Mixture Problems: Definition, Setup & Examples

Definitionnoun phrase

What are mixture problems in math?

Mixture problems are word problems in which two or more ingredients with different values, prices or strengths are combined. Each ingredient adds amount × value to the total, and the parts add up: 3 lb of 4-dollar nuts plus 2 lb of 9-dollar nuts make 5 lb worth 12 + 18 = 30 dollars, which is 6 dollars per pound.

In Spanish: problemas de mezclas
Also called: mixture word problems, solution problems, mixture applications
Grade level: Algebra 1

Algebra 1CCSS HSA-CED.A.1CCSS HSA-CED.A.3SATACTGEDASVABHiSETACCUPLACER

Mixture Problems in a picture

Track the pure acid, not the percents: 2 L + 18 L = 20 L of acid in 40 L, which is 50%. Averaging 20% and 60% would give the wrong answer, 40%.

Formal definition and notation

In a mixture problem, each component i has an amount a_i and a value per unit v_i (a price, a percent concentration or a coin value). The model is a₁v₁ + a₂v₂ + ⋯ = (a₁ + a₂ + ⋯)v_mix, where v_mix is the value per unit of the mixture. The unknown amounts are found by solving this equation, often together with an equation for the total amount.

Key facts

Every row follows number × value = total value: 12 dimes are worth 12 × 0.10 = 1.20 dollars.
Percents become decimals: 35% acid is 0.35 of the liquid.
The strength of a mix always lands between the strengths of its ingredients, closer to the one you use more of. It is a weighted average, not a plain mean.
Adding pure water adds amount but 0% acid; adding pure acid counts as 100%, or 1.
With two unknown amounts, write one equation for the amounts and one for the values, then solve the system.

Mixture equation

a₁v₁ + a₂v₂ = (a₁ + a₂)v_mix

a₁, a₂: the amounts of each ingredient (pounds, liters, tickets, coins)
v₁, v₂: the value of one unit of each (price per pound, percent as a decimal, cents per coin)
v_mix: the value per unit of the finished mixture

Organize the facts in a table with columns amount, value and total (amount × value). The totals column always adds up.

Mixture Problems examples, step by step

Example 1: Price of a blend

Warm-up

A shop mixes 4 lb of coffee costing 6 dollars per pound with 6 lb costing 11 dollars per pound. What is the price per pound of the blend?

Total value: 4 × 6 + 6 × 11 = 24 + 66 = 90 dollars.
Total amount: 4 + 6 = 10 lb.
Price per pound: 90 ÷ 10 = 9 dollars.

Answer: The blend costs 9 dollars per pound.

Example 2: How much to add?

Core

How many liters of a 10% acid solution must be added to 20 L of a 40% acid solution to make a 25% solution?

Let x be the liters of 10% solution. Acid: 0.10x + 0.40(20) = 0.25(x + 20).
Simplify: 0.10x + 8 = 0.25x + 5.
Subtract: 3 = 0.15x, so x = 20.
Check: 2 + 8 = 10 L of acid in 40 L, which is 25%.

Answer: Add 20 liters of the 10% solution.

Example 3: Ticket sales

Test-style

A school sold 300 tickets to a play. Adult tickets cost 12 dollars and student tickets cost 7 dollars. Sales totaled 2,900 dollars. How many adult tickets were sold?

Let a be adult and s student tickets: a + s = 300 and 12a + 7s = 2900.
Substitute s = 300 − a: 12a + 7(300 − a) = 2900.
12a + 2100 − 7a = 2900, so 5a = 800 and a = 160.
Then s = 140. Check: 160 × 12 + 140 × 7 = 1920 + 980 = 2900.

Answer: 160 adult tickets were sold (and 140 student tickets).

Common mistakes

✗ MistakeMixing 10 L of 20% acid with 30 L of 60% acid gives 20% + 60%2 = 40%.

✓ InsteadAcid: 0.20(10) + 0.60(30) = 20 L in 40 L, so the mix is 50%.

The two solutions are used in different amounts, so their strengths do not count equally. Track the pure ingredient instead.

✗ Mistake40(20) + 10x = 25(x + 20) with percents as whole numbers, then treating the result as liters of acid.

✓ InsteadUse decimals: 0.40(20) + 0.10x = 0.25(x + 20).

