Three-Solution Mixture: Acid Concentration

Question

PROBLEM

A chemist has three different acid solutions: 20% acid, 30% acid, and 75% acid. He wants to mix all three to obtain 48 liters of 55% acid solution, using 2 times as much of the 75% solution as the 30% solution. How many liters of each solution should be used?

What This Problem Teaches

Setting up systems of equations with three unknowns and a constraint relationship
Converting percentage concentrations to pure ingredient amounts
Using substitution to reduce complex systems to simpler forms
Interpreting constraint language ("2 times as much") in mathematical terms
Verifying solutions by checking both volume and concentration requirements

Picture This

A chemist has three different acid solutions: 20% acid, 30% acid, and 75% acid. He wants to mix all three to obtain...

Solution: Method 1 — The Three-Variable Setup

Step 1 — Define variables with the constraint

Let x = liters of 20% solution and y = liters of 30% solution. Since we use 2 times as much 75% solution as 30% solution, we'll use 2y liters of 75% solution.

Step 2 — Set up the volume equation

The total volume must equal 48 liters:

x + y + 2y = 48
x + 3y = 48

Step 3 — Calculate the required acid content

A 55% acid solution of 48 liters contains:

0.55 × 48 = 26.4 liters of pure acid

Step 4 — Set up the acid content equation

The pure acid from each solution must sum to 26.4 liters:

0.20x + 0.30y + 0.75(2y) = 26.4
0.20x + 0.30y + 1.50y = 26.4
0.20x + 1.80y = 26.4

Step 5 — Solve using substitution

From the volume equation: x = 48 - 3y

Substitute into the acid equation:

0.20(48 - 3y) + 1.80y = 26.4
9.6 - 0.60y + 1.80y = 26.4
9.6 + 1.20y = 26.4
1.20y = 16.8
y = 14

Step 6 — Find the remaining amounts

Now we can find x and the amount of 75% solution:

x = 48 - 3(14) = 48 - 42 = 6
Amount of 75% solution = 2y = 2(14) = 28

Solution: Method 2 — Systematic Table Approach

Step 1 — Organize the information in a table

Let's use the constraint to express all amounts in terms of one variable. Let y = liters of 30% solution.

Solution Type	Concentration	Volume (L)	Pure Acid (L)
20% acid	0.20	x	0.20x
30% acid	0.30	y	0.30y
75% acid	0.75	2y	1.50y
Final mixture	0.55	48	26.4

Step 2 — Express total volume in terms of y

From the volume constraint: x + y + 2y = 48, so x = 48 - 3y

Step 3 — Write the acid balance equation

Total pure acid equals sum of acid from each component:

0.20x + 0.30y + 1.50y = 26.4
0.20(48 - 3y) + 1.80y = 26.4
9.6 - 0.60y + 1.80y = 26.4
9.6 + 1.20y = 26.4
y = 14

Step 4 — Calculate all amounts

30% solution: y = 14 liters
75% solution: 2y = 28 liters
20% solution: x = 48 - 3(14) = 6 liters

Answer: Use 6 liters of the 20% solution, 14 liters of the 30% solution, and 28 liters of the 75% solution.

Verification

Check the volume requirement

6 + 14 + 28 = 48 liters ✓

Check the constraint relationship

75% amount = 28 liters
30% amount = 14 liters
28 = 2 × 14 ✓

Check the final concentration

Calculate the total pure acid:

From 20%: 0.20 × 6 = 1.2 liters
From 30%: 0.30 × 14 = 4.2 liters
From 75%: 0.75 × 28 = 21.0 liters
Total acid: 1.2 + 4.2 + 21.0 = 26.4 liters

Final concentration: 26.4 ÷ 48 = 0.55 = 55% ✓

Watch Out For These

✗ Misinterpreting the constraint

Writing "30% amount = 2 × 75% amount" instead of "75% amount = 2 × 30% amount." The problem states we use 2 times as much of the 75% solution, not that the 30% is twice the 75%. This reversal leads to completely wrong equations.

✗ Using concentrations instead of volumes in constraint

Setting up 0.75 = 2 × 0.30 instead of recognizing that the "2 times as much" refers to the volume amounts, not the concentration percentages. The constraint is about liquid quantities, not acid percentages.

