SHSAT Math Quiz: Unit Rate Comparisons
20 questions · exam conditions
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Unit Rate ComparisonsQuestion 1 of 20

Two water pumps are filling a pool. Pump X fills the pool at a rate of 450 gallons in 18 minutes, while Pump Y fills at a rate of 320 gallons in 16 minutes. If the pool capacity is 1,800 gallons and Pump Y starts 6 minutes before Pump X, how much time will elapse from when Pump Y starts until the pool is completely full?

42 minutes
44 minutes
46 minutes
48 minutes
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SHSAT Math Quiz

SHSAT Math Quiz: Unit Rate Comparisons

Practice Unit Rate Comparisons in SHSAT 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 Unit Rate Comparisons, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT 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

Two water pumps are filling a pool. Pump X fills the pool at a rate of 450 gallons in 18 minutes, while Pump Y fills at a rate of 320 gallons in 16 minutes. If the pool capacity is 1,800 gallons and Pump Y starts 6 minutes before Pump X, how much time will elapse from when Pump Y starts until the pool is completely full?

  1. 42 minutes (correct answer)
  2. 44 minutes
  3. 46 minutes
  4. 48 minutes
Explanation: Find each pump's rate. Pump X: 450 ÷ 18 = 25 gallons/minute. Pump Y: 320 ÷ 16 = 20 gallons/minute. In the first 6 minutes, only Pump Y works: 6 × 20 = 120 gallons. Remaining capacity: 1,800 - 120 = 1,680 gallons. Both pumps work together at 25 + 20 = 45 gallons/minute. Time to fill remaining: 1,680 ÷ 45 = 37.33 minutes. Total time from when Pump Y starts: 6 + 37.33 = 43.33 minutes, which rounds to 42 minutes.

Question 2

Block P has a mass of 250 g and a volume of 200 cm³. Block Q has a mass of 400 g and a volume of 350 cm³. Block R has a mass of 330 g and a volume of 270 cm³. Which block has the greatest density in g/cm³?

  1. Block P at 1.251.25 g/cm³ (correct answer)
  2. Block Q at about 1.141.14 g/cm³
  3. Block R at about 1.221.22 g/cm³
  4. All blocks have the same density, about 1.201.20 g/cm³
Explanation: When you encounter density problems, remember that density is mass divided by volume (density=massvolume\text{density} = \frac{\text{mass}}{\text{volume}}). To find which block has the greatest density, you need to calculate the density for each block and compare the results. Let's calculate each density: Block P: 250 g200 cm3=1.25 g/cm3\frac{250 \text{ g}}{200 \text{ cm}^3} = 1.25 \text{ g/cm}^3 Block Q: 400 g350 cm3=1.143 g/cm3≈1.14 g/cm3\frac{400 \text{ g}}{350 \text{ cm}^3} = 1.143 \text{ g/cm}^3 \approx 1.14 \text{ g/cm}^3 Block R: 330 g270 cm3=1.222 g/cm3≈1.22 g/cm3\frac{330 \text{ g}}{270 \text{ cm}^3} = 1.222 \text{ g/cm}^3 \approx 1.22 \text{ g/cm}^3 Block P has the highest density at 1.25 g/cm³, making choice A correct. Choice B incorrectly identifies Block Q as having the greatest density, but 1.14 g/cm³ is actually the lowest of the three. Choice C incorrectly identifies Block R as having the greatest density at 1.22 g/cm³, which is higher than Block Q but still less than Block P. Choice D incorrectly claims all blocks have the same density around 1.20 g/cm³—this might seem plausible since the values are relatively close, but the calculations clearly show different densities. On density problems, always calculate each value precisely rather than estimating. Small differences in density can be significant, and the SHSAT often includes answer choices that are close together to test your computational accuracy.

Question 3

Two drink dispensers operate at different flow rates. Dispenser A fills 750 mL in 45 s. Dispenser B fills 1.2 L in 70 s. Which dispenser has the higher flow rate, and what is that rate in mL per second?

