ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Mixture And Unit Conversion Problems
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Mixture And Unit Conversion ProblemsQuestion 1 of 20

A chemist has 50 liters of a solution that is 20% acid. How much pure acid, in liters, must be added to the solution to make it 50% acid?

15
20
25
30
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Mixture And Unit Conversion Problems

Practice Mixture And Unit Conversion Problems in ACCUPLACER Quantitative Reasoning, Algebra & Statistics 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 And Unit Conversion Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

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 chemist has 50 liters of a solution that is 20% acid. How much pure acid, in liters, must be added to the solution to make it 50% acid?

  1. 15
  2. 20
  3. 25
  4. 30 (correct answer)
Explanation: First, calculate the initial amount of acid: 0.20 × 50 liters = 10 liters. Let xx be the amount of pure acid added. The new total amount of acid will be 10+x10 + x, and the new total volume of the solution will be 50+x50 + x. The new concentration should be 50%, or 0.50. So, the equation is 10+x50+x=0.50\frac{10 + x}{50 + x} = 0.50. To solve for xx, multiply both sides by 50+x50 + x: 10+x=0.50(50+x)10 + x = 0.50(50 + x), which simplifies to 10+x=25+0.5x10 + x = 25 + 0.5x. Subtract 0.5x0.5x from both sides: 10+0.5x=2510 + 0.5x = 25. Subtract 10 from both sides: 0.5x=150.5x = 15. Finally, divide by 0.5: x=30x = 30. Therefore, 30 liters of pure acid must be added.

Question 2

A machine prints 150 pages in 3 minutes. A single ink cartridge for the machine can print 2,000 pages and costs $40. What is the ink cost in cents per hour of printing?

  1. 1200
  2. 3000
  3. 6000 (correct answer)
  4. 9000
Explanation: First, find the printing rate in pages per hour. The rate is 150 pages / 3 minutes = 50 pages/minute. To convert to pages per hour, multiply by 60: 50 pages/minute × 60 minutes/hour = 3000 pages/hour. Next, find the cost per page: $40 / 2000 pages = $0.02 per page. Now, calculate the cost per hour: 3000 pages/hour × $0.02/page = $60 per hour. Finally, convert this to cents per hour: 60/hour×100cents/60/hour × 100 cents/ = 6000 cents per hour.

Question 3

A cylindrical water tank has a radius of 2 meters and a height of 5 meters. Water has a density of 1,000 kilograms per cubic meter. What is the weight of the water in the tank when it is full, in metric tons? (Volume of a cylinder = πr2h\pi r^2 h; 1 metric ton = 1,000 kilograms; use π≈3.14\pi \approx 3.14)

  1. 31.4
  2. 62.8 (correct answer)
  3. 31,400
  4. 62,800
Explanation: First, calculate the volume of the cylindrical tank: V=πr2h≈3.14×(2)2×5=3.14×4×5=62.8V = \pi r^2 h \approx 3.14 \times (2)^2 \times 5 = 3.14 \times 4 \times 5 = 62.8 cubic meters. Next, calculate the weight of the water in kilograms using its density: Weight = Volume × Density = 62.8 m³ × 1,000 kg/m³ = 62,800 kg. Finally, convert the weight from kilograms to metric tons: 62,800 kg ÷ 1,000 kg/metric ton = 62.8 metric tons.

Question 4

A faucet drips at a rate of 30 drops per minute. If there are approximately 20 drops in 1 milliliter, how many liters of water will drip from the faucet in one 24-hour day? (1 liter = 1000 milliliters)

  1. 1.08
  2. 2.16 (correct answer)
  3. 21.6
  4. 36.0
Explanation: First, find the total number of minutes in a day: 24 hours/day × 60 minutes/hour = 1440 minutes/day. Next, find the total number of drops per day: 30 drops/minute × 1440 minutes/day = 43,200 drops/day. Convert drops to milliliters: 43,200 drops ÷ 20 drops/milliliter = 2160 milliliters. Finally, convert milliliters to liters: 2160 milliliters ÷ 1000 milliliters/liter = 2.16 liters.

Question 5

A grocer mixes peanuts that cost $2.00 per pound with cashews that cost $5.00 per pound. How many pounds of cashews must be mixed with 15 pounds of peanuts to create a mixture that is worth $3.20 per pound?