✗ Forgetting to calculate pure acid content

Setting up the equation as 0.20x + 0.30y + 0.75(2y) = 48 instead of = 26.4. The number 48 is the total volume, but the acid equation needs the pure acid content (55% of 48 = 26.4 liters).

✗ Losing track of the 2y factor

Writing the acid equation as 0.20x + 0.30y + 0.75y instead of 0.20x + 0.30y + 0.75(2y). Since we use 2y liters of 75% solution, the acid contribution is 0.75 × 2y = 1.50y, not just 0.75y.

The General Formula

For three-component mixture problems with one constraint relationship, the structure is:

Volume equation: x + y + (constraint factor)y = total volume
Content equation: c₁x + c₂y + c₃(constraint factor)y = target content
Where: c₁, c₂, c₃ are the concentrations of each component

In our case, the constraint factor was 2, target content was 0.55 × 48 = 26.4, and concentrations were 0.20, 0.30, and 0.75. The key insight is expressing one variable in terms of another using the constraint, reducing the system from three unknowns to two equations with two unknowns.

Important limitation: This approach works when the constraint gives a direct proportional relationship between two components. If the constraint involves all three components or non-linear relationships, you'll need different techniques like Lagrange multipliers or trial-and-error with boundary conditions.

How to Spot This Problem Type

"Mix all three" — signals a three-component system, not the simpler two-component mixture
"2 times as much of X as Y" — indicates a constraint relationship that reduces the variables
"Obtain [amount] liters of [concentration]%" — gives you both volume and content targets
Multiple percentage concentrations listed — distinguishes from simple dilution problems
"How many liters of each" — confirms you need to find all component amounts

Disguised versions: This same structure appears in alloy problems ("mix metals with 20%, 30%, 75% copper content"), investment problems ("combine accounts with different interest rates"), and even medication problems ("combine solutions with different drug concentrations"). The mathematical skeleton is identical regardless of the physical context.

What If?

1

Different Target Concentration

A chemist has the same three acid solutions: 20%, 30%, and 75% acid. He wants to mix all three to obtain 48 liters of 50% acid solution (instead of 55%), still using 2 times as much of the 75% solution as the 30% solution. How many liters of each solution should be used?

Step 1 — Set up with same constraint

Let y = liters of 30% solution. Then 75% solution = 2y liters, and 20% solution = 48 - 3y liters.

Step 2 — Calculate new target acid content

For 50% concentration: 0.50 × 48 = 24 liters of pure acid needed.

Step 3 — Write acid balance equation

0.20(48 - 3y) + 0.30y + 0.75(2y) = 24
9.6 - 0.60y + 0.30y + 1.50y = 24
9.6 + 1.20y = 24
y = 12

Step 4 — Find all amounts

30% solution: y = 12 liters
75% solution: 2y = 24 liters
20% solution: 48 - 3(12) = 12 liters

Step 5 — Verify

Volume: 12 + 12 + 24 = 48 ✓
Acid: 0.20(12) + 0.30(12) + 0.75(24) = 2.4 + 3.6 + 18 = 24 ✓
Concentration: 24 ÷ 48 = 50% ✓

Answer: 12 liters of 20%, 12 liters of 30%, 24 liters of 75%

2

Reversed Relationship

Now the chemist uses twice as much of the 30% solution as the 75% solution (the opposite of the original). Still making 48 L of 55% acid. How many liters of each solution should be used?

Step 1 — Redefine variables with new constraint

Let z = liters of 75% solution. Then 30% solution = 2z liters (twice as much 30% as 75%).

Step 2 — Set up volume equation

x + 2z + z = 48, so x = 48 - 3z

Step 3 — Set up acid equation

0.20x + 0.30(2z) + 0.75z = 26.4
0.20(48 - 3z) + 0.60z + 0.75z = 26.4
9.6 - 0.60z + 1.35z = 26.4
9.6 + 0.75z = 26.4
z = 22.4

Step 4 — Find all amounts

75% solution: z = 22.4 liters
30% solution: 2z = 44.8 liters
20% solution: 48 - 3(22.4) = -19.2 liters

Step 5 — Check feasibility

We get a negative amount for the 20% solution, which is impossible. This constraint combination cannot produce a 55% solution with these concentrations.

