Math Practice

Mixture Problems Practice Problems

Q: Percents or decimals in the equation?

Decimals. A $30\%$ solution contributes $0.30$ times its volume in pure substance.

Q: How do I handle pure water or pure substance?

Pure water is $0\%$ of the tracked substance, so its concentration is $0$. Pure substance is $100\%$, so its concentration is $1$.

Q: Do I still add to the total quantity when I add pure water?

Yes. Water's concentration term is $0$, but the mixture's total quantity is still the original plus the water added.

Twelve hand-built mixture problems with full worked solutions. Each tracks one quantity — pure acid, pure salt, dollars of value, grams of alcohol — and sets the total before mixing equal to the total after.

1.
How many liters of a $50\\%$ acid solution should be mixed with $6$ liters of a $20\\%$ solution to get a $30\\%$ solution?
Show full solution
$$ 0.50x + 0.20(6) = 0.30(x + 6) \quad\Longrightarrow\quad 0.50x + 1.2 = 0.30x + 1.8 \quad\Longrightarrow\quad 0.20x = 0.6 \quad\Longrightarrow\quad x = 3 $$
2.
How much pure water must be added to $8$ liters of a $40\\%$ salt solution to make it $25\\%$?
Show full solution
$$ 0.40(8) = 0.25(8 + x) \quad\Longrightarrow\quad 3.2 = 2 + 0.25x \quad\Longrightarrow\quad 0.25x = 1.2 \quad\Longrightarrow\quad x = 4.8 \text{ liters} $$
3.
How much pure antifreeze must be added to $12$ quarts of a $25\\%$ mixture to raise it to $40\\%$?
Show full solution
$$ 0.25(12) + 1 \cdot x = 0.40(12 + x) \quad\Longrightarrow\quad 3 + x = 4.8 + 0.40x \quad\Longrightarrow\quad 0.60x = 1.8 \quad\Longrightarrow\quad x = 3 \text{ quarts} $$
4.
A $6$/lb nut blend is made from cashews at $9$/lb and peanuts at $3$/lb. For $30$ lb of blend, how much of each?
Show full solution
Let $x$ be pounds of cashews; peanuts are $30 - x$.
$$ 9x + 3(30 - x) = 6(30) \quad\Longrightarrow\quad 6x + 90 = 180 \quad\Longrightarrow\quad x = 15 $$
$15$ lb cashews, $15$ lb peanuts.
5.
A $20\\%$ and a $50\\%$ alcohol solution are combined to make $30$ L of a $30\\%$ solution. How many liters of each?
Show full solution
Let $x$ be liters of $20\\%$; the $50\\%$ is $30 - x$.
$$ 0.20x + 0.50(30 - x) = 0.30(30) \quad\Longrightarrow\quad -0.30x + 15 = 9 \quad\Longrightarrow\quad x = 20 $$
$20$ L of $20\\%$, $10$ L of $50\\%$.
6.
How much of an $80\\%$ copper alloy must be melted with $60$ kg of a $20\\%$ alloy to get a $50\\%$ alloy?
Show full solution
$$ 0.80x + 0.20(60) = 0.50(x + 60) \quad\Longrightarrow\quad 0.80x + 12 = 0.50x + 30 \quad\Longrightarrow\quad 0.30x = 18 \quad\Longrightarrow\quad x = 60 \text{ kg} $$
7.
A grocer mixes $\$2.50$/lb candy with $\$4.00$/lb candy to make $24$ lb worth $\$3.00$/lb. How much of each?
Show full solution
Let $x$ be pounds of the $\$2.50$ candy; the other is $24 - x$.
$$ 2.5x + 4(24 - x) = 3(24) \quad\Longrightarrow\quad -1.5x + 96 = 72 \quad\Longrightarrow\quad x = 16 $$
$16$ lb at $\$2.50$, $8$ lb at $\$4.00$.
8.
How many liters of water must evaporate from $20$ L of a $12\\%$ salt solution to make it $15\\%$? (Salt stays.)
Show full solution
Salt is fixed at $0.12(20) = 2.4$ L. Let $x$ be liters removed.
$$ 2.4 = 0.15(20 - x) \quad\Longrightarrow\quad 2.4 = 3 - 0.15x \quad\Longrightarrow\quad 0.15x = 0.6 \quad\Longrightarrow\quad x = 4 \text{ liters} $$
9.
A jar of nickels and dimes has $50$ coins worth $\$4.10$. How many of each?
Show full solution
Let $n$ be nickels; dimes are $50 - n$.
$$ 0.05n + 0.10(50 - n) = 4.10 \quad\Longrightarrow\quad -0.05n + 5 = 4.10 \quad\Longrightarrow\quad n = 18 $$
$18$ nickels, $32$ dimes.
10.
A theater sells $150$ tickets, adults at $\$10$ and children at $\$6$, for $\$1{,}260$. How many of each?
Show full solution
Let $a$ be adult tickets; children are $150 - a$.
$$ 10a + 6(150 - a) = 1260 \quad\Longrightarrow\quad 4a + 900 = 1260 \quad\Longrightarrow\quad a = 90 $$
$90$ adult, $60$ child.
11.
$\$12{,}000$ is split between accounts paying $3\\%$ and $5\\%$. The total annual interest is $\$520$. How much is in each?
Show full solution
Let $x$ be the amount at $3\\%$; the rest is $12000 - x$.
$$ 0.03x + 0.05(12000 - x) = 520 \quad\Longrightarrow\quad -0.02x + 600 = 520 \quad\Longrightarrow\quad x = 4000 $$
$\$4{,}000$ at $3\\%$, $\$8{,}000$ at $5\\%$.
12.
A jar has twice as many dimes as quarters and is worth $\$4.05$. How many of each?
Show full solution
Let $q$ be quarters; dimes are $2q$.
$$ 0.25q + 0.10(2q) = 4.05 \quad\Longrightarrow\quad 0.45q = 4.05 \quad\Longrightarrow\quad q = 9 $$
$9$ quarters and $18$ dimes.

