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.
College Algebra Quiz: Mixture Problems
4 questions · exam conditions
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Mixture ProblemsQuestion 1 of 4
A pharmacist needs to prepare 200 mL of a 12% saline solution by mixing a 20% solution with distilled water (0% saline). Due to measurement constraints, the pharmacist can only measure in increments of 5 mL. What is the minimum amount of the 20% solution needed to achieve a concentration as close as possible to 12%?
A115 mL
B120 mL
C125 mL
D130 mL
College Algebra Quiz
College Algebra Quiz: Mixture Problems
Practice Mixture Problems in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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.
All questions
Question 1
A pharmacist needs to prepare 200 mL of a 12% saline solution by mixing a 20% solution with distilled water (0% saline). Due to measurement constraints, the pharmacist can only measure in increments of 5 mL. What is the minimum amount of the 20% solution needed to achieve a concentration as close as possible to 12%?
- 115 mL
- 120 mL (correct answer)
- 125 mL
- 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 2000.20(115)=20023=11.5%, and 125 mL gives concentration 2000.20(125)=20025=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
A paint store mixes two paints: Paint X contains 40% pigment and Paint Y contains 10% pigment. A customer orders 60 gallons of paint with 25% pigment content. After mixing, the store realizes they made an error and actually created a 28% pigment blend. If they used the correct total volume of 60 gallons, how many more gallons of Paint X did they use than intended?
- 4 gallons
- 6 gallons (correct answer)
- 8 gallons
- 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
A laboratory technician needs to create 400 mL of a 35% alcohol solution by mixing pure alcohol (100% alcohol) with a 20% alcohol solution. However, the technician accidentally uses a 15% alcohol solution instead of the 20% solution. To correct this and still achieve 400 mL of 35% alcohol solution, how much additional pure alcohol must be added to the incorrect mixture?
- 15 mL
- 20 mL
- 25 mL (correct answer)
- 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 400123.75=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: 400+z123.75+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
A chemistry student mixes Solution A (8% acid) with Solution B (20% acid) to create 150 mL of a 14% acid solution. The student then realizes that an additional 50 mL of 14% solution is needed for the experiment. If the student maintains the same ratio of Solution A to Solution B, what is the total amount of Solution A that will be used for both the original and additional mixtures combined?
- 75 mL
- 90 mL
- 100 mL (correct answer)
- 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 yx=7575=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.