Answer: No solution exists. The constraint makes the problem impossible.

3

Cost Optimization

The 20% solution costs $2/L, the 30% costs $5/L, and the 75% costs $12/L. You must make exactly 48 L of 55% acid using all three types, but you can choose any amounts (no fixed ratio). What mixture minimizes the total cost?

Step 1 — Set up the optimization problem

Minimize: Cost = 2x + 5y + 12z
Subject to: x + y + z = 48 and 0.20x + 0.30y + 0.75z = 26.4

Step 2 — Express in terms of one variable

From constraints: x = 48 - y - z
0.20(48 - y - z) + 0.30y + 0.75z = 26.4
9.6 - 0.20y - 0.20z + 0.30y + 0.75z = 26.4
0.10y + 0.55z = 16.8
y = 168 - 5.5z

Step 3 — Find the cost function

Cost = 2(48 - (168 - 5.5z) - z) + 5(168 - 5.5z) + 12z
= 2(-120 + 4.5z) + 5(168 - 5.5z) + 12z
= -240 + 9z + 840 - 27.5z + 12z
= 600 - 6.5z

Step 4 — Minimize cost

Since coefficient of z is negative, cost decreases as z increases. Maximum feasible z occurs when y = 0:
168 - 5.5z = 0, so z = 30.55 liters
x = 48 - 0 - 30.55 = 17.45 liters

Step 5 — Calculate minimum cost

Use 17.45 L of 20%, 0 L of 30%, 30.55 L of 75%
Minimum cost = 2(17.45) + 5(0) + 12(30.55) = $401.50

4

Fixed Component Amount

The chemist must use exactly 16 liters of the 75% solution. He can use any amounts of the 20% and 30% solutions to make 48 liters total of a 55% acid solution. How much of each should he use?

Step 1 — Account for the fixed component

75% solution contributes: 0.75 × 16 = 12 liters of pure acid
Remaining volume needed: 48 - 16 = 32 liters
Remaining acid needed: 26.4 - 12 = 14.4 liters

Step 2 — Set up two-component system

Let x = liters of 20% solution, y = liters of 30% solution
Volume: x + y = 32
Acid: 0.20x + 0.30y = 14.4

Step 3 — Solve by substitution

From volume equation: x = 32 - y
0.20(32 - y) + 0.30y = 14.4
6.4 - 0.20y + 0.30y = 14.4
0.10y = 8
y = 80

Step 4 — Check feasibility

y = 80 liters, but we only have 32 liters of volume left. This is impossible.

Step 5 — Analyze the problem

The 16 liters of 75% solution provides too much acid (12 L), requiring the remaining 32 L to have concentration 14.4/32 = 45%. But mixing 20% and 30% solutions cannot exceed 30% concentration.

Answer: No solution exists. 16 L of 75% provides too much acid.

Frequently Asked Questions

How do you set up equations for a three-component mixture problem? +

Write one equation for total volume and one for total acid content. In this problem: x + y + 2y = 48 (volume) and 0.20x + 0.30y + 0.75(2y) = 26.4 (acid). Use the given constraint to reduce variables—here, the 75% amount equals twice the 30% amount.

What's the difference between volume ratios and concentration ratios in mixture problems? +

Volume ratios tell you the proportions of the actual liquids being mixed. Concentration ratios involve the percentage of active ingredient. In this example, we use 2 times as much 75% solution by volume, but the acid content is 0.75 × (2y) = 1.5y liters of pure acid from that component.

How do you handle constraints like 'twice as much' in mixture problems? +

Express one variable in terms of another using the constraint relationship. Here, '2 times as much 75% as 30%' means if we use y liters of 30% solution, we use 2y liters of 75% solution. This reduces the three unknowns to just two variables.

NJ

Dr. Neven Jurkovic

Mathematician and educator with expertise in algebraic problem solving and chemistry applications. Specializes in making complex mixture problems accessible through systematic approaches.

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.

2026-08-06