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Mixture Problems Practice Problems

The Mixture Problems lesson covers percent solutions, dilutions, priced blends, and coins with a chart; this set is 12 problems to practice on, each fully solved.

How to use this set

These problems are hand-written, not auto-generated. For each problem, build a chart with a row per ingredient plus a mixture row, fill the pure-amount column as (quantity)(concentration), set the ingredient rows to sum to the mixture row, and solve. Then check.

Common mistakes to watch for

Using percents instead of decimals.

Setting the target concentration outside the range of the two ingredients.

Forgetting that added water still increases the total quantity.

Adding concentrations directly instead of multiplying each by its amount first.

Where to go next

The setup is the Word Problems method with a chart; split-investment problems overlap with simple interest. If you prefer to clear the decimals first, that is the Linear Equations with Fractions Practice Problems skill. For every mixture type, see the Mixture Problems lesson.

Frequently Asked Questions

What is the general setup for a mixture problem?+

Track the pure amount of one ingredient. For each part, that amount is (quantity)(concentration). The sum across all parts equals (total quantity)(final concentration): .

Percents or decimals in the equation?+

Decimals. A solution contributes times its volume in pure substance.

How do I handle pure water or pure substance?+

Pure water is of the tracked substance, so its concentration is . Pure substance is , so its concentration is .

Are coin and ticket problems mixture problems?+

Yes. The 'concentration' is the value of each item, and you balance total count and total value. Split-investment problems use an interest rate as the concentration.

Why does an answer sometimes come out negative or larger than the total?+

The final concentration must lie between the two starting concentrations. You cannot mix and to get ; an impossible target produces an impossible answer.

Do I still add to the total quantity when I add pure water?+

Yes. Water's concentration term is , but the mixture's total quantity is still the original plus the water added.

🧪Mixture Problems Practice Problems

Mixture Problems Practice Problems

How to use this set

Common mistakes to watch for

Where to go next

Frequently Asked Questions

More practice problems

Mixture Problems Practice Problems