6th Grade Math Quiz: Solve Unit Rate Problems
20 questions · exam conditions
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Solve Unit Rate ProblemsQuestion 1 of 20

A recipe uses 3 cups of flour to make 12 muffins. If you want to make 20 muffins at the same rate, how many cups of flour will you need?

8 cups
5 cups
7.5 cups
4 cups
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6th Grade Math Quiz

6th Grade Math Quiz: Solve Unit Rate Problems

Practice Solve Unit Rate Problems in 6th Grade Math 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 Solve Unit Rate Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for 6th Grade Math.

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 recipe uses 3 cups of flour to make 12 muffins. If you want to make 20 muffins at the same rate, how many cups of flour will you need?

  1. 8 cups
  2. 5 cups (correct answer)
  3. 7.5 cups
  4. 4 cups
Explanation: This problem is all about rates — keeping the same amount of flour per muffin no matter how many you bake. First, find how much flour goes into one muffin. The recipe uses 3 cups for 12 muffins, so divide:
3÷12=0.25 cups per muffin3 \div 12 = 0.25 \text{ cups per muffin} Now multiply by the 20 muffins you want:
0.25×20=5 cups0.25 \times 20 = 5 \text{ cups} So you need 5 cups of flour. Think of it like filling juice cups. If 3 pitchers fill 12 cups, then each cup needs the same pour. To fill more cups, you just keep pouring at that steady rate! Try this at home: Look at any recipe in your kitchen. Pick one ingredient and figure out how much you'd need to double or triple the batch. Find the amount for one serving first, then multiply!

Question 2

Maria can type 180 words in 4 minutes. At this rate, how long will it take her to type a 1,350-word essay?

  1. 25 minutes
  2. 30 minutes (correct answer)
  3. 35 minutes
  4. 40 minutes
Explanation: First find the unit rate: 180 words ÷ 4 minutes = 45 words per minute. Then divide the total words by the rate: 1,350 words ÷ 45 words/minute = 30 minutes. Choice A uses 54 words/minute (calculation error). Choice C uses 180 words in 5 minutes instead of 4. Choice D uses the original 4 minutes as the rate.

Question 3

A copying machine makes 84 copies in 6 minutes. At this rate, how many copies can it make in 45 minutes?

  1. 630 copies (correct answer)
  2. 645 copies
  3. 675 copies
  4. 720 copies
Explanation: First find the unit rate: 84 copies ÷ 6 minutes = 14 copies per minute. Then multiply by the new time: 14 copies/minute × 45 minutes = 630 copies. Choice B uses 14.33 copies per minute. Choice C uses 15 copies per minute. Choice D uses 16 copies per minute.

Question 4

A cyclist rides at a constant speed of 15 miles per hour. How long will it take the cyclist to ride 42 miles?

  1. 2.8 hours (correct answer)
  2. 3.5 hours
  3. 27 hours
  4. 630 hours
Explanation: This question tests solving unit rate problems: using the given speed to find time for a distance, assuming constant rate. Unit rate problems involve using rate to find time by dividing distance by rate (42 ÷ 15 = 2.8 hours), interpreting constant rate as same speed throughout. Speed formulas: time = distance ÷ rate (t = d/r), like reversing rate = d/t. Example: At 15 mph for 42 miles, 42 ÷ 15 = 2.8 hours. The correct calculation is accurate division. Common errors include multiplying (15 × 42 = 630), inverting, or units wrong. To solve: apply t = d/r (42 ÷ 15 = 2.8), verify units (hours) and reasonableness (15 × 2.8 = 42). Speed relations: d=rt, t=d/r, r=d/t.

Question 5

A school club buys 8 identical notebooks for $14.40. If each notebook costs the same amount, how much would 10 notebooks cost?

  1. $1.80
  2. $144.00
  3. $16.20
  4. $18.00 (correct answer)
Explanation: This question tests solving unit rate problems by finding cost per notebook from total for several, then using it for a different quantity. Unit rate problems involve finding rate ($14.40 ÷ 8 = $1.80 per notebook), then multiplying (1.80 × 10 = $18.00). Constant pricing means same cost per item. For example, similar to $6 for 3 lb at $2/lb, then $10 for 5 lb. The correct calculation divides then multiplies accurately. Common errors include multiplying totals directly like 14.40 × 10 = 144, or dividing wrong like 8 ÷ 14.40. To solve: (1) find unit rate (14.40 ÷ 8 = 1.80), (2) multiply by new quantity (1.80 × 10 = 18), (3) verify by proportion (14.40/8 = x/10, x = (14.40 × 10)/8 = 144/8 = 18), (4) check units are dollars.