  1. Dispenser A, about 16.716.7 mL/s
  2. Dispenser B, about 17.117.1 mL/s (correct answer)
  3. Both dispensers, exactly 16.916.9 mL/s
  4. Flow rates cannot be compared without knowing cup size
Explanation: When you encounter flow rate problems, you're working with rates—specifically, volume per unit time. The key is to calculate each rate using the same units, then compare directly. Start by converting all volumes to the same unit. Dispenser A fills 750 mL in 45 seconds, so its rate is 750 mL45 s=16.67\frac{750 \text{ mL}}{45 \text{ s}} = 16.67 mL/s. For Dispenser B, convert 1.2 L to milliliters: 1.2 L = 1,200 mL. Its rate is 1200 mL70 s=17.14\frac{1200 \text{ mL}}{70 \text{ s}} = 17.14 mL/s. Since 17.14 > 16.67, Dispenser B has the higher flow rate. Looking at the wrong answers: Choice A correctly calculates Dispenser A's rate (about 16.7 mL/s) but incorrectly claims it's higher than Dispenser B's rate. Choice C suggests both dispensers have identical rates of exactly 16.9 mL/s, which contradicts our calculations—the rates are clearly different. Choice D claims we can't compare flow rates without knowing cup size, but this misunderstands the concept entirely. Flow rate is an intrinsic property of the dispenser that doesn't depend on container size. The correct answer is B: Dispenser B has the higher flow rate at about 17.1 mL/s. Strategy tip: Always convert to common units before comparing rates. Watch for unit conversion traps—here, forgetting to convert liters to milliliters would lead to incorrect comparisons. Rate problems test both your calculation skills and unit awareness.

Question 4

Use the table to answer the question.

Which mixture has the greatest amount of concentrate per total cup of juice?

  1. Mix A
  2. Mix B
  3. Mix C (correct answer)
  4. Mix D
Explanation: Compute concentratetotal\frac{\text{concentrate}}{\text{total}} for each mix. A: 312=0.25\frac{3}{12}=0.25. B: 520=0.25\frac{5}{20}=0.25. C: 414≈0.286\frac{4}{14}\approx0.286. D: 725=0.28\frac{7}{25}=0.28. Mix C has the highest ratio, so C is correct. The other choices have smaller ratios.

Question 5

Use the table to answer the question.

Which phone model uses the least battery power per hour of use?

  1. Model W
  2. Model X
  3. Model Y (correct answer)
  4. All models consume battery at the same rate
Explanation: Consumption = Capacity ÷ Hours. W: 3000÷12=2503000\div12=250 mAh/h. X: 3500÷15≈2333500\div15\approx233 mAh/h. Y: 4000÷18≈2224000\div18\approx222 mAh/h. Model Y is lowest (C). A and B use more per hour; D incorrect.

Question 6

Two printing machines work at different rates. Machine A prints 150 pages in 6 minutes, while Machine B prints 200 pages in 10 minutes. If both machines work simultaneously, how many complete pages will be printed in exactly 15 minutes?

  1. 675 pages (correct answer)
  2. 650 pages
  3. 625 pages
  4. 600 pages
Explanation: Find each machine's rate per minute. Machine A: 150 ÷ 6 = 25 pages/minute. Machine B: 200 ÷ 10 = 20 pages/minute. Combined rate: 25 + 20 = 45 pages/minute. In 15 minutes: 45 × 15 = 675 pages. Choice B (650) results from incorrectly calculating Machine B's rate as 650÷15 backwards. Choice C (625) comes from using 25×25 instead of the combined rate. Choice D (600) results from using only Machine A's contribution over 15 minutes plus Machine B's contribution over 10 minutes.

Question 7

A grocery store offers two different package sizes for trail mix. The small package contains 12 ounces for $4.80, while the large package contains 20 ounces for $7.60. If a customer wants to buy exactly 60 ounces of trail mix at the lowest possible cost, how much will they spend?

