GMAT Quiz: Mixture Problems
8 questions · exam conditions
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Mixture ProblemsQuestion 1 of 8

Bartender A has 15% alcohol punch, and Bartender B has 35% alcohol punch. They will combine the two to make 10 liters of a punch that is exactly 25% alcohol. How many liters must Bartender A contribute?

2.5 liters
5 liters
7.5 liters
8 liters
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GMAT Quiz

GMAT Quiz: Mixture Problems

Practice Mixture Problems in GMAT with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 GMAT.

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

Bartender A has 15% alcohol punch, and Bartender B has 35% alcohol punch. They will combine the two to make 10 liters of a punch that is exactly 25% alcohol. How many liters must Bartender A contribute?

  1. 2.5 liters
  2. 5 liters (correct answer)
  3. 7.5 liters
  4. 8 liters
Explanation: When you encounter mixture problems on the GMAT, you're dealing with weighted averages. The key insight is that the final concentration will be a weighted average of the individual concentrations, where the weights are the volumes contributed. Let's set up the equation systematically. If Bartender A contributes xx liters of 15% alcohol punch, then Bartender B must contribute (10−x)(10-x) liters of 35% alcohol punch to reach our target of 10 liters total. The total alcohol content must equal 25% of 10 liters, which is 2.5 liters of pure alcohol. So we can write: 0.15x+0.35(10−x)=2.50.15x + 0.35(10-x) = 2.5 Expanding: 0.15x+3.5−0.35x=2.50.15x + 3.5 - 0.35x = 2.5 Combining like terms: −0.20x=−1-0.20x = -1 Therefore: x=5x = 5 Bartender A must contribute 5 liters, making (B) correct. Let's check the wrong answers: (A) 2.5 liters would create 0.15(2.5)+0.35(7.5)=3.00.15(2.5) + 0.35(7.5) = 3.0 liters of alcohol, giving 30% concentration—too strong. (C) 7.5 liters would yield 0.15(7.5)+0.35(2.5)=2.00.15(7.5) + 0.35(2.5) = 2.0 liters of alcohol for 20% concentration—too weak. (D) 8 liters would produce 0.15(8)+0.35(2)=1.90.15(8) + 0.35(2) = 1.9 liters of alcohol for 19% concentration—also too weak. Study tip: In mixture problems, always check that your volumes add up to the target total, and verify your answer by calculating the final concentration. The algebra will guide you to the right setup every time.

Question 2

A fertilizer company sells two soil mixes. Mix R contains 10% peat moss by weight, and Mix S contains 25% peat moss by weight. If a landscaper wants exactly 40 kilograms of soil that is 16% peat moss, how many kilograms of Mix R should be used?

  1. 28 kg
  2. 32 kg
  3. 36 kg
  4. 24 kg (correct answer)
Explanation: This is a classic mixture problem that tests your ability to set up and solve weighted average equations. When you see questions about combining substances with different concentrations to achieve a target concentration, think in terms of the total amount of the key component (here, peat moss) coming from each source. Let's say you use xx kg of Mix R and (40−x)(40-x) kg of Mix S. The total peat moss must equal 16% of 40 kg, which is 6.4 kg. Setting up the equation: 0.10x+0.25(40−x)=6.40.10x + 0.25(40-x) = 6.4 Solving: 0.10x+10−0.25x=6.40.10x + 10 - 0.25x = 6.4, so −0.15x=−3.6-0.15x = -3.6, which gives us x=24x = 24 kg of Mix R. You can verify: 24 kg of Mix R contributes 24×0.10=2.424 \times 0.10 = 2.4 kg of peat moss, and 16 kg of Mix S contributes 16×0.25=4.016 \times 0.25 = 4.0 kg of peat moss, totaling exactly 6.4 kg. Choice A (28 kg) would give you too little peat moss overall since you're using more of the lower-concentration mix. Choice B (32 kg) makes the same error but more severely. Choice C (36 kg) would leave you with only 4 kg of the high-concentration Mix S, resulting in insufficient peat moss content. Strategy tip: In mixture problems, always check that your amounts add up to the total quantity required, and verify your answer by calculating the actual concentration achieved. Setting up the equation systematically—total of desired component equals sum from each source—prevents algebraic errors.

Question 3

An investment firm manages two portfolios with different average annual returns: Portfolio A averages 8% annually, while Portfolio B averages 12% annually. At the beginning of the year, the firm had a total of 2,000,000investedacrossbothportfolios.Duringtheyear,theytransferred2,000,000 invested across both portfolios. During the year, they transferred 200,000 from Portfolio A to Portfolio B. At year-end, the weighted average return across all investments was 9.6%. What was the initial amount invested in Portfolio A?

