- A cyclist rides at a constant speed of 12 miles per hour. How long will it take to ride 30 miles?
Answer: 2.5 hours
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÷3=60 mph), use rate: speed×time=distance (60×5=300 miles in 5 hours). Unit pricing: total÷quantity gives per-unit cost (6÷3 lb=2/lb), use: rate×quantity=total (2/lb \times 5$ lb=10). Time: distance÷speed=30÷12=2.5 hours. Error: multiplying (30×12=360, choice D) or inverting wrongly (12÷30=0.4, not matching). Solving: (1) find unit rate (given 12 mph), (2) identify what's asked (time for 30 miles), (3) apply rate (divide distance by rate: 30/12=2.5), (4) verify units (hours) and reasonableness (12 mph for 2 hours=24 miles, plus 0.5 hour=6 more, total 30, yes). - A grocery store sells 6 apples for $3.00. If the apples cost the same per apple, how much does 1 apple cost?
Answer: $0.50 per apple
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.
- A grocery store sells 6 apples for $3.00. If the apples all cost the same, what is the unit price per apple?
Answer: $0.50 per apple
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÷3 lb=$2/lb), use: rate×quantity=total ($2/lb×5 lb=$10). In this case, divide total cost by number of apples: $3.00÷6=$0.50 per apple, which is the unit price. A common error is multiplying instead of dividing (3×6=$18, like choice D) or inverting (6÷3=$2, choice C). Solving: (1) find unit rate (total÷quantity: $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).
- Jake earns $126 for working 18 hours. At this rate, how much will he earn for working 25 hours?
Answer: $175.00
When you see a problem asking "at this rate," you're dealing with a unit rate problem. You need to find how much Jake earns per hour, then use that rate to calculate earnings for a different number of hours.
First, find Jake's hourly rate by dividing his total earnings by the hours worked: $126 ÷ 18 = $7 per hour. Now multiply this rate by 25 hours: $7 × 25 = $175.
Let's examine why the other answers are incorrect. Choice A ($168.50) gives you $6.74 per hour, which doesn't match our calculation. This might result from rounding errors or miscalculating the division. Choice B ($189.75) suggests an hourly rate of $7.59, which is too high and could come from incorrectly adding rather than using pure multiplication. Choice C ($182.25) represents $7.29 per hour, another calculation error that might occur if you mishandle the decimal placement when dividing $126 by 18.
The correct answer is D ($175.00), which gives us exactly $7 per hour when we work backwards: $175 ÷ 25 = $7, and $7 × 18 = $126, confirming our original information.
Remember this two-step strategy for rate problems: first find the unit rate (divide to get "per one unit"), then multiply by the new quantity. Always check your work by seeing if your unit rate makes sense when applied back to the original situation.
- A grocery store sells 6 apples for $3.00. If the apples cost the same per apple, how much does 1 apple cost?
Answer: $0.50 per apple
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.
- A cyclist rides at a constant speed of 15 miles per hour. How long will it take the cyclist to ride 42 miles?
Answer: 2.8 hours
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.
- Which is the better deal (lower cost per ounce)?
- Option 1: 12 ounces of trail mix for $4.20
- Option 2: 18 ounces of trail mix for $5.94
Answer: Option 2, because it costs $0.33 per ounce
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÷3 lb=$2/lb), use: rate×quantity=total ($2/lb×5 lb=$10). Option 1: $4.20÷12= $0.35/oz; Option 2: $5.94÷18=0.33/oz,soOption2better(lowerrate).Mistake:calculatingtotalinsteadofperounce(4.20vs5.94,ignoringsize)orinverting(12÷4.20≈2.86oz/, not helpful). Solving: (1) find unit rates ($/oz for each), (2) identify better deal (lower /oz),(3)compare(0.33<0.35),(4)verifyunits( per oz) and reasonableness (18 oz for $5.94 is like 12 oz for (5.94/18)*12≈$3.96, cheaper than $4.20). - 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?
Answer: 24 miles
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÷3 lb=$2/lb), use: rate×quantity=total ($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).
- A factory produces 240 widgets in 8 hours. At this rate, how many complete widgets can be produced in a 5-day work week if the factory operates 6 hours per day?
Answer: 900 widgets
First find the unit rate: 240 widgets ÷ 8 hours = 30 widgets per hour. Then calculate total work hours: 5 days × 6 hours/day = 30 hours. Finally: 30 widgets/hour × 30 hours = 900 widgets. Choice B uses 8 hours/day instead of 6. Choice C uses 4 days instead of 5. Choice D uses only 1 day of work.
- 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?
Answer: 5 hours
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÷3 lb=$2/lb), use: rate×quantity=total ($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).
