Fruit Juice Mixture Problem: Blending Two Concentrations

Mixture Concentration 9th-10th Grade
PROBLEM

The Royal Fruit Company produces two types of fruit drinks. The first type is 60% pure fruit juice, and the second type is 85% pure fruit juice. The company is attempting to produce a fruit drink that contains 65% pure fruit juice. How many pints of each of the two existing types of drink must be used to make 170 pints of a mixture that is 65% pure fruit juice?

What This Problem Teaches

  • Setting up and organizing mixture problems with a systematic table approach
  • Translating percentage concentrations into decimal form for calculations
  • Building and solving systems of linear equations from real-world constraints
  • Understanding the relationship between total amount, concentration, and pure substance
  • Recognizing that mixture problems always involve conservation of both total volume and pure substance

Let's Organize the Information

Mixture problems become much clearer when we organize the information in a table. Let's define our variables and set up what we know:

Type of DrinkAmount (pints)ConcentrationPure Juice (pints)
First type (60%)x0.600.6x
Second type (85%)y0.850.85y
Final mixture (65%)1700.65110.5

The key insight is that pure juice = amount × concentration. For the final mixture: 170 × 0.65 = 110.5 pints of pure juice.

Solution: Method 1 — The System of Equations Approach

From our table, we can write two equations based on what must be conserved in the mixture.

Step 1 — Set up the total amount equation

The total amount of the two drinks must equal the final mixture amount:

x + y = 170

Step 2 — Set up the pure juice equation

The pure juice from both drinks must equal the pure juice in the final mixture:

0.6x + 0.85y = 110.5

Step 3 — Solve using substitution

From the first equation, solve for one variable: x = 170 - y

Substitute this into the second equation:

0.6(170 - y) + 0.85y = 110.5
102 - 0.6y + 0.85y = 110.5
102 + 0.25y = 110.5
0.25y = 8.5
y = 34

Step 4 — Find the other variable

Now substitute back: x = 170 - 34 = 136

Solution: Method 2 — The Weighted Average Approach

We can think of this as finding the right "weighted average" of 60% and 85% that gives us 65%.

Step 1 — Calculate the distances from target

The target concentration (65%) is closer to 60% than to 85%:

  • Distance from 60% to 65%: 65 - 60 = 5 percentage points
  • Distance from 85% to 65%: 85 - 65 = 20 percentage points

Step 2 — Apply the inverse ratio principle

The amounts should be in the inverse ratio of these distances. If we need 5 parts "pull" from the 85% juice and 20 parts "pull" from the 60% juice, then:

Ratio of x : y = 20 : 5 = 4 : 1

Step 3 — Set up proportions

If x : y = 4 : 1, then x = 4k and y = k for some value k.

Since the total must be 170 pints:

4k + k = 170
5k = 170
k = 34

Step 4 — Find the final amounts

Therefore: x = 4(34) = 136 and y = 34

The company needs 136 pints of the 60% juice and 34 pints of the 85% juice.

Verification

Let's check both constraints to make sure our answer is correct:

Total amount check:

136 + 34 = 170 ✓

Pure juice concentration check:

Pure juice from first type: 136 × 0.6 = 81.6 pints
Pure juice from second type: 34 × 0.85 = 28.9 pints
Total pure juice: 81.6 + 28.9 = 110.5 pints
Final concentration: 110.5 ÷ 170 = 0.65 = 65% ✓

Both checks confirm our answer is correct.

Common Pitfalls

✗ Mistake 1: Forgetting to convert percentages to decimals

Writing 60x + 85y = 65(170) instead of 0.6x + 0.85y = 0.65(170). This gives completely wrong numbers because you're treating percentages as if they're whole numbers.

✗ Mistake 2: Setting up the equation incorrectly

Writing x + y = 65 instead of x + y = 170. Students sometimes confuse the target concentration (65%) with the total amount (170 pints).

✗ Mistake 3: Adding concentrations directly

Thinking that mixing equal amounts gives (60% + 85%) ÷ 2 = 72.5%. This only works if you use equal amounts, but mixture problems specifically ask for unequal amounts to hit a target concentration.

Recognizing This Problem Type

Watch for these telltale phrases that signal a mixture concentration problem:

  • "Two types of solutions with different concentrations..."
  • "How much of each ingredient to make a mixture that is X% of something..."
  • "Blending," "mixing," or "combining" substances with known percentages
  • A target total amount AND a target concentration
  • Problems involving alcohol content, acid solutions, alloys, or juice concentrations

The structure is always the same: you have two unknowns (the amounts) and two constraints (total amount and target concentration).

The Underlying Pattern

All mixture concentration problems follow this template:

Amount equation: x + y = Total
Pure substance equation: c₁x + c₂y = c₃(Total)

Where:

  • x and y are the amounts of each component
  • c₁ and c₂ are the concentrations (as decimals) of each component
  • c₃ is the target concentration (as decimal)
  • Total is the desired final amount

This pattern works for any two-component mixture: coffee blends, metal alloys, chemical solutions, investment portfolios, or even mixing groups of people with different characteristics.

Does the Answer Make Sense?

Our answer shows we need much more of the weaker juice (136 pints) than the stronger juice (34 pints). This makes intuitive sense.