Writing 40 for 40% makes each acid amount 100 times too big. It still gives the right x here, but any acid amount you report from it will be wrong.

✗ MistakeAdding x liters to 20 L gives a mix of 20 L: 0.10x + 8 = 0.25(20).

✓ InsteadThe mixture has x + 20 liters: 0.10x + 8 = 0.25(x + 20).

Whatever you pour in becomes part of the total amount, so the total must include x.

Not to be confused with

Mixture Problemsvs.Weighted average

A mixture problem is a weighted average problem: the mix value is a₁v₁ + a₂v₂a₁ + a₂. The word problem just hides one of the amounts.

Mixture Problemsvs.System of Equations

A system of equations is a tool; many mixture problems use one (an equation for amounts and one for values), but simple ones need only one equation.

Mixture Problemsvs.Work rate problems

Work rate problems add rates of doing a job (parts per hour). Mixture problems add amounts times values. Both add parts to make a whole, but the parts are different.

Quick check: 3 questions

Question 1 of 3

5 lb of candy at 3 dollars per pound is mixed with 3 lb at 7 dollars per pound. What is the price per pound of the mix?

Show answer and explanation

Answer: C. Value: 15 + 21 = 36 dollars for 8 lb, so 36 ÷ 8 = 4.50 dollars. 5.00 averages the prices, 4.00 is wrong arithmetic, and 36.00 is the total value, not per pound.

Question 2 of 3

How much pure water must be added to 12 L of a 30% salt solution to make a 20% solution?

Show answer and explanation

Answer: B. Salt stays 3.6 L, so 3.6 = 0.20(12 + w), 18 = 12 + w, w = 6 L. 18 L is the new total, 2 L and 4 L do not dilute enough.

Question 3 of 3

A jar has 40 coins, all nickels and dimes, worth 3.10 dollars. How many dimes are there?

Show answer and explanation

Answer: A. n + d = 40 and 0.05n + 0.10d = 3.10. Then 0.05(40 − d) + 0.10d = 3.10, so 0.05d = 1.10 and d = 22. 18 is the number of nickels, 31 is 3.10 divided by 0.10, and 20 splits the coins evenly.

Mixture Problems: frequently asked questions

How do you set up a mixture problem?

Make a table with a row for each ingredient and for the mixture, and columns for amount, value per unit and total value. Fill in what you know, use a variable for the unknown amount, and multiply across. Then write the equation: the total values of the ingredients add up to the total value of the mixture.

Why can you not just average the percents?

A plain average treats every ingredient as if you used the same amount. If you mix 1 liter of 10% acid with 9 liters of 50% acid, the result is much closer to 50%: 0.1 + 4.510 = 46%, not 30%. The amounts act as weights, so you must multiply before you add.

What kinds of problems count as mixture problems?

Any problem that combines items with different values per unit: coffee or nut blends by price, chemical solutions by percent, coins by value, tickets by price, and money split between accounts at different interest rates. They all use the same model, amount times value equals total value.

Practice and tools

Systems of equations word problemsPractice on Effortless Math Weighted Average CalculatorFind the value of a mix from amounts and values System of Equations CalculatorSolve the two-equation set-up

Sources

Updated November 2026·Reviewed by the Effortless Math team·Standards: CCSS HSA-CED.A.1, CCSS HSA-CED.A.3

Cite this entry

Effortless Math Team. “Mixture Problems.” Effortless Math Dictionary. Updated November 2026. https://www.effortlessmath.com/math-dictionary/mixture-problems/

Flashcard

TermMixture Problemsamount × value = total value

DefinitionWord problems that combine ingredients of different values. 10 L of 20% acid plus 30 L of 60% acid: 2 + 18 = 20 L acid in 40 L, so 50%.

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Mixture Problems

What are mixture problems in math?

Mixture Problems in a picture

Formal definition and notation

Key facts

Mixture equation

Mixture Problems examples, step by step

Example 1: Price of a blend

Example 2: How much to add?

Example 3: Ticket sales

Common mistakes

Not to be confused with

Quick check: 3 questions

Mixture Problems: frequently asked questions

How do you set up a mixture problem?

Why can you not just average the percents?

What kinds of problems count as mixture problems?

Related terms

Practice and tools

Sources

Flashcard

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