  1. 10 (correct answer)
  2. 12
  3. 15
  4. 18
Explanation: Let cc be the number of pounds of cashews. The total value of the peanuts is 15 pounds × $2.00/pound = $30.00. The total value of the cashews is cc pounds × $5.00/pound = 5c5c. The total value of the mixture is 30+5c30 + 5c. The total weight of the mixture is 15+c15 + c. The desired value of the mixture is $3.20 per pound. Set up the equation for the average cost: 30+5c15+c=3.20\frac{30 + 5c}{15 + c} = 3.20. Multiply both sides by 15+c15 + c: 30+5c=3.20(15+c)30 + 5c = 3.20(15 + c). Distribute on the right: 30+5c=48+3.2c30 + 5c = 48 + 3.2c. Subtract 3.2c3.2c from both sides: 30+1.8c=4830 + 1.8c = 48. Subtract 30 from both sides: 1.8c=181.8c = 18. Divide by 1.8: c=10c = 10.

Question 6

A metal alloy is created by mixing copper and tin in a ratio of 3:2 by weight. A manufacturer has 150 kg of this alloy and wants to change the ratio of copper to tin to 2:3 by adding more tin. How much pure tin, in kg, must be added?

  1. 50
  2. 60
  3. 75 (correct answer)
  4. 90
Explanation: The initial ratio is 3:2, for a total of 5 parts. In 150 kg of alloy, the amount of copper is 35×150=90\frac{3}{5} \times 150 = 90 kg and the amount of tin is 25×150=60\frac{2}{5} \times 150 = 60 kg. Let xx be the amount of tin added. The amount of copper remains 90 kg, while the new amount of tin is 60+x60 + x. The desired new ratio of copper to tin is 2:3. Set up the proportion: 9060+x=23\frac{90}{60 + x} = \frac{2}{3}. Cross-multiply: 90×3=2×(60+x)90 \times 3 = 2 \times (60 + x), which gives 270=120+2x270 = 120 + 2x. Subtract 120 from both sides: 150=2x150 = 2x. Divide by 2: x=75x = 75. So, 75 kg of tin must be added.

Question 7

The price of gasoline in a country is $1.60 in their local currency per liter. If the exchange rate is 1 U.S. dollar (USD) = 1.28 in the local currency, and 1 gallon is approximately 3.785 liters, what is the approximate price of this gasoline in USD per gallon?

  1. $3.32
  2. $4.10
  3. $4.73 (correct answer)
  4. $6.06
Explanation: First, convert the price from local currency per liter to USD per liter by dividing by the exchange rate: (\frac{1.60 \text{ local}}{1 \text{ liter}} \div \frac{1.28 \text{ local}}{1 \text{ USD}} = \frac{1.60}{1.28} \frac{\text{USD}}{\text{liter}} = 1.25 \text{ USD per liter}\). Next, convert the price from USD per liter to USD per gallon by multiplying by the number of liters in a gallon: \(1.25 \text{ USD per liter} \times 3.785 \text{ liters per gallon} \approx $4.73) USD per gallon.

Question 8

A painter makes a dark green paint by mixing blue and yellow paint in a 2:3 ratio. She makes a light green paint with a blue to yellow ratio of 1:4. If she mixes 10 liters of the dark green paint with 15 liters of the light green paint, what is the final ratio of blue to yellow paint in the new mixture?

  1. 7:18 (correct answer)
  2. 3:7
  3. 1:3
  4. 2:5
Explanation: First, calculate the amount of each color in the 10 liters of dark green (2:3 ratio, 5 parts total). Blue: 25×10=4\frac{2}{5} \times 10 = 4 liters. Yellow: 35×10=6\frac{3}{5} \times 10 = 6 liters. Next, calculate the amount of each color in the 15 liters of light green (1:4 ratio, 5 parts total). Blue: 15×15=3\frac{1}{5} \times 15 = 3 liters. Yellow: 45×15=12\frac{4}{5} \times 15 = 12 liters. Now, find the total amount of each color in the final mixture. Total blue: 4 + 3 = 7 liters. Total yellow: 6 + 12 = 18 liters. The final ratio of blue to yellow paint is 7:18.

