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.

This question tests solving unit rate problems: finding the flour rate per muffin, then using that to predict flour for more muffins, assuming constant recipe rate. Unit rate problems involve finding the rate by dividing flour by muffins (3 ÷ 12 = 0.25 cups per muffin), then multiplying by new quantity (0.25 × 20 = 5 cups), interpreting constant rate as same flour per muffin. For recipes, total ÷ quantity gives per-unit, and rate × new quantity = new total, like pricing or work rates. Example: 3 cups for 12 muffins is 0.25 per muffin; for 20, 0.25 × 20 = 5 cups; or proportion: (3/12) × 20 = 5. The correct calculation uses division and multiplication. Common errors include multiplying originals without rate (3 × 20 = 60), or wrong fraction (12/3 × 20 = 80). To solve: find rate (3/12 = 0.25), apply to 20 (0.25 × 20 = 5), verify (20 is about 1.67 times 12, 3 × 1.67 ≈ 5). Mistakes: inverted rate or wrong operations.

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.

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).

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).

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.

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, unit price is $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).

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.

This question tests solving unit rate problems: finding the typing rate in words per minute, then using that rate to predict words for a different time, assuming constant rate. Unit rate problems involve finding the rate by dividing words by time (120 ÷ 5 = 24 words per minute), then using the rate to find words for new time by multiplying (24 × 12 = 288 words), interpreting constant rate as the same words per minute throughout. For work rates like typing, total ÷ time gives rate, and rate × new time gives new total, similar to speed or pricing. Example: For 120 words in 5 minutes, rate is 24 per minute; for 12 minutes, 24 × 12 = 288 words. The correct calculation ensures proper division for rate and multiplication for prediction. Common errors include adding instead of multiplying (120 + 5 + 12), not calculating rate, or wrong arithmetic (120 ÷ 5 = 20). To solve: find unit rate (120 ÷ 5 = 24), identify what's asked (words in 12 minutes), apply rate (24 × 12 = 288), verify reasonableness (12 is 2.4 times 5, 120 × 2.4 = 288). Mistakes: inverted rate, wrong units, or reversed operations.

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.