Ethanol Fuel Mixture: Find the Right Blend

Mixture Problems 9th-10th Grade
PROBLEM
Sofia owns a car that runs on a mixture of gasoline and ethanol. She can buy fuels that have 85% ethanol or 25% ethanol. How much of each type of fuel should she mix if she wants to fill her 20-gallon tank with a mixture of fuel that contains 50% ethanol?

What This Problem Teaches

Mixture equation setup: Track the pure substance (ethanol) from each source rather than just volumes
Variable definition strategy: Choose one unknown that automatically defines the other through constraints
Percentage-to-decimal conversion: Apply concentrations as multipliers to find actual amounts
Fraction arithmetic in context: Work confidently with answers that aren't whole numbers
Alternative solution methods: Recognize when visual proportion methods (alligation) offer elegant shortcuts

Picture This

Fuel TypeEthanol %Amount (gallons)Pure Ethanol (gallons)
E85 (high ethanol)85%x0.85x
E25 (low ethanol)25%20 - x0.25(20 - x)
Final mixture50%2010

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:

Ethanol from E85 + Ethanol from E25 = Total ethanol needed
0.85x + 0.25(20 - x) = 0.50 × 20

Step 3 — Solve the equation

First, simplify the right side:

0.85x + 0.25(20 - x) = 10

Distribute on the left:

0.85x + 5 - 0.25x = 10

Combine like terms:

0.60x + 5 = 10
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:

E85: 85% \
X 50% (target)
E25: 25% /

Step 2 — Calculate the differences

Find how far each concentration is from the target:

E85 is 85% - 50% = 35 percentage points above 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:

Ratio E85 : E25 = 25 : 35 = 5 : 7

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

E85: (5/12) × 20 = 100/12 = 8⅓ gallons
E25: (7/12) × 20 = 140/12 = 11⅔ gallons
Sofia should mix 8⅓ gallons of E85 fuel with 11⅔ gallons of E25 fuel.

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:

(concentration₁)(amount₁) + (concentration₂)(amount₂) = (target concentration)(total amount)

With the volume constraint: amount₁ + amount₂ = total amount

For this specific problem:

0.85x + 0.25(20-x) = 0.50(20)

The general alligation formula: When mixing concentrations A and B to get target T, the ratio is:

Amount of A : Amount of B = |T - B| : |A - T|

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?

1
Different Target Concentration
Sofia decides she wants a mixture that is 60% ethanol instead. Her tank still holds 20 gallons. How much E85 and E25 should she mix now?
Step 1 — Set up with new target

Let x = gallons of E85. Then 20-x = gallons of E25. Target is now 60% ethanol.

Step 2 — Write the equation

0.85x + 0.25(20-x) = 0.60(20)

Step 3 — Solve

0.85x + 5 - 0.25x = 12
0.60x = 7
x = 7/0.60 = 35/3 = 11⅔

Step 4 — Find amounts

E85: 11⅔ gallons, E25: 8⅓ gallons

Step 5 — Verify

Pure ethanol: 0.85(35/3) + 0.25(25/3) = 35/3 + 25/12 = 12 gallons
Concentration: 12/20 = 0.60 = 60%

2
Partial Pre-fill
Sofia's tank already contains 6 gallons of E25 fuel. She wants to fill the remaining 14 gallons with E85 to achieve some final ethanol percentage. What will be the ethanol percentage of her full 20-gallon tank?
Step 1 — Identify the amounts

E25: 6 gallons (already in tank)
E85: 14 gallons (to be added)
Total: 20 gallons

Step 2 — Calculate pure ethanol from each

From E25: 0.25 × 6 = 1.5 gallons
From E85: 0.85 × 14 = 11.9 gallons

Step 3 — Find total pure ethanol

Total pure ethanol: 1.5 + 11.9 = 13.4 gallons

Step 4 — Calculate final concentration

Final concentration: 13.4/20 = 0.67 = 67%

Step 5 — Verify

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%.

3
Three-Fuel Mixture
A gas station offers E85 (85% ethanol), E50 (50% ethanol), and E15 (15% ethanol). Sofia wants 30 gallons of a 45% ethanol mixture. She decides to use equal amounts of E85 and E50. How much of each fuel should she use?
Step 1 — Define variables

Let x = gallons of E85 = gallons of E50 (equal amounts)
Let y = gallons of E15
Total: x + x + y = 30, so y = 30 - 2x

Step 2 — Set up ethanol equation

0.85x + 0.50x + 0.15(30-2x) = 0.45(30)

Step 3 — Solve

1.35x + 4.5 - 0.30x = 13.5
1.05x = 9
x = 9/1.05 = 60/7 ≈ 8.57

Step 4 — Find all amounts

E85: 60/7 gallons ≈ 8.57 gallons
E50: 60/7 gallons ≈ 8.57 gallons
E15: 30 - 2(60/7) = 90/7 ≈ 12.86 gallons

Step 5 — Verify

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%

4
Reverse Engineering
Sofia mixed some amount of E85 with 15 gallons of E25 and ended up with exactly 40% ethanol in her final mixture. How much E85 did she use, and what's the total volume of her mixture?
Step 1 — Define the unknowns

Let x = gallons of E85 used
E25 used: 15 gallons (given)
Total volume: x + 15 gallons

Step 2 — Set up the ethanol equation

Pure ethanol from E85 + Pure ethanol from E25 = 40% of total volume
0.85x + 0.25(15) = 0.40(x + 15)

Step 3 — Solve for x

0.85x + 3.75 = 0.40x + 6
0.45x = 2.25
x = 5

Step 4 — Find total volume

E85 used: 5 gallons
Total mixture: 5 + 15 = 20 gallons

Step 5 — Verify

Pure ethanol: 0.85(5) + 0.25(15) = 4.25 + 3.75 = 8 gallons
Final concentration: 8/20 = 0.40 = 40%

Frequently Asked Questions

How do you set up an ethanol mixture equation?+
Track the pure ethanol from each source. If x gallons come from 85% fuel, it contributes 0.85x gallons of pure ethanol. The remaining (20-x) gallons from 25% fuel contribute 0.25(20-x) gallons. Set their sum equal to the target: 0.85x + 0.25(20-x) = 0.50(20).
Why multiply percentages by the amounts in mixture problems?+
Percentages tell you the concentration, but you need actual amounts of the pure substance. In this problem, 85% ethanol means 0.85 gallons of pure ethanol per gallon of fuel. So x gallons of E85 contains 0.85x gallons of actual ethanol.
What's the alligation method for mixture problems?+
Alligation finds the ratio by measuring how far each concentration is from the target. Here, E85 (85%) is 35 points above the 50% target, and E25 (25%) is 25 points below. The ratio needed is inversely proportional: 25:35 or 5:7, giving 8⅓ gallons of E85 and 11⅔ gallons of E25.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-05-27