  1. $$1,200,000 was initially in Portfolio A
  2. $$1,300,000 was initially in Portfolio A
  3. $$1,400,000 was initially in Portfolio A (correct answer)
  4. $$1,500,000 was initially in Portfolio A
Explanation: Let x = initial amount in Portfolio A. Then initial amount in Portfolio B = $2,000,000 - x. After transfer: Portfolio A has (x - 200,000),PortfolioBhas(200,000), Portfolio B has (2,000,000 - x + 200,000)=(200,000) = (2,200,000 - x). Returns: Portfolio A generates (x - 200,000)(0.08), Portfolio B generates (2,200,000 - x)(0.12). Total return: (x - 200,000)(0.08) + (2,200,000 - x)(0.12) = 2,000,000(0.096) = $192,000. Expanding: 0.08x - 16,000 + 264,000 - 0.12x = 192,000. Simplifying: -0.04x + 248,000 = 192,000. So -0.04x = -56,000, giving x = $1,400,000. Verification: Initial A = $1,400,000, Initial B = $600,000. After transfer: A = $1,200,000, B = $800,000. Returns: $1,200,000(0.08) + $800,000(0.12) = $96,000 + $96,000 = $192,000. Weighted average: 192,000/192,000/2,000,000 = 9.6%. ✓

Question 4

A coffee shop blends three types of beans: Colombian beans at 12perpound,Brazilianbeansat12 per pound, Brazilian beans at 8 per pound, and Ethiopian beans at 16perpound.Theshopcreatesa50−poundblendusingtwiceasmanypoundsofColombianbeansasBrazilianbeans,andtheremainingweightisEthiopianbeans.Iftheblend′saveragecostperpoundis16 per pound. The shop creates a 50-pound blend using twice as many pounds of Colombian beans as Brazilian beans, and the remaining weight is Ethiopian beans. If the blend's average cost per pound is 12.80, how many pounds of Brazilian beans are used?

  1. 10 pounds of Brazilian beans (correct answer)
  2. 12 pounds of Brazilian beans
  3. 15 pounds of Brazilian beans
  4. 18 pounds of Brazilian beans
Explanation: Let x = pounds of Brazilian beans. Then Colombian = 2x pounds, and Ethiopian = 50 - 3x pounds. The cost equation: 12(2x) + 8(x) + 16(50 - 3x) = 50(12.80). Simplifying: 24x + 8x + 800 - 48x = 640. This gives: -16x = -160, so x = 10. Choice B (12) results from incorrectly using equal weights of Colombian and Brazilian. Choice C (15) comes from setting up the constraint as x + 2x + Ethiopian = 50 but solving incorrectly. Choice D (18) assumes Colombian beans are 1.5 times Brazilian beans instead of twice.

Question 5

A chemistry lab has two acid solutions: Solution A contains 25% acid by volume, and Solution B contains 60% acid by volume. If 40 liters of Solution A is mixed with some amount of Solution B to create a mixture that is 45% acid, how many liters of Solution B were added?

  1. 28 liters
  2. 32 liters (correct answer)
  3. 36 liters
  4. 40 liters
Explanation: Let x = liters of Solution B. The equation is: (40)(0.25) + (x)(0.60) = (40 + x)(0.45). Simplifying: 10 + 0.6x = 18 + 0.45x. Solving: 0.15x = 8, so x = 32 liters. Choice A (28) results from incorrectly setting up 10 + 0.6x = 0.45(40), ignoring the variable volume. Choice C (36) comes from solving 10 + 0.6x = 0.45x + 18 with a sign error. Choice D (40) assumes equal volumes are needed.

Question 6

Coffee blend X costs $5 per pound and blend Y costs $9 per pound. How many pounds of blend X must be mixed with 6 pounds of blend Y to obtain a mixture costing exactly $7 per pound?

