SSAT Upper Level Quantitative Quiz: Scale And Rate Problems

6 questions · exam conditions

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Scale And Rate ProblemsQuestion 1 of 6

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?

1

hour

20

minutes

1

hour

40

minutes

2

hours

10

minutes

2

hours

40

minutes

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SSAT Upper Level Quantitative Quiz

SSAT Upper Level Quantitative Quiz: Scale And Rate Problems

Practice Scale And Rate Problems in SSAT Upper Level Quantitative 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 Scale And Rate Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for SSAT Upper Level Quantitative.

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

$1$ hour $20$ minutes (correct answer)
$1$ hour $40$ minutes
$2$ hours $10$ minutes
$2$ hours $40$ minutes

Explanation: 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.

Question 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?

$15$ people
$13$ people
$14$ people
$12$ people (correct answer)

Explanation: 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.

Question 3

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?

$55$ mph
$60$ mph (correct answer)
$65$ mph
$70$ mph

Explanation: 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.

Question 4

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?

$18$ days
$15$ days
$12$ days (correct answer)
$21$ days

Explanation: 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.

Question 5

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?

$300$ widgets
$360$ widgets (correct answer)
$540$ widgets
$600$ widgets

Explanation: 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.

Question 6

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?

$8$ inches by $12$ inches
$12$ inches by $18$ inches
$10$ inches by $15$ inches (correct answer)
$15$ inches by $10$ inches

Explanation: When you encounter problems involving proportional enlargement or reduction, you're working with similar figures and scale factors. The key insight is that when dimensions change by a scale factor, the area changes by the square of that scale factor. Start by finding the original area:

4 \times 6 = 24

square inches. The enlarged photograph has an area of

150

square inches, so the area scale factor is

\frac{150}{24} = 6.25

. Since area scales by the square of the linear scale factor, the linear scale factor is

\sqrt{6.25} = 2.5

. This means each dimension is multiplied by

2.5

. The new dimensions are:

4 \times 2.5 = 10

inches and

6 \times 2.5 = 15

inches. Let's verify:

10 \times 15 = 150

square inches. ✓ Now examine why the other choices fail. Choice A gives

8 \times 12 = 96

square inches, which is too small and represents a scale factor of only

2

. Choice B yields

12 \times 18 = 216

square inches, which exceeds our target area and represents a scale factor of

3

. Choice D lists the same dimensions as C but in reverse order (

15 \times 10

), which gives the correct area but doesn't follow the conventional format of listing the smaller dimension first when maintaining the original orientation. Remember: when dealing with similar figures, always work backwards from the area ratio to find the linear scale factor by taking the square root. This prevents calculation errors and helps you systematically approach enlargement problems.