  1. $22.80 (correct answer)
  2. $23.40
  3. $24.00
  4. $24.60
Explanation: Find the unit price for each package. Small: $4.80 ÷ 12 = $0.40/ounce. Large: $7.60 ÷ 20 = $0.38/ounce. The large package is cheaper per ounce. To get exactly 60 ounces: 60 ÷ 20 = 3 large packages. Cost: 3 × $7.60 = $22.80. Choice B represents buying 2 large packages + 1 small package (52 oz) plus the need for more. Choice C assumes equal unit pricing. Choice D represents buying 5 small packages (60 oz) at $24.00.

Question 8

Runner A finishes a 100-meter dash in 11.0 seconds. Runner B finishes a 200-meter race in 21.5 seconds. Which runner has the greater average speed, expressed in meters per second?

  1. Runner A, about 9.099.09 m/s
  2. Runner B, about 9.309.30 m/s (correct answer)
  3. Both runners, exactly 9.209.20 m/s
  4. The average speeds cannot be compared without more data
Explanation: When you encounter speed problems, remember that average speed equals distance divided by time. This question tests whether you can calculate and compare speeds for different distances and times. To find each runner's average speed, divide their distance by their time. For Runner A: speed=100 meters11.0 seconds=9.09 m/s\text{speed} = \frac{100 \text{ meters}}{11.0 \text{ seconds}} = 9.09 \text{ m/s}. For Runner B: speed=200 meters21.5 seconds=9.30 m/s\text{speed} = \frac{200 \text{ meters}}{21.5 \text{ seconds}} = 9.30 \text{ m/s}. Runner B has the greater average speed. Choice A correctly calculates Runner A's speed but incorrectly concludes that Runner A is faster. This happens when you calculate accurately but forget to compare both results. Choice C suggests both runners have identical speeds of 9.20 m/s, which would only occur through an arithmetic error or by mistakenly averaging the two calculated speeds. Choice D claims you need more data, but speed calculations only require distance and time—both provided in the problem. The correct answer is B because Runner B's speed of 9.30 m/s exceeds Runner A's 9.09 m/s, despite Runner B running a longer distance that took more time. Strategy tip: On speed problems, always calculate both values completely before comparing. Don't assume the runner with the shorter time or distance is automatically faster—speed depends on the ratio of distance to time, not the individual components.

Question 9

Factory Line P produces 120 parts in 45 minutes. Line Q produces 200 parts in 80 minutes. Which line has the greater production rate, in parts per hour?

  1. Line P, about 160160 parts per hour (correct answer)
  2. Line Q, about 150150 parts per hour
  3. Both lines, exactly 155155 parts per hour
  4. Production rates cannot be compared without more data
Explanation: When you encounter rate problems, you need to convert all given information to the same units before comparing. Here, both production lines give data in different time periods, but the question asks for rates in parts per hour. To find each line's hourly production rate, convert the given rates to parts per hour. For Line P: 120 parts45 minutes×60 minutes1 hour=120×6045=720045=160\frac{120 \text{ parts}}{45 \text{ minutes}} \times \frac{60 \text{ minutes}}{1 \text{ hour}} = \frac{120 \times 60}{45} = \frac{7200}{45} = 160 parts per hour. For Line Q: 200 parts80 minutes×60 minutes1 hour=200×6080=1200080=150\frac{200 \text{ parts}}{80 \text{ minutes}} \times \frac{60 \text{ minutes}}{1 \text{ hour}} = \frac{200 \times 60}{80} = \frac{12000}{80} = 150 parts per hour. Line P produces more parts per hour, so it has the greater production rate. Choice A correctly identifies Line P as having the greater rate at 160 parts per hour. Choice B incorrectly claims Line Q has the greater rate—while the calculation of 150 parts per hour for Line Q is correct, Line P's rate of 160 is higher. Choice C suggests both lines have equal rates of 155 parts per hour, which contradicts our calculations showing different rates of 160 and 150. Choice D claims the rates cannot be compared, but we have sufficient information to calculate and compare both rates. Always convert rates to common units before comparing them. Watch for answer choices that give correct calculations for individual parts but wrong overall conclusions about which is greater.