Question 6

A movie theater charges $9 per ticket. How much will 7 tickets cost?

  1. $16
  2. $49
  3. $63 (correct answer)
  4. $72
Explanation: This question tests solving unit rate problems: finding rate (cost per item, speed mph, work rate), using rate to predict (at constant rate, how much in given time/quantity), interpreting in contexts. Unit rate problems: (1) find rate (4 lawns in 7 hours gives rate 4/7 lawns per hour, dividing quantity by time), (2) use rate (at 4/7 lawns/hr, in 35 hours: multiply rate×time=(4/7)×35=20 lawns), (3) interpret (constant rate means same amount per unit throughout: 4/7 per hour maintained). Speed: distance÷time gives rate (180 miles÷3 hours=60 mph), use rate: speed×time=distance (60×5=300 miles in 5 hours). Unit pricing: total÷quantity gives per-unit cost (6÷3lb=6÷3 lb=2/lb), use: rate×quantity=total (2/lb×5lb=2/lb×5 lb=10). Here, unit price is 9perticket,for7:9×7=9 per ticket, for 7: 9×7=63. Mistake: adding instead (9+7=16, choice A) or wrong multiplication (9×8=72, choice D). Solving: (1) find unit rate (given as $9 per ticket), (2) identify what's asked (cost for 7), (3) apply rate (multiply rate by quantity: 9×7=63), (4) verify units (dollars) and reasonableness (7 at $9: 5×9=45, 2×9=18, total 63, yes).

Question 7

A school club sells tickets for $8 each. If the club collects $104 from ticket sales, how many tickets did they sell?

  1. 12 tickets
  2. 13 tickets (correct answer)
  3. 96 tickets
  4. 832 tickets
Explanation: This question tests solving unit rate problems: using the given rate per ticket to find the quantity from total collected, assuming constant price. Unit rate problems involve using the rate to predict by dividing total by rate (104÷8=13104 \div 8 = 13) tickets, interpreting constant rate as same price per ticket. For pricing, total ÷\div rate gives quantity, reverse of usual rate finding, like finding time from distance and speed (t=d/rt = d/r). Example: At $8 each, for $104, (104÷8=13104 \div 8 = 13) tickets. The correct calculation is division with units (tickets). Common errors include multiplying instead (8×104=8328 \times 104 = 832), or wrong division. To solve: identify rate ($8 per ticket), divide total by rate (104÷8=13104 \div 8 = 13), verify reasonableness (13×8=10413 \times 8 = 104). Mistakes: wrong operation or units missing.

Question 8

A school club buys 8 identical notebooks for $14.40. If each notebook costs the same amount, how much would 10 notebooks cost?​

  1. $18.00 (correct answer)
  2. $16.20
  3. $144.00
  4. $1.80
Explanation: This question tests solving unit rate problems by finding cost per notebook from total for several, then using it for a different quantity. Unit rate problems involve finding rate ($14.40 ÷ 8 = $1.80 per notebook), then multiplying (1.80 × 10 = $18.00). Constant pricing means same cost per item. For example, similar to $6 for 3 lb at $2/lb, then $10 for 5 lb. The correct calculation divides then multiplies accurately. Common errors include multiplying totals directly like 14.40 × 10 = 144, or dividing wrong like 8 ÷ 14.40. To solve: (1) find unit rate (14.40 ÷ 8 = 1.80), (2) multiply by new quantity (1.80 × 10 = 18), (3) verify by proportion (14.40/8 = x/10, x = (14.40 × 10)/8 = 144/8 = 18), (4) check units are dollars.

Question 9

A car travels 180 miles in 3 hours at a constant speed. What is the car's speed in miles per hour (mph)?

  1. 54 mph
  2. 60 miles
  3. 60 mph (correct answer)
  4. 90 mph
Explanation: This question tests solving unit rate problems: finding the speed in miles per hour from distance and time, assuming constant speed. Unit rate problems involve finding the rate by dividing distance by time (180 ÷ 3 = 60 mph), interpreting constant rate as the same speed per hour. Speed is distance ÷ time for the rate (like 180 miles ÷ 3 hours = 60 mph), and this can be used further for predictions like distance or time. Example: For 180 miles in 3 hours, rate is 60 mph. The correct calculation is straightforward division with units (mph). Common errors include multiplying instead of dividing (180 × 3), inverting (3 ÷ 180), or omitting units. To solve: identify rate as distance ÷ time (180 ÷ 3 = 60), verify units (mph) and reasonableness (180 in 3 hours means 60 each hour). Speed formulas: rate = d/t, and check choices for matching calculation.