Think about it this way: if we used equal amounts (85 pints each), we'd get (60% + 85%) ÷ 2 = 72.5% concentration—too strong. To get down to 65%, we need to use more of the weaker 60% juice and less of the stronger 85% juice.

The 4:1 ratio (136:34) tells us we need four times as much weak juice as strong juice. This aligns with our intuition that we're much closer to the weak concentration (60%) than the strong one (85%).

Real Applications

  • Food and beverage industry: Blending different grape varieties to achieve target sugar content in wine, or mixing coffee beans with different caffeine levels.
  • Manufacturing: Creating metal alloys with specific properties by combining metals with different purities or characteristics.
  • Pharmacy: Preparing medication solutions with precise active ingredient concentrations by diluting or combining stock solutions.
  • Finance: Portfolio management where you mix investments with different risk levels or expected returns to achieve target portfolio characteristics.

Four "What-If?" Problems

1
Change the Target
The same company wants to make 200 pints of juice that is 70% pure fruit juice using the same two types (60% and 85%). How many pints of each type are needed?
Step 1 — Set up the equations

Amount: x + y = 200
Pure juice: 0.6x + 0.85y = 0.70(200) = 140

Step 2 — Solve by substitution

From first equation: x = 200 - y
Substitute: 0.6(200 - y) + 0.85y = 140

Step 3 — Simplify and solve

120 - 0.6y + 0.85y = 140
120 + 0.25y = 140
0.25y = 20
y = 80

Step 4 — Find x and verify

x = 200 - 80 = 120
Check: 120(0.6) + 80(0.85) = 72 + 68 = 140 ✓

Answer: 120 pints of 60% juice and 80 pints of 85% juice

2
Limited Supply
The company has only 80 pints of the 85% juice available. If they use all of it, how much of the 60% juice do they need to add to create a 65% mixture? What will be the total volume?
Step 1 — Set up with known amount

We know y = 80 (pints of 85% juice)
Unknown: x (pints of 60% juice)

Step 2 — Write the pure juice equation

Total pure juice in final mixture must equal sum from components:
0.65(x + 80) = 0.6x + 0.85(80)

Step 3 — Solve for x

0.65x + 52 = 0.6x + 68
0.05x = 16
x = 320

Step 4 — Find total and verify

Total volume: 320 + 80 = 400 pints
Check: (320(0.6) + 80(0.85)) ÷ 400 = 260 ÷ 400 = 0.65 ✓

Answer: 320 pints of 60% juice, total volume of 400 pints

3
Reverse Problem
The company mixed 90 pints of 60% juice with 60 pints of 85% juice. What is the concentration of pure fruit juice in the resulting mixture?
Step 1 — Calculate pure juice from each type

From 60% juice: 90 × 0.60 = 54 pints pure juice
From 85% juice: 60 × 0.85 = 51 pints pure juice

Step 2 — Find total amounts

Total pure juice: 54 + 51 = 105 pints
Total mixture: 90 + 60 = 150 pints

Step 3 — Calculate concentration

Concentration = pure juice ÷ total mixture
105 ÷ 150 = 0.70 = 70%

Step 4 — Verify reasonableness

70% is between 60% and 85% ✓
Since we used more 60% juice (90 vs 60 pints), the result should be closer to 60%—but 70% is reasonable given the amounts.

Answer: The mixture is 70% pure fruit juice

4
Three-Way Mixture
The company introduces a third juice that is 40% pure. They want to make 180 pints of 65% juice using all three types, with equal amounts of the 60% and 85% juices. How much of each type is needed?
Step 1 — Define variables

Let x = amount of each of 60% and 85% juices (equal amounts)
Let z = amount of 40% juice

Step 2 — Set up equations

Total amount: x + x + z = 180, so 2x + z = 180
Pure juice: 0.6x + 0.85x + 0.4z = 0.65(180) = 117

Step 3 — Simplify pure juice equation

1.45x + 0.4z = 117
From first equation: z = 180 - 2x

Step 4 — Substitute and solve

1.45x + 0.4(180 - 2x) = 117
1.45x + 72 - 0.8x = 117
0.65x = 45
x = 69.23 (approximately)

Step 5 — Find z and verify

z = 180 - 2(69.23) = 41.54
For exact values: x = 900/13 ≈ 69.23, z = 540/13 ≈ 41.54

Answer: About 69.2 pints each of 60% and 85% juice, and 41.6 pints of 40% juice

Frequently Asked Questions

How do you set up a mixture concentration problem? +
Set up a table with columns for amount, concentration, and pure substance. Write one equation for total amount and another for pure substance. In this problem: x + y = 170 (total pints) and 0.6x + 0.85y = 110.5 (pure juice).
What's the difference between amount and concentration in mixture problems? +
Amount is the total volume or quantity. Concentration is the percentage of the desired substance. Pure substance = amount × concentration. Here, 170 pints is the amount, 65% is the concentration, so pure juice = 170 × 0.65 = 110.5 pints.
When do you use substitution versus elimination for mixture problems? +
Use whichever method feels more natural. Substitution works well when one equation easily solves for a variable (like x = 170 - y). Elimination works when coefficients align nicely. Both methods give the same answer.
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.

2026-10-18