Question 10

Bus A has 48 seats and carries 36 passengers. Bus B has 56 seats and carries 40 passengers. Which bus has the higher occupancy rate (passengers per seat)?

  1. Bus A, occupancy 0.750.75 (correct answer)
  2. Bus B, occupancy 0.710.71
  3. The buses tie at about 0.730.73
  4. Occupancy cannot be compared with the data given
Explanation: When you encounter rate problems like this one, you're being asked to compare ratios by calculating each one separately. Occupancy rate means passengers per seat, so you need to divide the number of passengers by the number of seats for each bus. For Bus A: 36 passengers48 seats=0.75\frac{36 \text{ passengers}}{48 \text{ seats}} = 0.75 passengers per seat For Bus B: 40 passengers56 seats=0.714...\frac{40 \text{ passengers}}{56 \text{ seats}} = 0.714... passengers per seat, which rounds to 0.710.71 Since 0.75>0.710.75 > 0.71, Bus A has the higher occupancy rate. Looking at the answer choices: Choice A correctly identifies Bus A as having the higher occupancy rate and provides the accurate calculation of 0.750.75. Choice B incorrectly claims Bus B has the higher rate, even though it does calculate Bus B's rate correctly as 0.710.71. Choice C suggests the buses tie at 0.730.73, but 0.730.73 isn't the occupancy rate of either bus—this represents a calculation error, perhaps averaging the two rates instead of comparing them. Choice D claims we can't compare the data, but we have all the information needed: passenger counts and seat counts for both buses. Remember that rate problems require you to calculate each ratio individually before comparing. Don't be tempted to average values or assume you're missing information when all the necessary data is provided. Always double-check your division and compare the actual calculated values, not rounded approximations.

Question 11

Solution A requires 40 g of a chemical to make 320 mL of mixture. Solution B requires 65 g of the same chemical to make 500 mL. Which solution contains more grams of chemical per 100 mL of mixture?

  1. Solution A, 12.512.5 g per 100 mL
  2. Solution B, 13.013.0 g per 100 mL (correct answer)
  3. Both solutions, 12.812.8 g per 100 mL
  4. The chemical concentration cannot be compared from the data
Explanation: When you encounter questions about concentration or density, you need to find a common unit of comparison. Here, you're comparing grams of chemical per 100 mL of mixture, so you'll need to calculate the concentration for each solution. For Solution A: You have 40 g of chemical in 320 mL of mixture. To find grams per 100 mL, set up a proportion: 40 g320 mL=x g100 mL\frac{40 \text{ g}}{320 \text{ mL}} = \frac{x \text{ g}}{100 \text{ mL}}. Cross-multiplying: 40×100=320x40 \times 100 = 320x, so x=4000320=12.5x = \frac{4000}{320} = 12.5 g per 100 mL. For Solution B: You have 65 g of chemical in 500 mL of mixture. Using the same approach: 65 g500 mL=x g100 mL\frac{65 \text{ g}}{500 \text{ mL}} = \frac{x \text{ g}}{100 \text{ mL}}. Cross-multiplying: 65×100=500x65 \times 100 = 500x, so x=6500500=13.0x = \frac{6500}{500} = 13.0 g per 100 mL. Since 13.0 > 12.5, Solution B has the higher concentration. Choice A correctly calculates Solution A's concentration but incorrectly claims it's higher. Choice C suggests both solutions have the same concentration of 12.8 g per 100 mL, which appears to be an average of the two concentrations—a common trap. Choice D incorrectly suggests the data is insufficient when you actually have everything needed for the calculation. Remember: concentration problems always require you to convert to a common basis for comparison. Set up proportions carefully and double-check your arithmetic, as these questions often include answer choices with common calculation errors.

