Mixture Problem: Antifreeze Solution

Mixture & Concentration 9th-10th Grade
PROBLEM
A radiator contains 13 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze?

What This Problem Teaches

  • How to track the amount of a substance (not just percentages) through mixture changes
  • Setting up equations where the same unknown appears in both "removal" and "addition" terms
  • Understanding that draining removes substance proportionally to current concentration
  • Recognizing drain-and-replace as a fundamental mixture operation in real applications
  • Converting between percentage concentrations and actual quantities

Solution: Method 1 — The Antifreeze Balance Approach

The key insight is to track the actual amount of antifreeze, not just the percentages. We'll follow where the antifreeze goes: some gets removed when we drain, and some gets added when we replace.

Step 1 — Calculate the current antifreeze amount

The radiator currently holds:

Current antifreeze = 0.30 × 13 = 3.9 quarts of pure antifreeze

Step 2 — Set up what happens when we drain and replace

Let x = quarts of fluid drained and replaced.

When we drain x quarts of the 30% mixture, we remove:

Antifreeze removed = 0.30x quarts

When we replace those x quarts with pure antifreeze, we add:

Antifreeze added = x quarts (since pure antifreeze is 100%)

Step 3 — Write the equation for final antifreeze amount

The final mixture should be 60% antifreeze out of 13 quarts total:

Target antifreeze = 0.60 × 13 = 7.8 quarts

The balance equation is:

(Original antifreeze) - (Antifreeze removed) + (Antifreeze added) = (Target antifreeze) 3.9 - 0.30x + x = 7.8

Step 4 — Solve for x

Simplify the left side:

3.9 - 0.30x + x = 7.8 3.9 + 0.70x = 7.8 0.70x = 7.8 - 3.9 0.70x = 3.9 x = 3.9 ÷ 0.70 = 5.57 quarts (approximately)

Let's be more precise:

x = 3.9 ÷ 0.7 = 39/7 ≈ 5.57 quarts

Solution: Method 2 — The Final Composition Setup

Instead of tracking the changes, we can set up an equation based on what the final mixture contains.

Step 1 — Identify what's in the final mixture

After draining x quarts and replacing with pure antifreeze, we still have 13 quarts total. This final mixture contains antifreeze from two sources:

  • Antifreeze remaining from the original mixture: (13-x) × 0.30 quarts
  • Pure antifreeze added: x quarts

Step 2 — Set up the final concentration equation

The total antifreeze in the final 13 quarts should equal 60% of 13:

(13-x) × 0.30 + x = 0.60 × 13

Step 3 — Solve the equation

Expand and simplify:

3.9 - 0.30x + x = 7.8 3.9 + 0.70x = 7.8 0.70x = 3.9 x = 39/7 ≈ 5.57 quarts

This confirms our answer using a different setup approach.

Answer: Approximately 5.57 quarts (or exactly 39/7 quarts) of fluid should be drained and replaced with pure antifreeze.

Verification

Let's check our answer by calculating the final concentration:

Starting mixture after draining 5.57 quarts:

Remaining fluid = 13 - 5.57 = 7.43 quarts at 30% concentration Antifreeze remaining = 7.43 × 0.30 = 2.229 quarts

After adding 5.57 quarts of pure antifreeze:

Total antifreeze = 2.229 + 5.57 = 7.799 quarts Total fluid = 13 quarts Final concentration = 7.799 ÷ 13 = 0.5999 ≈ 60% ✓

Using the exact fraction 39/7:

Final antifreeze = (13 - 39/7) × 0.30 + 39/7 = (52/7) × 0.30 + 39/7 = 15.6/7 + 39/7 = 54.6/7 = 7.8 quarts Final concentration = 7.8 ÷ 13 = 0.60 = 60% exactly ✓

Common Pitfalls

✗ MISTAKE #1: Averaging the concentrations

Wrong thinking: "We're mixing 30% with 100%, so the answer should be around (30% + 100%) ÷ 2 = 65%."

Why it's wrong: Percentages can't be averaged directly because they depend on the amounts being mixed. We're not mixing equal quantities—we're replacing a specific amount that we need to calculate.

✗ MISTAKE #2: Forgetting that draining removes antifreeze proportionally

Wrong setup: "3.9 + x = 7.8, so x = 3.9 quarts"

Why it's wrong: This assumes draining removes no antifreeze at all. In reality, when you drain x quarts of 30% mixture, you lose 0.30x quarts of antifreeze.

✗ MISTAKE #3: Setting up the equation with final volume wrong

Wrong thinking: "After draining x and adding x, we have 13-x+x = 13 quarts, but let me write the volume as 13-x somewhere in my equation."

Why it's wrong: The total volume stays at 13 quarts because we drain and replace the same amount. The equation should reflect that we want 60% of those same 13 quarts.

The General Formula

For any drain-and-replace mixture problem, if we drain and replace x units:

(Original amount of substance) - (Concentration₁ × x) + (Concentration₂ × x) = (Target amount)

Where:

  • Concentration₁ = current concentration (what's being drained)
  • Concentration₂ = concentration of replacement fluid
  • Target amount = desired concentration × total volume

This simplifies to:

x = (Target amount - Original amount) ÷ (Concentration₂ - Concentration₁)

Important limitation: This formula only works when Concentration₂ ≠ Concentration₁. If they're equal, you're replacing with the same concentration, so nothing changes!