Question 10

A grocery store sells 6 apples for $3.00. If the apples all cost the same, what is the unit price per apple?

  1. $0.20 per apple
  2. $0.50 per apple (correct answer)
  3. $2.00 per apple
  4. $18.00 per apple
Explanation: This question tests solving unit rate problems: finding rate (cost per item, speed mph, work rate), using rate to predict (at constant rate, how much in given time/quantity), interpreting in contexts. Unit rate problems: (1) find rate (4 lawns in 7 hours gives rate 4/7 lawns per hour, dividing quantity by time), (2) use rate (at 4/7 lawns/hr, in 35 hours: multiply rate×time=(4/7)×35=20 lawns), (3) interpret (constant rate means same amount per unit throughout: 4/7 per hour maintained). Speed: distance÷time gives rate (180 miles÷3 hours=60 mph), use rate: speed×time=distance (60×5=300 miles in 5 hours). Unit pricing: total÷quantity gives per-unit cost (6÷3lb=6÷3 lb=2/lb), use: rate×quantity=total (2/lb×5lb=2/lb×5 lb=10). In this case, divide total cost by number of apples: 3.00÷6=3.00÷6=0.50 per apple, which is the unit price. A common error is multiplying instead of dividing (3×6=18,likechoiceD)orinverting(6÷3=18, like choice D) or inverting (6÷3=2, choice C). Solving: (1) find unit rate (total÷quantity: 3÷6=3÷6=0.50 per apple), (2) identify what's asked (price per apple), (3) apply rate (division gives unit cost), (4) verify units (dollars per apple) and reasonableness (6 apples at $0.50 each total $3, checks out).

Question 11

A lawn care team mows 4 lawns in 7 hours at a steady rate. If they keep working at that same rate, how many lawns can they mow in 35 hours?

  1. 5 lawns
  2. 20 lawns (correct answer)
  3. 39 lawns
  4. 140 lawns
Explanation: This question tests solving unit rate problems by finding a work rate from lawns and hours, then using it to predict lawns in more hours at constant rate. Unit rate problems involve (1) finding rate (4 lawns ÷ 7 hours = 4/7 per hour), (2) using rate ((4/7) × 35 = 20 lawns), (3) interpreting constant rate as same fraction per hour. For example, similar to 120 words in 5 minutes scaling to 288 in 12. The correct calculation simplifies (4 × 35)/7 = 140/7 = 20. Common errors include inverting to 7/4 × 35 ≈ 61.25, or adding 4 + 35 = 39. To solve: (1) find unit rate (4/7), (2) multiply by time (35), (3) verify (35 ÷ 7 = 5 intervals, 5 × 4 = 20 lawns), (4) ensure units are lawns.

Question 12

A recipe calls for 3 cups of flour to make 24 cookies. If you want to make 40 cookies, how many cups of flour will you need?

  1. 4 cups
  2. 4.5 cups
  3. 5 cups (correct answer)
  4. 6 cups
Explanation: This is a proportion problem where you need to find how ingredients scale up when making more of something. The key is recognizing that the ratio of flour to cookies stays constant - if you make more cookies, you need proportionally more flour. Start by finding the relationship: 3 cups of flour makes 24 cookies. You can set up a proportion: 3 cups24 cookies=x cups40 cookies\frac{3 \text{ cups}}{24 \text{ cookies}} = \frac{x \text{ cups}}{40 \text{ cookies}} Cross multiply: 3×40=24×x3 \times 40 = 24 \times x, so 120=24x120 = 24x. Dividing both sides by 24 gives you x=5x = 5 cups. Another way to think about it: First find how much flour you need per cookie. 3÷24=0.1253 \div 24 = 0.125 cups per cookie. Then multiply by 40 cookies: 0.125×40=50.125 \times 40 = 5 cups. Looking at the wrong answers: A) 4 cups would only make 32 cookies (since 4÷0.125=324 \div 0.125 = 32), which is too few. B) 4.5 cups would make 36 cookies, still not enough. D) 6 cups is what you might get if you incorrectly thought "40 is about twice 24, so double the flour," but 40 isn't twice 24 - it's 4024=53\frac{40}{24} = \frac{5}{3} times as much. The answer is C) 5 cups. Study tip: For proportion problems, always check if your answer makes sense by asking "does this give me the right ratio?" You can verify: 5 cups ÷ 40 cookies = 0.125 cups per cookie, which matches the original recipe.