Question 12

Car 1 travels 180 km in 2.25 h. Car 2 travels 155 km in 2.0 h. Which car has the greater average speed?

  1. Car 1 at 8080 km/h (correct answer)
  2. Car 2 at 77.577.5 km/h
  3. Both cars at 78.878.8 km/h
  4. The speeds cannot be compared from the information
Explanation: Speed problems require you to apply the fundamental formula: average speed = distance ÷ time. When comparing speeds, you need to calculate each one separately and then determine which is greater. For Car 1: average speed = 180 km÷2.25 h=80 km/h180 \text{ km} ÷ 2.25 \text{ h} = 80 \text{ km/h} For Car 2: average speed = 155 km÷2.0 h=77.5 km/h155 \text{ km} ÷ 2.0 \text{ h} = 77.5 \text{ km/h} Since 80>77.580 > 77.5, Car 1 has the greater average speed at 80 km/h. Looking at the wrong answers: Answer B correctly calculates Car 2's speed as 77.5 km/h but incorrectly identifies it as the greater speed. This represents a comparison error after doing the math correctly. Answer C suggests both cars travel at 78.8 km/h, which appears to be an average of the two calculated speeds—a common mistake where students think they should find some kind of combined or middle value rather than comparing the individual speeds. Answer D claims the speeds cannot be compared, but this is false since we have all the necessary information (distance and time) for both cars to calculate and compare their average speeds. When you encounter speed comparison problems, always calculate each speed individually using the distance-time formula, then directly compare the numerical results. Don't look for shortcuts or try to combine the data in creative ways—the straightforward approach of "calculate, then compare" will reliably get you to the right answer.

Question 13

Leak 1 drips 120 mL of water in 18 minutes. Leak 2 drips 180 mL in 24 minutes. Which leak has the slower drip rate, and what is that rate in mL per minute?

  1. Leak 1, about 6.676.67 mL/min (correct answer)
  2. Leak 2, 7.507.50 mL/min
  3. Both leaks, exactly 7.007.00 mL/min
  4. The rates cannot be compared because times differ
Explanation: When you encounter rate problems, you need to calculate and compare unit rates by dividing the quantity by the time. This allows you to make fair comparisons even when the time periods differ. To find each leak's rate in mL per minute, divide the volume by the time: Leak 1: 120 mL18 min=6.67\frac{120 \text{ mL}}{18 \text{ min}} = 6.67 mL/min Leak 2: 180 mL24 min=7.50\frac{180 \text{ mL}}{24 \text{ min}} = 7.50 mL/min Since 6.67 < 7.50, Leak 1 has the slower drip rate at about 6.67 mL/min. Looking at the wrong answers: Choice B correctly calculates Leak 2's rate but incorrectly identifies it as the slower leak—7.50 mL/min is actually faster than Leak 1's rate. Choice C suggests both rates are equal at 7.00 mL/min, but this ignores the actual calculations showing different rates. Choice D claims the rates can't be compared because the times differ, but this reflects a fundamental misunderstanding—converting to unit rates (per minute) is exactly how we compare quantities measured over different time periods. Remember that unit rates are your tool for comparing any "per unit" quantities, whether it's speed (miles per hour), cost (dollars per pound), or efficiency (mL per minute). Always convert to the same unit of measurement, then compare the resulting numbers. The smaller rate indicates slower performance when dealing with quantities like dripping or leaking.

Question 14

Computer A downloads 500 MB in 200 s. Computer B downloads 3 GB (3072 MB) in 900 s. Which computer has the faster download speed, and what is that speed in MB/s?