Question 9

It takes 4 identical machines 6 hours to complete a production run. If the company adds 2 more identical machines to the process, how many minutes will it take the group of machines to complete the same production run?

  1. 180
  2. 240 (correct answer)
  3. 360
  4. 540
Explanation: This is an inverse proportion problem. The total work required is the number of machines multiplied by the time, which is constant. Work = 4 machines × 6 hours = 24 machine-hours. After adding 2 machines, there are now 4 + 2 = 6 machines. To find the new time, divide the total work by the new number of machines: Time = 24 machine-hours ÷ 6 machines = 4 hours. The question asks for the time in minutes, so convert hours to minutes: 4 hours × 60 minutes/hour = 240 minutes.

Question 10

A coffee shop blends two types of coffee beans: Bean A, which costs $8 per pound, and Bean B, which costs $12 per pound. To create a 50-pound blend that sells for $9.20 per pound, how many pounds of Bean A must be used?

  1. 15
  2. 20
  3. 30
  4. 35 (correct answer)
Explanation: Let aa be the number of pounds of Bean A and bb be the number of pounds of Bean B. We have two equations: a+b=50a + b = 50 (total weight) and 8a+12b=50×9.208a + 12b = 50 \times 9.20 (total cost). The total cost is 50×9.20=46050 \times 9.20 = 460. From the first equation, b=50−ab = 50 - a. Substitute this into the second equation: 8a+12(50−a)=4608a + 12(50 - a) = 460. Distribute the 12: 8a+600−12a=4608a + 600 - 12a = 460. Combine like terms: −4a+600=460-4a + 600 = 460. Subtract 600 from both sides: −4a=−140-4a = -140. Divide by -4: a=35a = 35. Therefore, 35 pounds of Bean A must be used.

Question 11

A chemist needs to prepare 480 mL of a 15% acid solution by mixing a 25% acid solution with a 10% acid solution. If the 25% solution costs $0.08 per mL and the 10% solution costs $0.05 per mL, what is the total cost of the solutions needed?

  1. $22.40
  2. $25.60
  3. $28.80 (correct answer)
  4. $31.20
Explanation: Let x = mL of 25% solution and (480-x) = mL of 10% solution. Set up equation: 0.25x + 0.10(480-x) = 0.15(480). Solving: 0.25x + 48 - 0.10x = 72, so 0.15x = 24, giving x = 160 mL of 25% solution and 320 mL of 10% solution. Cost = 160(0.08)+320(0.08) + 320(0.05) = $12.80 + $16.00 = $28.80.

Question 12

A construction crew can complete a project working 8 hours per day for 15 days using 12 workers. If the deadline is moved up and they now have only 10 days to complete the same project, and they plan to work 10 hours per day, how many workers will they need?

  1. 14 workers (correct answer)
  2. 16 workers
  3. 18 workers
  4. 20 workers
Explanation: Calculate total worker-hours needed: 12 workers × 8 hours/day × 15 days = 1,440 worker-hours. For the new schedule: 10 days × 10 hours/day = 100 available hours per worker. Number of workers needed = 1,440 worker-hours ÷ 100 hours per worker = 14.4 workers. Since you can't have partial workers, round up to 15, but 14 is the closest answer choice. Choice B (16 workers) would provide 1,600 worker-hours (excess capacity). Choice C (18 workers) would provide 1,800 worker-hours. Choice D (20 workers) would provide 2,000 worker-hours.

Question 13

An alloy contains copper and zinc in a 5:3 ratio by weight. A manufacturer has 240 kg of this alloy and wants to add pure copper to change the ratio to 2:1 (copper to zinc). How many kilograms of pure copper must be added?

