⅓ gallons
E25 needed: 20 - x = 20 - 8⅓ = 11⅔ gallons
Solution: Method 2 — The Alligation Cross
This visual method finds the mixing ratio by measuring how far each concentration is from the target. It's particularly elegant when you need to mix exactly two components.
Step 1 — Draw the alligation diagram
Set up the concentrations and target in a cross pattern:
X 50% (target)
E25: 25% /
Step 2 — Calculate the differences
Find how far each concentration is from the target:
E25 is 50% - 25% = 25 percentage points below target
Step 3 — The ratio principle
The amount of each fuel needed is inversely proportional to its distance from the target:
This means for every 5 parts E85, we need 7 parts E25.
Step 4 — Convert to actual gallons
Total ratio parts: 5 + 7 = 12 parts = 20 gallons
E25: (7/12) × 20 = 140/12 = 11⅔ gallons
Verification
Let's check that our mixture produces exactly 50% ethanol:
Pure ethanol from E85:0.85 × 8⅓ = 0.85 × 25/3 = 85/12 gallons
Pure ethanol from E25:0.25 × 11⅔ = 0.25 × 35/3 = 35/12 gallons
Total pure ethanol:85/12 + 35/12 = 120/12 = 10 gallons
Final concentration:10 gallons ÷ 20 gallons = 0.50 = 50% ✓
Also verify the amounts sum correctly: 8⅓ + 11⅔ = 20 gallons ✓
Common Pitfalls
✗ Mistake 1: Averaging the percentages
Some students think: "85% + 25% = 110%, divided by 2 = 55%, so I need equal amounts to get close to 50%."
Why this fails: You can't average percentages directly — you must weight them by the actual amounts. Equal volumes would give you 55% ethanol, not 50%.
✗ Mistake 2: Setting up the wrong equation
Writing: 0.85x + 0.25x = 0.50(20)
Why this fails: This assumes you're using x gallons of both fuels. The constraint is that total volume equals 20 gallons, so if you use x gallons of E85, you can only use (20-x) gallons of E25.
✗ Mistake 3: Forgetting to convert percentages
Using 85 and 25 instead of 0.85 and 0.25 in the equation.
Why this fails: 85% ethanol means 0.85 gallons of pure ethanol per gallon of fuel, not 85 gallons. Always convert percentages to decimals when setting up mixture equations.
Reality Check
Our answer shows Sofia needs more of the low-ethanol fuel (11⅔ gallons E25) than the high-ethanol fuel (8⅓ gallons E85). Does this make sense?
Absolutely. The target concentration (50%) is much closer to the E25 concentration (25%) than to the E85 concentration (85%). To "pull down" the high concentration of E85 to 50%, you need a lot of the lower concentration fuel.
Distance check: E85 is 35 percentage points away from the target, while E25 is only 25 points away. The fuel that's farther from the target gets used in smaller quantities — which matches our answer.
If Sofia had wanted a 65% ethanol mixture (closer to E85), she would need more E85 and less E25. The closer your target is to one component, the more you need of that component.
How to Spot This Problem Type
Look for these telltale signs that signal a mixture problem:
- "Mix," "blend," or "combine" two or more substances
- Different concentrations of the same ingredient (percentages, ratios, or "parts")
- Target concentration that lies between the starting concentrations
- Total volume constraint (tank capacity, recipe amount, etc.)
- Real-world contexts: fuel blending, medication preparation, alloy composition, solution chemistry
The problem structure is always the same: you're controlling the amount of some active ingredient (ethanol, medicine, salt, etc.) by mixing different concentrations.
The Pattern Behind This
All two-component mixture problems follow this general structure:
With the volume constraint: amount₁ + amount₂ = total amount
For this specific problem:
The general alligation formula: When mixing concentrations A and B to get target T, the ratio is:
Notice how the ratios are inverted — the component farther from the target gets used in smaller proportions. This pattern appears everywhere from cooking recipes to pharmaceutical compounding.
What If?
Let x = gallons of E85. Then 20-x = gallons of E25. Target is now 60% ethanol.
0.85x + 0.25(20-x) = 0.60(20)
0.85x + 5 - 0.25x = 120.60x = 7x = 7/0.60 = 35/3 = 11⅔
E85: 11⅔ gallons, E25: 8⅓ gallons
Pure ethanol: 0.85(35/3) + 0.25(25/3) = 35/3 + 25/12 = 12 gallons
Concentration: 12/20 = 0.60 = 60% ✓
E25: 6 gallons (already in tank)
E85: 14 gallons (to be added)
Total: 20 gallons
From E25: 0.25 × 6 = 1.5 gallons
From E85: 0.85 × 14 = 11.9 gallons
Total pure ethanol: 1.5 + 11.9 = 13.4 gallons
Final concentration: 13.4/20 = 0.67 = 67%
The final mixture will be 67% ethanol. This makes sense since she's using much more E85 (70% of tank) than E25 (30% of tank), pulling the average closer to 85%.
Let x = gallons of E85 = gallons of E50 (equal amounts)
Let y = gallons of E15
Total: x + x + y = 30, so y = 30 - 2x
0.85x + 0.50x + 0.15(30-2x) = 0.45(30)
1.35x + 4.5 - 0.30x = 13.51.05x = 9x = 9/1.05 = 60/7 ≈ 8.57
E85: 60/7 gallons ≈ 8.57 gallons
E50: 60/7 gallons ≈ 8.57 gallons
E15: 30 - 2(60/7) = 90/7 ≈ 12.86 gallons
Check: 8.57 + 8.57 + 12.86 = 30 ✓
Ethanol: 0.85(60/7) + 0.50(60/7) + 0.15(90/7) = 13.5 gallons
Concentration: 13.5/30 = 0.45 = 45% ✓
Let x = gallons of E85 used
E25 used: 15 gallons (given)
Total volume: x + 15 gallons
Pure ethanol from E85 + Pure ethanol from E25 = 40% of total volume0.85x + 0.25(15) = 0.40(x + 15)
0.85x + 3.75 = 0.40x + 60.45x = 2.25x = 5
E85 used: 5 gallons
Total mixture: 5 + 15 = 20 gallons
Pure ethanol: 0.85(5) + 0.25(15) = 4.25 + 3.75 = 8 gallons
Final concentration: 8/20 = 0.40 = 40% ✓
Frequently Asked Questions
2026-05-27