College Algebra Quiz: Mixture Problems

What this quiz covers

This quiz focuses on Mixture Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 115 mL
2. 120 mL (correct answer)
3. 125 mL
4. 130 mL

Explanation: Let $x$ = mL of 20% solution. Then $(200-x)$ = mL of water. The equation is: $0.20x + 0(200-x) = 0.12(200)$, so $0.20x = 24$, giving $x = 120$ mL exactly. Since 120 is divisible by 5, this is achievable with the measurement constraint. Checking nearby values in 5 mL increments: 115 mL gives concentration $\frac{0.20(115)}{200} = \frac{23}{200} = 11.5\%$, and 125 mL gives concentration $\frac{0.20(125)}{200} = \frac{25}{200} = 12.5\%$. The differences from 12% are: 115 mL gives $|11.5 - 12| = 0.5\%$, 120 mL gives $|12 - 12| = 0\%$, 125 mL gives $|12.5 - 12| = 0.5\%$. Therefore 120 mL is optimal and achievable.

Question 2

1. 4 gallons
2. 6 gallons (correct answer)
3. 8 gallons
4. 10 gallons

Explanation: First find the intended mixture for 25% pigment: Let $x$ = gallons of Paint X and $y$ = gallons of Paint Y. We have $x + y = 60$ and $0.40x + 0.10y = 0.25(60) = 15$. From the first equation: $y = 60 - x$. Substituting: $0.40x + 0.10(60 - x) = 15$, so $0.40x + 6 - 0.10x = 15$, giving $0.30x = 9$, thus $x = 30$ gallons of Paint X and $y = 30$ gallons of Paint Y. Now find the actual mixture that produced 28% pigment: Let $a$ = gallons of Paint X actually used and $b$ = gallons of Paint Y actually used. We have $a + b = 60$ and $0.40a + 0.10b = 0.28(60) = 16.8$. From the first equation: $b = 60 - a$. Substituting: $0.40a + 0.10(60 - a) = 16.8$, so $0.40a + 6 - 0.10a = 16.8$, giving $0.30a = 10.8$, thus $a = 36$ gallons. The difference is $36 - 30 = 6$ gallons more Paint X than intended.

Question 3

1. 15 mL
2. 20 mL
3. 25 mL (correct answer)
4. 30 mL

Explanation: First, find what the correct mixture should have been: Let $x$ = mL of pure alcohol and $y$ = mL of 20% solution. We have $x + y = 400$ and $1.00x + 0.20y = 0.35(400) = 140$. From the first equation: $y = 400 - x$. Substituting: $x + 0.20(400 - x) = 140$, so $x + 80 - 0.20x = 140$, giving $0.80x = 60$, thus $x = 75$ mL pure alcohol and $y = 325$ mL of 20% solution. Instead, the technician used 75 mL pure alcohol and 325 mL of 15% solution, creating: $75(1.00) + 325(0.15) = 75 + 48.75 = 123.75$ mL of pure alcohol in 400 mL total, giving $\frac{123.75}{400} = 30.9375\%$ concentration. To get 35% in 400 mL, we need 140 mL of pure alcohol total. Currently have 123.75 mL, so need $140 - 123.75 = 16.25$ mL more. But this changes the total volume. Let $z$ = additional pure alcohol added. New total volume is $400 + z$, pure alcohol amount is $123.75 + z$. For 35% concentration: $\frac{123.75 + z}{400 + z} = 0.35$. Solving: $123.75 + z = 0.35(400 + z) = 140 + 0.35z$, so $123.75 + z = 140 + 0.35z$, giving $0.65z = 16.25$, thus $z = 25$ mL.

Question 4

1. 75 mL
2. 90 mL
3. 100 mL (correct answer)
4. 112.5 mL

Explanation: First, find the amounts needed for the original 150 mL mixture: Let $x$ = mL of Solution A and $y$ = mL of Solution B. We have $x + y = 150$ and $0.08x + 0.20y = 0.14(150) = 21$. From the first equation: $y = 150 - x$. Substituting: $0.08x + 0.20(150 - x) = 21$, so $0.08x + 30 - 0.20x = 21$, giving $-0.12x = -9$, thus $x = 75$ mL of Solution A and $y = 75$ mL of Solution B. The ratio is $\frac{x}{y} = \frac{75}{75} = 1:1$. For the additional 50 mL of 14% solution maintaining the same 1:1 ratio: 25 mL of Solution A and 25 mL of Solution B. Verification: $0.08(25) + 0.20(25) = 2 + 5 = 7$ mL pure acid in 50 mL total = 14% ✓. Total Solution A used: $75 + 25 = 100$ mL.