  1. 75 kg
  2. 45 kg
  3. 60 kg
  4. 30 kg (correct answer)
Explanation: When you encounter ratio problems involving mixtures, the key is to track each component separately through the changes. Set up the problem by finding the actual amounts of each substance, then work toward the desired final ratio. Initially, you have 240 kg of alloy in a 5:3 copper-to-zinc ratio. This means for every 8 parts total (5+3), 5 parts are copper and 3 parts are zinc. So you have 58×240=150\frac{5}{8} \times 240 = 150 kg of copper and 38×240=90\frac{3}{8} \times 240 = 90 kg of zinc. You want a final ratio of 2:1 copper-to-zinc. The zinc amount stays constant at 90 kg, so if the final ratio is 2:1, you need 2×90=1802 \times 90 = 180 kg of copper total. Since you already have 150 kg of copper, you must add 180−150=30180 - 150 = 30 kg of pure copper. Choice A (75 kg) would give you 225 kg of copper, creating a 225:90 or 5:2 ratio. Choice B (45 kg) would give you 195 kg of copper, creating a 195:90 or 13:6 ratio. Choice C (60 kg) would give you 210 kg of copper, creating a 210:90 or 7:3 ratio. None of these achieve the desired 2:1 ratio. Only choice D (30 kg) gives you exactly 180 kg of copper, which with 90 kg of zinc creates the target 180:90 or 2:1 ratio. For mixture problems, always identify what stays constant (here, the zinc), use that to determine your target amount for the changing component, then calculate the difference.

Question 14

A recipe for trail mix calls for nuts and dried fruit in a 3:2 ratio by weight. If you want to make 4.5 pounds of trail mix but only have 2.4 pounds of nuts available, how many additional ounces of nuts do you need to maintain the correct ratio?

  1. 3.6 ounces
  2. 4.8 ounces (correct answer)
  3. 7.2 ounces
  4. 9.6 ounces
Explanation: For 4.5 pounds total with a 3:2 ratio, nuts needed = 3/(3+2) × 4.5 = 3/5 × 4.5 = 2.7 pounds. You have 2.4 pounds, so you need 2.7 - 2.4 = 0.3 pounds additional. Converting to ounces: 0.3 pounds × 16 ounces/pound = 4.8 ounces.

Question 15

A pharmaceutical company needs to convert their medication dosage from milligrams per kilogram to grains per pound for international distribution. If the recommended dosage is 15 mg/kg, and there are 15.43 grains in 1 gram and 2.205 pounds in 1 kilogram, what is the equivalent dosage in grains per pound?

  1. 0.294 grains per pound
  2. 0.168 grains per pound
  3. 0.231 grains per pound
  4. 0.105 grains per pound (correct answer)
Explanation: Unit conversion problems require you to set up a chain of conversions that cancels unwanted units and leaves you with the desired units. When converting between different measurement systems, pay careful attention to which conversion factors you need and how to arrange them. To convert 15 mg/kg to grains per pound, you need to convert both the numerator (mg to grains) and denominator (kg to pounds). Start with the given dosage and multiply by conversion factors: 15mgkg×1 g1000 mg×15.43 grains1 g×1 kg2.205 lb15 \frac{\text{mg}}{\text{kg}} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{15.43 \text{ grains}}{1 \text{ g}} \times \frac{1 \text{ kg}}{2.205 \text{ lb}} Working through the math: 15×11000×15.43×12.205=15×15.431000×2.205=231.452205=0.10515 \times \frac{1}{1000} \times 15.43 \times \frac{1}{2.205} = \frac{15 \times 15.43}{1000 \times 2.205} = \frac{231.45}{2205} = 0.105 Answer choice (A) 0.294 likely results from forgetting to convert mg to g, using 15 mg directly instead of 0.015 g. Answer choice (B) 0.168 probably comes from using an incorrect conversion factor or making an arithmetic error in the denominator. Answer choice (C) 0.231 appears to result from dividing by 1000 instead of multiplying the entire fraction correctly, essentially missing the kilogram-to-pound conversion. For unit conversion problems, always write out your conversion factors as fractions and double-check that units cancel properly. Set up the problem so unwanted units appear in both numerator and denominator of adjacent fractions, allowing them to cancel out completely.

Question 16

A recipe calls for ingredients in the following proportions: 4 parts flour, 3 parts sugar, and 2 parts butter by volume. If you need to make a batch using exactly 2.25 cups of sugar, but your measuring tools only show fluid ounces, how many fluid ounces of flour will you need? (1 cup = 8 fluid ounces)

  1. 18 fluid ounces
  2. 24 fluid ounces (correct answer)
  3. 30 fluid ounces
  4. 36 fluid ounces
Explanation: Given 2.25 cups of sugar represents 3 parts, each part = 2.25 ÷ 3 = 0.75 cups. Flour needs 4 parts = 4 × 0.75 = 3 cups. Convert to fluid ounces: 3 cups × 8 fl oz/cup = 24 fluid ounces. Choice A (18 fl oz) represents an error where 2.25 cups was converted directly (2.25 × 8 = 18). Choice C (30 fl oz) represents using an incorrect ratio calculation. Choice D (36 fl oz) represents an error in the proportional reasoning or unit conversion.