  1. 3 pounds
  2. 4.5 pounds
  3. 6 pounds (correct answer)
  4. 9 pounds
Explanation: This is a weighted average mixture problem, where you're combining two items with different costs to achieve a target average cost. The key insight is setting up an equation where the total cost of ingredients equals the total cost of the final mixture. Let's call the unknown pounds of blend X as xx. You have xx pounds at $5/pound plus 6 pounds at $9/pound, and you want the entire mixture to cost $7/pound. Setting up the equation: $5x+9(6)=7(x+6)5x + 9(6) = 7(x + 6) $ Solving: 5x + 54 = 7x + 42 54 - 42 = 7x - 5x 12 = 2x x = 6 You can verify: 6 pounds of blend X (30)+6poundsofblendY(30) + 6 pounds of blend Y (54) = 12 pounds total costing $84, which is indeed $7 per pound. Answer A (3 pounds) gives you a mixture costing $7.50 per pound—too expensive because you're using too little of the cheaper blend. Answer B (4.5 pounds) yields approximately $7.14 per pound, still too high. Answer D (9 pounds) would create a mixture costing about $6.40 per pound—too cheap because you're using too much of the less expensive blend. Strategy tip: In mixture problems, always check if your answer makes intuitive sense. Since $7 is exactly halfway between $5 and $9, you need equal amounts of each blend to hit that target. This "halfway" pattern is a quick way to spot the answer when the target average falls exactly in the middle of your two values.

Question 7

A chemist needs 20 liters of a 30% saline solution. She has an unlimited supply of a 25% saline solution and a 40% saline solution. How many liters of the 25% solution should she use?

  1. 8 liters
  2. 12 liters
  3. 16 liters (correct answer)
  4. 18 liters
Explanation: When you encounter mixture problems on the GMAT, you're dealing with weighted averages. The key insight is that you're combining two solutions to create a target concentration that falls between the original concentrations. Let's set up the problem systematically. If you use xx liters of the 25% solution, then you'll need (20−x)(20-x) liters of the 40% solution to reach 20 total liters. The amount of pure salt from each solution must equal the pure salt in the final mixture: 0.25x+0.40(20−x)=0.30(20)0.25x + 0.40(20-x) = 0.30(20) Expanding: 0.25x+8−0.40x=60.25x + 8 - 0.40x = 6 Simplifying: −0.15x=−2-0.15x = -2 Therefore: x=20.15=20015=13.33...x = \frac{2}{0.15} = \frac{200}{15} = 13.33... Wait - let me recalculate more carefully: x=20.15=201.5=13.33x = \frac{2}{0.15} = \frac{20}{1.5} = 13.33 liters. Actually, let me verify with answer choice C) 16 liters: If you use 16 liters of 25% solution and 4 liters of 40% solution, you get 16(0.25)+4(0.40)=4+1.6=5.616(0.25) + 4(0.40) = 4 + 1.6 = 5.6 liters of pure salt, giving 5.620=0.28=28%\frac{5.6}{20} = 0.28 = 28\%. Let me recalculate: 0.25x+0.40(20−x)=60.25x + 0.40(20-x) = 6 gives us 0.25x+8−0.40x=60.25x + 8 - 0.40x = 6, so −0.15x=−2-0.15x = -2, meaning x=13.33x = 13.33 liters. The closest answer is B) 12 liters. Choice A) 8 liters would create too strong a solution, while D) 18 liters would create too weak a solution. Remember: in mixture problems, always verify your algebra by checking that the pure substance amounts balance correctly.

Question 8

Coffee blend X costs $5 per pound and blend Y costs $9 per pound. How many pounds of blend X must be mixed with 6 pounds of blend Y to obtain a mixture costing exactly $7 per pound?

  1. 3 pounds
  2. 4.5 pounds
  3. 6 pounds (correct answer)
  4. 9 pounds
Explanation: This is a weighted average problem, where you're combining two items with different costs to achieve a target average cost. The key insight is that the total cost of the mixture equals the sum of the costs of individual components. Let's call the pounds of blend X needed "x". You can set up an equation based on total costs: Cost of X + Cost of Y = Cost of mixture 5x+9(6)=7(x+6)5x + 9(6) = 7(x + 6) Solving: 5x+54=7x+425x + 54 = 7x + 42 54−42=7x−5x54 - 42 = 7x - 5x 12=2x12 = 2x x=6x = 6 So you need 6 pounds of blend X. You can verify this makes sense: 6 pounds at $5 plus 6 pounds at $9 gives you 12 pounds costing $84 total, which is $7 per pound. Choice A (3 pounds) would create a mixture that's too expensive. With only 3 pounds of the cheaper blend X mixed with 6 pounds of expensive blend Y, the average would be closer to Y's price of $9. Choice B (4.5 pounds) also doesn't provide enough of the cheaper blend to bring the average down to $7. Choice D (9 pounds) overshoots by adding too much of the cheaper blend, which would make the mixture cost less than $7 per pound. Remember: in mixture problems, when the target average is exactly halfway between two prices ($7 is halfway between $5 and $9), you need equal quantities of each component.