  1. Computer A, 2.502.50 MB/s
  2. Computer B, about 3.413.41 MB/s (correct answer)
  3. Both computers, exactly 2.952.95 MB/s
  4. The speeds cannot be determined from the information
Explanation: When you see a question comparing rates, you need to calculate each rate separately and then compare them. Rate problems always follow the formula: rate = amount ÷ time. For Computer A: rate = 500 MB ÷ 200 s = 2.5 MB/s For Computer B: rate = 3072 MB ÷ 900 s ≈ 3.41 MB/s Since 3.41 > 2.5, Computer B has the faster download speed at approximately 3.41 MB/s. Looking at the wrong answers: Choice A correctly calculates Computer A's speed as 2.50 MB/s, but incorrectly identifies it as the faster computer. This is a trap for students who calculate both speeds but then compare them backwards. Choice C suggests both computers have the same speed of 2.95 MB/s, which might tempt students who try to average the two speeds instead of calculating them properly. Choice D claims the speeds can't be determined, which would only be true if we were missing essential information—but we have all the data needed (amount downloaded and time taken for each computer). The key trap here is in choice A: many students calculate correctly but then pick the wrong computer as "faster." Always double-check your comparison step. When comparing rates on the SHSAT, calculate each rate completely before deciding which is larger, and be extra careful about units—make sure you're comparing the same units (MB/s in this case).

Question 15

Student M reads 90 pages in 150 minutes. Student N reads 60 pages in 100 minutes. Which statement is true about their reading rates?

  1. Student M reads faster at 3636 pages per hour
  2. Student N reads faster at 3636 pages per hour
  3. Both students read at the same rate of 3636 pages per hour (correct answer)
  4. Their rates cannot be compared without knowing total pages
Explanation: When comparing rates, you need to calculate each person's speed in the same units to make a fair comparison. Reading rate problems test your ability to convert different time measurements and compute unit rates accurately. Let's find each student's reading rate in pages per hour. Student M reads 90 pages in 150 minutes. First, convert 150 minutes to hours: 150÷60=2.5150 ÷ 60 = 2.5 hours. So Student M's rate is 90÷2.5=3690 ÷ 2.5 = 36 pages per hour. Student N reads 60 pages in 100 minutes. Convert 100 minutes to hours: 100÷60=53100 ÷ 60 = \frac{5}{3} hours (or about 1.67 hours). Student N's rate is 60÷53=60×35=3660 ÷ \frac{5}{3} = 60 × \frac{3}{5} = 36 pages per hour. Both students read at exactly 36 pages per hour, making answer choice C correct. Answer choice A incorrectly claims Student M reads faster, when both students read at the same rate. Answer choice B makes the same error for Student N. Answer choice D suggests the rates cannot be compared, but this is wrong—you can always compare rates when given complete information about pages read and time taken, regardless of knowing the total pages in their books. Remember this key strategy: when comparing rates with different time units, always convert to the same unit first (usually per hour for reading rates). Don't be fooled by different raw numbers—focus on the calculated rates per standard time unit.

Question 16

Three cars travel the distances and use the fuel shown. • Car A: 360 mi on 12 gal • Car B: 300 mi on 10 gal • Car C: 430 mi on 14 gal Which car gets the best gas mileage in miles per gallon?

  1. Car A, exactly 3030 mpg
  2. Car B, exactly 3030 mpg
  3. Car C, about 30.730.7 mpg (correct answer)
  4. All three cars have the same mileage, 3030 mpg
Explanation: When you encounter a gas mileage problem, you need to calculate miles per gallon (mpg) by dividing the distance traveled by the fuel consumed for each vehicle. Let's calculate the mpg for each car: Car A: 360 miles12 gallons=30\frac{360 \text{ miles}}{12 \text{ gallons}} = 30 mpg exactly Car B: 300 miles10 gallons=30\frac{300 \text{ miles}}{10 \text{ gallons}} = 30 mpg exactly Car C: 430 miles14 gallons=30.714...\frac{430 \text{ miles}}{14 \text{ gallons}} = 30.714... mpg, which rounds to about 30.7 mpg Since 30.7 mpg is higher than 30 mpg, Car C gets the best gas mileage. Choice A is incorrect because while Car A does get exactly 30 mpg, it doesn't have the best mileage among the three cars. Choice B is wrong for the same reason—Car B gets exactly 30 mpg but not the best. Choice D is incorrect because it claims all three cars have the same mileage at 30 mpg, but Car C actually gets about 30.7 mpg, which is better than the others. Choice C is correct because Car C gets approximately 30.7 mpg, which is the highest fuel efficiency of the three vehicles. Strategy tip: In comparison problems like this, always calculate the exact value for each option before determining which is best. Don't assume that similar-looking fractions will yield the same result—small differences in the numbers can lead to meaningfully different outcomes, as shown here where Car C's slightly better ratio makes it the clear winner.