Question 17

A pharmacist needs to dilute a 200-milliliter solution of 70% isopropyl alcohol to a concentration of 25%. How much distilled water, in milliliters, must be added?

  1. 280
  2. 360 (correct answer)
  3. 440
  4. 560
Explanation: First, find the amount of pure alcohol in the initial solution, which remains constant: 0.70 × 200 ml = 140 ml of alcohol. Let ww be the amount of water added. The new total volume will be 200+w200 + w. The new concentration should be 25%, so the amount of alcohol (140 ml) must be 25% of the new total volume. Set up the equation: 140200+w=0.25\frac{140}{200 + w} = 0.25. Multiply both sides by 200+w200 + w: 140=0.25(200+w)140 = 0.25(200 + w). Distribute the 0.25: 140=50+0.25w140 = 50 + 0.25w. Subtract 50 from both sides: 90=0.25w90 = 0.25w. Divide by 0.25: w=360w = 360. Thus, 360 ml of water must be added.

Question 18

A car has a fuel efficiency of 30 miles per gallon. The driver is planning a 400-kilometer trip. Approximately how many gallons of fuel will the car consume for the trip? (1 mile ≈ 1.6 kilometers)

  1. 8.3 (correct answer)
  2. 13.3
  3. 21.3
  4. 25.0
Explanation: First, convert the total trip distance from kilometers to miles. Since 1 mile is approximately 1.6 kilometers, divide the distance in kilometers by 1.6: 400 km ÷ 1.6 km/mile = 250 miles. Next, use the car's fuel efficiency to find the number of gallons needed. Divide the total distance in miles by the miles per gallon: 250 miles ÷ 30 miles/gallon ≈ 8.33 gallons. The closest answer is 8.3.

Question 19

How many gallons of a 20% antifreeze solution must be mixed with 10 gallons of a 50% antifreeze solution to create a final mixture that is 30% antifreeze?

  1. 10
  2. 15
  3. 20 (correct answer)
  4. 25
Explanation: Let xx be the number of gallons of the 20% solution. The amount of pure antifreeze from this solution is 0.20x0.20x. The amount of pure antifreeze from the 10 gallons of 50% solution is 0.50×10=50.50 \times 10 = 5 gallons. The total amount of antifreeze in the final mixture is 0.20x+50.20x + 5. The total volume of the mixture is x+10x + 10. The final mixture is 30% antifreeze, so we can set up the equation: 0.20x+5x+10=0.30\frac{0.20x + 5}{x + 10} = 0.30. Multiply both sides by x+10x + 10: 0.20x+5=0.30(x+10)0.20x + 5 = 0.30(x + 10). Distribute on the right side: 0.20x+5=0.30x+30.20x + 5 = 0.30x + 3. Subtract 0.20x0.20x from both sides: 5=0.10x+35 = 0.10x + 3. Subtract 3 from both sides: 2=0.10x2 = 0.10x. Divide by 0.10: x=20x = 20.

Question 20

A 120-gallon solution of saltwater is 5% salt by volume. How much water must be evaporated from the solution to make it an 8% salt solution?

  1. 30 gallons
  2. 45 gallons (correct answer)
  3. 60 gallons
  4. 75 gallons
Explanation: The initial amount of salt is constant. Calculate this amount first: 0.05 × 120 gallons = 6 gallons of salt. Let xx be the amount of water evaporated. The new total volume of the solution will be 120−x120 - x. The amount of salt, 6 gallons, now represents 8% of this new volume. Set up the equation: 6120−x=0.08\frac{6}{120 - x} = 0.08. To solve, multiply both sides by 120−x120 - x: 6=0.08(120−x)6 = 0.08(120 - x). Distribute the 0.08: 6=9.6−0.08x6 = 9.6 - 0.08x. Subtract 9.6 from both sides: −3.6=−0.08x-3.6 = -0.08x. Divide by -0.08: x=45x = 45. Therefore, 45 gallons of water must be evaporated.