- A snack mix recipe uses 3 cups of pretzels for every 5 cups of cereal. If you use 20 cups of cereal, how many cups of pretzels are needed to keep the same ratio?
Answer: 12 cups
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÷3 lb=$2/lb), use: rate×quantity=total ($2/lb×5 lb=$10). Ratio pretzels to cereal 3:5, rate 3/5 pretzels per cereal cup, for 20 cereal: (3/5)×20=12 pretzels. Common mistake: inverting ratio (5/3×20≈33.3, not an option) or adding (3+5+20=28, no). Solving: (1) find unit rate (pretzels÷cereal=3÷5=0.6 per cup), (2) identify what's asked (pretzels for 20 cereal), (3) apply rate (0.6×20=12), (4) verify units (cups) and reasonableness (5 cereal need 3 pretzels, 20 is 4 times, 3×4=12, yes).
- A runner travels 9 miles in 1.5 hours at a constant speed. At this speed, how far will the runner travel in 4 hours?
Answer: 24 miles
This question tests solving unit rate problems: finding the speed in miles per hour, then using that rate to predict distance for a different time, assuming constant speed. Unit rate problems involve finding the rate by dividing distance by time (9 ÷ 1.5 = 6 mph), then using the rate to find distance for new time by multiplying (6 × 4 = 24 miles), interpreting constant rate as the same speed throughout. Speed is distance ÷ time for the rate (like 180 miles ÷ 3 hours = 60 mph), and using rate × time = distance (60 × 5 = 300 miles). Example: For 9 miles in 1.5 hours, rate is 6 mph; for 4 hours, 6 × 4 = 24 miles. The correct calculation is accurate rate division and multiplication for prediction. Common errors include inverting the rate (time ÷ distance), wrong operation like subtraction, or arithmetic errors like 9 ÷ 1.5 = 5. To solve: find unit rate (9 ÷ 1.5 = 6 mph), identify what's asked (distance in 4 hours), apply rate (6 × 4 = 24), verify units (miles) and reasonableness (4 is about 2.67 times 1.5, 9 × 2.67 ≈ 24). Speed formulas: distance = rate × time, time = distance ÷ rate, rate = distance ÷ time.
- A school club can pack 15 care kits in 6 hours at a constant rate. If the club works at the same rate, how many care kits can they pack in 18 hours?
Answer: 45 kits
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÷3 lb=$2/lb), use: rate×quantity=total ($2/lb×5 lb=$10). Rate: 15 kits÷6 hours=2.5 per hour, for 18 hours: 2.5×18=45 kits. Error: dividing hours by kits (6÷15=0.4, then 0.4×18=7.2, not matching) or multiplying directly (15×18=270, too big). Solving: (1) find unit rate (kits÷hours=15÷6=2.5 per hour), (2) identify what's asked (kits in 18 hours), (3) apply rate (multiply by 18: 45), (4) verify units (kits) and reasonableness (6 hours=15, 18 is 3 times, 15×3=45, yes).
- A bike rider travels 24 miles at a constant speed of 12 miles per hour. How long does the trip take?
Answer: 2 hours
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÷12=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). - A car travels 180 miles in 3 hours at a constant speed. What is the car's speed in miles per hour (mph)?
Answer: 60 mph
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.
- 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?
Answer: 5 cups
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 muffin
Now multiply by the 20 muffins you want:
0.25×20=5 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! - 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?
Answer: 20 lawns
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.
- A student types 120 words in 5 minutes at a steady rate. At this rate, how many words will the student type in 12 minutes?
Answer: 288 words
This question tests solving unit rate problems by finding a typing rate in words per minute and using it to predict words typed in a different time at a constant rate. Unit rate problems involve (1) finding the rate by dividing words by minutes, like 120 words in 5 minutes gives 24 words per minute, (2) using the rate by multiplying by the new time, such as 24 times 12 = 288 words. Interpreting constant rate means the same words per minute throughout. For example, at 24 words per minute, in 12 minutes it's 288 words. The correct calculation applies the rate multiplication accurately. Common errors include adding instead of multiplying, like 120 + 5 + 12, or inverting the rate to 5/120. To solve: (1) find unit rate (120 ÷ 5 = 24), (2) identify the new time (12 minutes), (3) multiply rate by time (24 × 12 = 288), (4) verify reasonableness, like double the time from 5 to 10 would be 240, plus 2 more minutes at 24 each is 48, totaling 288.
- A copying machine makes 84 copies in 6 minutes. At this rate, how many copies can it make in 45 minutes?
Answer: 630 copies
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.
- A car travels 150 miles in 3 hours at a constant speed. What is the car's speed in miles per hour (mph)?
Answer: 50 mph
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÷3 lb=$2/lb), use: rate×quantity=total ($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).