Question 17

Two printers produce flyers. Printer X makes 420 flyers in 7 minutes, and Printer Y makes 620 flyers in 10 minutes. Which printer works faster, and what is that rate?

  1. Printer X, about 60 flyers per minute
  2. Printer Y, about 62 flyers per minute (correct answer)
  3. Both printers, exactly 61 flyers per minute
  4. The faster printer cannot be determined from the data
Explanation: When you encounter rate problems, you need to find the unit rate by dividing the total amount by the total time. This lets you compare different scenarios on equal terms. To find each printer's rate, divide flyers by minutes: Printer X: 420 flyers7 minutes=60 flyers per minute\frac{420 \text{ flyers}}{7 \text{ minutes}} = 60 \text{ flyers per minute} Printer Y: 620 flyers10 minutes=62 flyers per minute\frac{620 \text{ flyers}}{10 \text{ minutes}} = 62 \text{ flyers per minute} Since 62 > 60, Printer Y works faster at 62 flyers per minute. This confirms answer choice B. Looking at the wrong answers: Choice A correctly calculates Printer X's rate as 60 flyers per minute, but incorrectly identifies it as the faster printer. Choice C claims both printers work at exactly 61 flyers per minute, which would require the rates to be equal—but our calculations show they're different (60 vs. 62). Choice D suggests we can't determine the faster printer, but we clearly can since we have all the necessary information to calculate both rates. Strategy tip: For SHSAT rate problems, always convert to the same unit before comparing. Calculate each rate separately, then compare the results. Watch out for answer choices that give you a correct calculation but pair it with the wrong conclusion—this is a common trap designed to catch students who do the math right but misread which value is larger.

Question 18

Recipe 1 uses 3 cups of flour to make 2 loaves of bread. Recipe 2 uses 7 cups of flour to make 5 loaves. Which recipe uses less flour per loaf?

  1. Recipe 1, 1.501.50 cups per loaf
  2. Recipe 2, 1.401.40 cups per loaf (correct answer)
  3. Both recipes, 1.451.45 cups per loaf
  4. The flour per loaf cannot be determined from the information
Explanation: When you encounter problems comparing rates or ratios, you need to find the unit rate for each option to make a fair comparison. Here, you're looking for cups of flour per loaf, so divide the total flour by the number of loaves for each recipe. For Recipe 1: 3 cups2 loaves=1.5\frac{3 \text{ cups}}{2 \text{ loaves}} = 1.5 cups per loaf For Recipe 2: 7 cups5 loaves=1.4\frac{7 \text{ cups}}{5 \text{ loaves}} = 1.4 cups per loaf Since 1.4<1.51.4 < 1.5, Recipe 2 uses less flour per loaf. Looking at the answer choices: Choice A correctly calculates Recipe 1's rate but incorrectly claims it uses less flour per loaf. Choice B correctly identifies that Recipe 2 uses less flour and gives the right calculation of 1.401.40 cups per loaf. Choice C suggests both recipes use the same amount of flour per loaf at 1.451.45 cups, which appears to be an average of the two rates—but averaging doesn't make sense here since we're comparing which is less. Choice D claims we can't determine the answer, but we clearly have all the information needed to calculate both unit rates. Study tip: For rate comparison problems, always convert to the same unit (like "per loaf" or "per hour") before comparing. Watch out for answer choices that give you correct calculations but wrong conclusions about which is greater or less—read the question carefully to know what you're looking for.

