Fruit Juice Mixture Problem: Blending Two Concentrations
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 Drink | Amount (pints) | Concentration | Pure Juice (pints) |
|---|---|---|---|
| First type (60%) | x | 0.60 | 0.6x |
| Second type (85%) | y | 0.85 | 0.85y |
| Final mixture (65%) | 170 | 0.65 | 110.5 |
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:
Step 2 — Set up the pure juice equation
The pure juice from both drinks must equal the pure juice in the final mixture:
Step 3 — Solve using substitution
From the first equation, solve for one variable: x = 170 - y
Substitute this into the second equation:
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 = 5percentage points - Distance from 85% to 65%:
85 - 65 = 20percentage 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:
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:
5k = 170
k = 34
Step 4 — Find the final amounts
Therefore: x = 4(34) = 136 and y = 34
Verification
Let's check both constraints to make sure our answer is correct:
Total amount check:
Pure juice concentration check:
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:
Pure substance equation: c₁x + c₂y = c₃(Total)
Where:
xandyare the amounts of each componentc₁andc₂are the concentrations (as decimals) of each componentc₃is the target concentration (as decimal)Totalis 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
Amount: x + y = 200
Pure juice: 0.6x + 0.85y = 0.70(200) = 140
From first equation: x = 200 - y
Substitute: 0.6(200 - y) + 0.85y = 140
120 - 0.6y + 0.85y = 140120 + 0.25y = 1400.25y = 20y = 80
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
We know y = 80 (pints of 85% juice)
Unknown: x (pints of 60% juice)
Total pure juice in final mixture must equal sum from components:0.65(x + 80) = 0.6x + 0.85(80)
0.65x + 52 = 0.6x + 680.05x = 16x = 320
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
From 60% juice: 90 × 0.60 = 54 pints pure juice
From 85% juice: 60 × 0.85 = 51 pints pure juice
Total pure juice: 54 + 51 = 105 pints
Total mixture: 90 + 60 = 150 pints
Concentration = pure juice ÷ total mixture105 ÷ 150 = 0.70 = 70%
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
Let x = amount of each of 60% and 85% juices (equal amounts)
Let z = amount of 40% juice
Total amount: x + x + z = 180, so 2x + z = 180
Pure juice: 0.6x + 0.85x + 0.4z = 0.65(180) = 117
1.45x + 0.4z = 117
From first equation: z = 180 - 2x
1.45x + 0.4(180 - 2x) = 1171.45x + 72 - 0.8x = 1170.65x = 45x = 69.23 (approximately)
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
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2026-10-18