Question 13

A runner jogs 9 miles in 1.5 hours at a constant speed. At this same speed, how many miles will the runner jog in 4 hours?

  1. 13.5 miles
  2. 6 miles
  3. 24 miles (correct answer)
  4. 36 miles
Explanation: This question tests solving unit rate problems: finding rate (cost per item, speed mph, work rate), using rate to predict (at constant rate, how much in given time/quantity), interpreting in contexts. Unit rate problems: (1) find rate (4 lawns in 7 hours gives rate 4/7 lawns per hour, dividing quantity by time), (2) use rate (at 4/7 lawns/hr, in 35 hours: multiply rate×time=(4/7)×35=20 lawns), (3) interpret (constant rate means same amount per unit throughout: 4/7 per hour maintained). Speed: distance÷time gives rate (180 miles÷3 hours=60 mph), use rate: speed×time=distance (60×5=300 miles in 5 hours). Unit pricing: total÷quantity gives per-unit cost (6÷3lb=6÷3 lb=2/lb), use: rate×quantity=total (2/lb×5lb=2/lb×5 lb=10). Here, find speed: 9 miles÷1.5 hours=6 mph, then predict for 4 hours: 6×4=24 miles. Errors include dividing time by distance (1.5÷9=0.166, not useful) or adding (9+1.5+4=14.5, close to B but wrong). Solving: (1) find unit rate (distance÷time=9÷1.5=6 mph), (2) identify what's asked (miles in 4 hours), (3) apply rate (multiply rate by time: 6×4=24), (4) verify units (miles) and reasonableness (in 1.5 hours 9 miles, so double time to 3 hours=18 miles, plus another hour at 6 mph=24, yes).

Question 14

A car travels 168 miles and uses 6 gallons of gas. At the same rate, how many gallons will be needed to travel 420 miles?

  1. 12 gallons
  2. 15 gallons (correct answer)
  3. 18 gallons
  4. 21 gallons
Explanation: First find the unit rate: 168 miles ÷ 6 gallons = 28 miles per gallon. Then divide the target distance by the rate: 420 miles ÷ 28 miles/gallon = 15 gallons. Choice A assumes 35 miles/gallon (using 7 gallons instead of 6). Choice C uses 168 ÷ 420 × 6 (incorrect proportion setup). Choice D assumes 20 miles/gallon.

Question 15

A train travels 240 kilometers in 3 hours. At this constant speed, how far will it travel in 7.5 hours?

  1. 560 kilometers
  2. 580 kilometers
  3. 600 kilometers (correct answer)
  4. 620 kilometers
Explanation: First find the unit rate: 240 km ÷ 3 hours = 80 km/hour. Then multiply by the new time: 80 km/hour × 7.5 hours = 600 km. Choice A uses 74.67 km/hour (calculation error). Choice B uses 77.33 km/hour. Choice D uses 82.67 km/hour.

Question 16

A grocery store sells 6 apples for $3.00. If the apples cost the same per apple, how much does 1 apple cost?

  1. $0.25 per apple
  2. $0.50 per apple (correct answer)
  3. $2.00 per apple
  4. $18.00 per apple
Explanation: This question tests solving unit rate problems by finding the cost per item, specifically the price per apple when given a total cost for multiple apples. Unit rate problems involve finding the rate by dividing the total cost by the quantity, such as $3.00 divided by 6 apples gives $0.50 per apple. You can use this rate to confirm consistency, but here it's directly asking for the unit rate. For example, if 6 apples cost $3.00, the rate is 3/6 = 0.5 dollars per apple. The correct calculation is dividing the total cost by the number of apples to get the unit price. A common error is multiplying instead of dividing, like 3 times 6 equaling $18, or inverting to 6/3 = $2. To solve: (1) identify the total and quantity, (2) divide total by quantity for the rate, (3) verify units are dollars per apple, and (4) check reasonableness, like 6 apples at $0.50 each totaling $3.00.

Question 17

A car travels 150 miles in 3 hours at a constant speed. What is the car's speed in miles per hour (mph)?