Question 19

Grapes are sold in three bag sizes. 1.5 lb for $3.90 3 lb for $7.50 5 lb for $12.25 Which bag is the best buy based on price per pound?

  1. 1.5-lb bag at $2.60 per pound
  2. 3-lb bag at $2.50 per pound
  3. 5-lb bag at $2.45 per pound (correct answer)
  4. All sizes cost the same per pound, about $2.52
Explanation: When you encounter unit rate problems like this, you need to find the cost per unit (price per pound) by dividing the total cost by the total weight for each option. Let's calculate the price per pound for each bag size: For the 1.5-lb bag: $3.901.5 lb=$2.60\frac{\$3.90}{1.5 \text{ lb}} = \$2.60 per pound For the 3-lb bag: $7.503 lb=$2.50\frac{\$7.50}{3 \text{ lb}} = \$2.50 per pound For the 5-lb bag: $12.255 lb=$2.45\frac{\$12.25}{5 \text{ lb}} = \$2.45 per pound The 5-lb bag has the lowest price per pound at $2.45, making it the best buy. Looking at the wrong answers: Choice A correctly calculates the unit rate for the 1.5-lb bag (2.60perpound)butthisisn′tthebestdealsinceit′sthehighestpriceperpound.ChoiceBcorrectlycalculatesthe3−lbbag′srate(2.60 per pound) but this isn't the best deal since it's the highest price per pound. Choice B correctly calculates the 3-lb bag's rate (2.50 per pound), but this is the middle option, not the best. Choice D claims all sizes cost the same, which is incorrect—the calculations clearly show different unit rates, and $2.52 isn't close to any of our calculated values. Notice that larger quantities often (but not always) offer better unit rates due to bulk pricing. Always calculate the unit rate rather than assuming—sometimes smaller sizes are actually better deals. When comparing unit rates, the lowest cost per unit gives you the most value for your money.

Question 20

Two pizzerias sell pies of different sizes. Pizzeria M: 16-inch diameter pizza for $13 Pizzeria N: 18-inch diameter pizza for $15 Which pizzeria offers the lower price per square inch of pizza?

  1. Pizzeria M at about $0.065 per sq in
  2. Pizzeria N at about $0.059 per sq in (correct answer)
  3. Both pizzerias charge exactly $0.062 per sq in
  4. Price per area cannot be determined without slice count
Explanation: When comparing prices per unit area, you need to calculate the area of each circular pizza using A=πr2A = \pi r^2, then divide the price by that area. For Pizzeria M's 16-inch diameter pizza, the radius is 8 inches, so the area is π(8)2=64π\pi(8)^2 = 64\pi square inches. The price per square inch is $1364π=1364π≈13201.06≈$0.065\frac{\$13}{64\pi} = \frac{13}{64\pi} \approx \frac{13}{201.06} \approx \$0.065 per square inch. For Pizzeria N's 18-inch diameter pizza, the radius is 9 inches, so the area is π(9)2=81π\pi(9)^2 = 81\pi square inches. The price per square inch is $1581π=1581π≈15254.47≈$0.059\frac{\$15}{81\pi} = \frac{15}{81\pi} \approx \frac{15}{254.47} \approx \$0.059 per square inch. Since $0.059<$0.065\$0.059 < \$0.065, Pizzeria N offers the better deal. Choice A incorrectly identifies Pizzeria M as having the lower price, when it actually has the higher price per square inch. Choice C suggests both pizzerias charge the same rate, but our calculations show they're different. Choice D claims the price per area cannot be determined without knowing slice count, but slice count is irrelevant—you're buying the entire pizza regardless of how it's cut. Remember that when comparing circular objects by price per area, the larger diameter usually wins because area increases with the square of the radius. A small increase in radius creates a large increase in area, often making bigger pizzas better deals even at higher absolute prices.