Scale and Rate Problems - SSAT Upper Level Quantitative

1. A machine produces widgets at a rate that varies throughout the day. In the first $3$ hours, it produces $450$ widgets. In the next $2$ hours, its rate increases by $20\%$. How many widgets does it produce during those $2$ hours?
   Answer: $360$ widgets Initial rate: $\frac{450 \text{ widgets}}{3 \text{ hours}} = 150$ widgets per hour. The rate increases by $20\%$, so new rate $= 150 \times 1.20 = 180$ widgets per hour. In $2$ hours at this rate: $180 \times 2 = 360$ widgets. Choice A incorrectly uses the original rate for $2$ hours: $150 \times 2 = 300$. Choice C incorrectly calculates $450 \times 1.20 = 540$. Choice D uses an incorrect rate of $300$ widgets per hour.
2. A recipe that serves $6$ people calls for $2\frac{1}{4}$ cups of flour. If you want to serve $14$ people but only have $4\frac{3}{4}$ cups of flour available, what is the maximum number of people you can actually serve with the flour you have?
   Answer: $12$ people When you encounter recipe scaling problems, you need to work with ratios and proportions. The key insight is finding how much flour is needed per person, then determining how many people your available flour can serve. First, let's find the flour needed per person. The recipe serves 6 people with $2\frac{1}{4}$ cups of flour. Converting to an improper fraction: $2\frac{1}{4} = \frac{9}{4}$ cups. So each person needs $\frac{9/4}{6} = \frac{9}{4} \times \frac{1}{6} = \frac{9}{24} = \frac{3}{8}$ cups of flour. Now, with $4\frac{3}{4}$ cups available, let's see how many people we can serve. Converting: $4\frac{3}{4} = \frac{19}{4}$ cups. Dividing by the per-person amount: $\frac{19/4}{3/8} = \frac{19}{4} \times \frac{8}{3} = \frac{152}{12} = 12\frac{2}{3}$ people. Since you can't serve a fraction of a person, you can serve a maximum of 12 people. Choice (A) 15 people would require $15 \times \frac{3}{8} = \frac{45}{8} = 5\frac{5}{8}$ cups—more than you have. Choice (B) 13 people would need $13 \times \frac{3}{8} = \frac{39}{8} = 4\frac{7}{8}$ cups—still too much. Choice (C) 14 people would require $14 \times \frac{3}{8} = \frac{42}{8} = 5\frac{1}{4}$ cups—way more than available. Choice (D) 12 people needs exactly $12 \times \frac{3}{8} = 4\frac{1}{2}$ cups, which is less than your $4\frac{3}{4}$ cups. Remember: in "maximum" problems, always round down to the nearest whole number when dealing with people or discrete items.
3. A map has a scale where $1.5$ inches represents $12$ miles. If two cities are $8.5$ inches apart on the map, and a car travels between them at an average speed of $51$ miles per hour, how long will the trip take?
   Answer: $1$ hour $20$ minutes First, find the actual distance: $\frac{8.5 \text{ inches}}{1.5 \text{ inches}} \times 12 \text{ miles} = \frac{8.5 \times 12}{1.5} = \frac{102}{1.5} = 68$ miles. Then find the time: $\frac{68 \text{ miles}}{51 \text{ mph}} = \frac{68}{51} = \frac{4}{3} = 1\frac{1}{3}$ hours $= 1$ hour $20$ minutes. Choice B uses incorrect scale calculation ($8.5 \times 12 = 102$ miles directly). Choice C incorrectly calculates $\frac{68}{52}$ instead of $\frac{68}{51}$. Choice D uses $8.5 \times 1.5 \times 12 = 153$ miles.
4. A photograph measuring $4$ inches by $6$ inches is enlarged so that its area becomes $150$ square inches. If the enlargement maintains the same proportions, what are the dimensions of the enlarged photograph?
   Answer: $10$ inches by $15$ inches The original area is $4 \times 6 = 24$ square inches. The area grows by a factor of $\frac{150}{24} = 6.25$, and since area scales with the square of the length factor, each side is multiplied by $\sqrt{6.25} = 2.5$. The new dimensions are $4 \times 2.5 = 10$ inches and $6 \times 2.5 = 15$ inches, and $10 \times 15 = 150$ square inches. $6$ by $25$ also has an area of 150, but $\frac{6}{25}$ does not equal $\frac{4}{6}$, so the proportions change. $8$ by $12$ doubles each side, for an area of only 96. $12$ by $18$ triples each side, for an area of 216. Strategy tip: When an enlargement keeps its proportions, find the length factor from the square root of the area factor, then check both the area and the ratio of the sides.
5. A model train is built to a scale of $1:87$. If the model locomotive is $6.2$ inches long, and the real locomotive travels $280$ miles in $4$ hours $40$ minutes, what is the speed of the real locomotive in miles per hour?
   Answer: $60$ mph The scale information and model length are red herrings - they don't affect the speed calculation. The real locomotive travels $280$ miles in $4$ hours $40$ minutes. Convert time: $4$ hours $40$ minutes $= 4\frac{40}{60} = 4\frac{2}{3} = \frac{14}{3}$ hours. Speed $= \frac{280}{\frac{14}{3}} = 280 \times \frac{3}{14} = \frac{840}{14} = 60$ mph. Choice A incorrectly uses $5$ hours instead of $4\frac{2}{3}$ hours. Choice C uses $4.5$ hours instead of $4\frac{2}{3}$ hours. Choice D uses $4$ hours, ignoring the extra $40$ minutes.
6. A contractor estimates that $8$ workers can complete a job in $15$ days. After $6$ days, $2$ workers are reassigned to another project. Assuming all workers work at the same rate, how many additional days will it take the remaining workers to finish the job?
   Answer: $12$ days When you encounter work rate problems, think in terms of total work units and how they're distributed over time. The key insight is that work rate remains constant per worker. First, calculate the total work needed. If 8 workers complete the job in 15 days, the total work equals $8 \times 15 = 120$ worker-days. Next, determine how much work gets completed in the first 6 days. With 8 workers working for 6 days, that's $8 \times 6 = 48$ worker-days of completed work. This leaves $120 - 48 = 72$ worker-days of remaining work. After 6 days, 2 workers are reassigned, leaving $8 - 2 = 6$ workers. To find how long these 6 workers need to complete the remaining 72 worker-days of work: $\frac{72}{6} = 12$ days. Looking at the wrong answers: Choice (A) 18 days likely comes from incorrectly calculating $\frac{72}{4}$ workers instead of 6. Choice (B) 15 days represents the original timeline, ignoring both the work already completed and the reduced workforce. Choice (D) 21 days might result from adding the original 15 days to the 6 days already worked, which double-counts the time. The correct answer is (C) 12 days. Strategy tip: In work rate problems, always convert to total work units first (worker-days, worker-hours, etc.), then track what's completed versus what remains. This systematic approach prevents confusion when workforce size changes mid-project.

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