  1. 0.02 mph
  2. 50 mph (correct answer)
  3. 450 mph
  4. 2 mph
Explanation: This question tests solving unit rate problems: finding rate (cost per item, speed mph, work rate), using rate to predict (at constant rate, how much in given time/quantity), interpreting in contexts. Unit rate problems: (1) find rate (4 lawns in 7 hours gives rate 4/7 lawns per hour, dividing quantity by time), (2) use rate (at 4/7 lawns/hr, in 35 hours: multiply rate×time=(4/7)×35=20 lawns), (3) interpret (constant rate means same amount per unit throughout: 4/7 per hour maintained). Speed: distance÷time gives rate (180 miles÷3 hours=60 mph), use rate: speed×time=distance (60×5=300 miles in 5 hours). Unit pricing: total÷quantity gives per-unit cost (6÷3lb=6÷3 lb=2/lb), use: rate×quantity=total (2/lb×5lb=2/lb×5 lb=10). Calculate speed: 150 miles÷3 hours=50 mph. Error like multiplying (150×3=450, choice B) or inverting wrongly (3÷150=0.02, choice D). Solving: (1) find unit rate (distance÷time=150÷3=50 mph), (2) identify what's asked (speed in mph), (3) apply rate (division gives rate), (4) verify units (mph) and reasonableness (50 mph for 3 hours covers 150 miles, yes).

Question 18

A water bottle holds 2.5 liters of water. If a hiker drinks water at a constant rate of 0.5 liters per hour, how long will the bottle last?

  1. 3 hours
  2. 5 hours (correct answer)
  3. 2 hours
  4. 1.25 hours
Explanation: This question tests solving unit rate problems: finding rate (cost per item, speed mph, work rate), using rate to predict (at constant rate, how much in given time/quantity), interpreting in contexts. Unit rate problems: (1) find rate (4 lawns in 7 hours gives rate 4/7 lawns per hour, dividing quantity by time), (2) use rate (at 4/7 lawns/hr, in 35 hours: multiply rate×time=(4/7)×35=20 lawns), (3) interpret (constant rate means same amount per unit throughout: 4/7 per hour maintained). Speed: distance÷time gives rate (180 miles÷3 hours=60 mph), use rate: speed×time=distance (60×5=300 miles in 5 hours). Unit pricing: total÷quantity gives per-unit cost (6÷3lb=6÷3 lb=2/lb), use: rate×quantity=total (2/lb×5lb=2/lb×5 lb=10). Time: total water÷rate=2.5÷0.5=5 hours. Common error: multiplying (2.5×0.5=1.25, choice A) instead of dividing. Solving: (1) find unit rate (given 0.5 liters per hour), (2) identify what's asked (time for 2.5 liters), (3) apply rate (divide total by rate: 2.5/0.5=5), (4) verify units (hours) and reasonableness (0.5 per hour, 2.5 liters: 5 times 0.5=2.5, yes).

Question 19

A bike rider travels 24 miles at a constant speed of 12 miles per hour. How long does the trip take?​

  1. 12 hours
  2. 288 hours
  3. 0.5 hour
  4. 2 hours (correct answer)
Explanation: This question tests solving unit rate problems by using a given speed to find the time for a distance, rearranging distance = rate × time to time = distance ÷ rate. Unit rate problems include finding time as 24 miles ÷ 12 mph = 2 hours. Constant speed means same rate throughout. For example, similar to finding time for 300 miles at 60 mph: 300 ÷ 60 = 5 hours. The correct approach divides distance by rate. Common mistakes include multiplying 24 × 12 = 288, or inverting to 12 ÷ 24 = 0.5. To solve: (1) identify formula t = d/r, (2) plug in 24 ÷ 12 = 2, (3) verify units are hours, (4) check by multiplying back (12 × 2 = 24 miles).

Question 20

A bike rider travels 24 miles at a constant speed of 12 miles per hour. How long does the trip take?

  1. 2 hours (correct answer)
  2. 12 hours
  3. 0.5 hour
  4. 288 hours
Explanation: This question tests solving unit rate problems by using a given speed to find the time for a distance, rearranging distance=rate×timedistance = rate \times time to time=distance÷ratetime = distance \div rate. Unit rate problems include finding time as 24÷12=224 \div 12 = 2 hours. Constant speed means same rate throughout. For example, similar to finding time for 300 miles at 60 mph: 300÷60=5300 \div 60 = 5 hours. The correct approach divides distance by rate. Common mistakes include multiplying 24×12=28824 \times 12 = 288, or inverting to 12÷24=0.512 \div 24 = 0.5. To solve: (1) identify formula t=d/rt = d/r, (2) plug in 24÷12=224 \div 12 = 2, (3) verify units are hours, (4) check by multiplying back (12×2=2412 \times 2 = 24 miles).