Real Applications

This exact calculation appears frequently in practical situations:

  • Automotive maintenance: Adjusting antifreeze concentration for different climate zones or seasonal changes
  • Swimming pool chemistry: Adjusting chlorine or pH buffer concentrations by partial water changes
  • Industrial manufacturing: Adjusting solution concentrations in chemical processes, plating baths, or cleaning solutions
  • Medical applications: Diluting or concentrating IV solutions, preparing topical treatments with specific active ingredient percentages

Does the Answer Make Sense?

Let's check whether 5.57 quarts seems reasonable:

Boundary check: We need to go from 30% to 60% antifreeze. That's doubling the concentration, which intuitively should require replacing a significant portion of the mixture with pure antifreeze.

Scale check: We're replacing 5.57 out of 13 quarts = about 43% of the total volume. Since we're roughly doubling the antifreeze concentration by adding 100% antifreeze, replacing nearly half makes sense.

Limit check: If we wanted 100% antifreeze, we'd replace all 13 quarts. If we wanted just slightly more than 30%, we'd replace very little. At 60%, replacing about 43% falls nicely between these extremes.

What If? — Try These Variations

1
Higher Target Concentration
A radiator contains 13 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 80% antifreeze?
Step 1 — Calculate current and target antifreeze

Current antifreeze: 0.30 × 13 = 3.9 quarts
Target antifreeze: 0.80 × 13 = 10.4 quarts

Step 2 — Set up the balance equation

Let x = quarts drained and replaced
3.9 - 0.30x + x = 10.4

Step 3 — Solve for x

3.9 + 0.70x = 10.4
0.70x = 6.5
x = 6.5 ÷ 0.70 = 65/7 ≈ 9.29 quarts

Step 4 — Verify

Remaining original: (13 - 65/7) × 0.30 = 26/7 quarts antifreeze
Added pure: 65/7 quarts antifreeze
Total: 26/7 + 65/7 = 91/7 = 13 × 0.80 ✓

Answer

Drain and replace approximately 9.29 quarts

2
Reverse the Unknown
A radiator contains 13 quarts of 30% antifreeze. If 4 quarts are drained and replaced with pure antifreeze, what is the new percentage of antifreeze?
Step 1 — Calculate antifreeze remaining

After draining 4 quarts, 13 - 4 = 9 quarts of original mixture remain
Antifreeze in remaining mixture: 9 × 0.30 = 2.7 quarts

Step 2 — Add pure antifreeze

Pure antifreeze added: 4 quarts
Total antifreeze: 2.7 + 4 = 6.7 quarts

Step 3 — Calculate new percentage

Total volume remains: 13 quarts
New concentration: 6.7 ÷ 13 = 0.515 = 51.5%

Verification

Check: 9 × 0.30 + 4 × 1.00 = 2.7 + 4 = 6.7 quarts antifreeze ✓

Answer

The new mixture is 51.5% antifreeze

3
Two-Stage Process
Start with 13 quarts of 30% antifreeze. First, 3 quarts are drained and replaced with water. Then, from this new mixture, some amount is drained and replaced with pure antifreeze to get a final 60% mixture. How much is drained in the second step?
Step 1 — Find concentration after first stage

After draining 3 quarts and adding water:
Remaining antifreeze: (13-3) × 0.30 = 10 × 0.30 = 3.0 quarts
New concentration: 3.0 ÷ 13 = 23.08%

Step 2 — Set up second stage equation

Let x = quarts drained and replaced in second stage
Current antifreeze: 3.0 quarts
Target antifreeze: 0.60 × 13 = 7.8 quarts

Step 3 — Solve for second drainage

3.0 - (3.0/13)x + x = 7.8
3.0 + x(1 - 3.0/13) = 7.8
3.0 + x(10/13) = 7.8
x = 4.8 ÷ (10/13) = 4.8 × 13/10 = 6.24 quarts

Answer

Drain and replace 6.24 quarts in the second step

4
Limited Pure Supply
You have only 6 quarts of pure antifreeze available. Starting with 13 quarts of 30% antifreeze, what is the maximum concentration you can achieve by draining some fluid and replacing it with all 6 quarts of pure antifreeze?
Step 1 — Set up with x = 6 quarts

We must drain and replace exactly 6 quarts
Original antifreeze: 13 × 0.30 = 3.9 quarts
Antifreeze removed: 6 × 0.30 = 1.8 quarts

Step 2 — Calculate final antifreeze amount

Final antifreeze = Original - Removed + Added
= 3.9 - 1.8 + 6 = 8.1 quarts

Step 3 — Find maximum concentration

Total volume remains: 13 quarts
Maximum concentration: 8.1 ÷ 13 = 0.623 = 62.3%

Verification

Check: 7 × 0.30 + 6 × 1.00 = 2.1 + 6 = 8.1 quarts antifreeze ✓

Answer

The maximum achievable concentration is 62.3%

Frequently Asked Questions

How do you solve mixture problems with percentage concentrations? +
Track the amount of pure substance (not the total fluid) before and after the change. Set up an equation where: original amount + amount added = final amount. In this problem: 3.9 quarts (original antifreeze) + x quarts (pure antifreeze added) = 0.60 × 13 quarts (antifreeze in final mixture).
What's the key insight for drain-and-replace mixture problems? +
When you drain x quarts of mixture, you remove antifreeze proportionally to the current concentration. When you replace with pure antifreeze, you add 100% antifreeze. The net effect changes the total antifreeze content while keeping total volume constant.
Why can't you just average the concentrations in mixture problems? +
Concentrations are ratios, not quantities. You must track the actual amounts of the substance. In this problem, averaging 30% and 100% gives 65%, but the correct answer is 60% because we're replacing a specific amount, not mixing equal volumes.
NJ
Neven Jurkovic, PhD

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

Developer of Algebrator

Contact

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

2026-09-07