Problem-Solving & Data Analysis Domain Test 1

Calculator is allowed on all questions. Figures are not necessarily drawn to scale.

Question 1

A recipe calls for 3 cups of oats for every 2 cups of flour. If a baker wants to use 9 cups of oats, how many cups of flour are needed?




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The ratio of oats to flour is \(3 : 2\). Set up a proportion:

\[\frac{3}{2} = \frac{9}{f} \Rightarrow f = \frac{9 \times 2}{3} = 6\]

    1. \(4\): used \(9 - 5 = 4\) or applied the ratio incorrectly
    1. \(8\): added 6 to 2 (added the oats increase to flour) instead of scaling proportionally
    1. \(13.5\): multiplied \(9 \times \frac{3}{2}\) — inverted the ratio

Answer: B

Question 2

A car travels 240 miles in 4 hours. At the same rate, how many miles will it travel in 7 hours?




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Rate \(= \dfrac{240}{4} = 60\) miles per hour.

Distance in 7 hours \(= 60 \times 7 = 420\) miles.

    1. \(380\): added \(240 + (7-4) \times \frac{240}{4} - \text{error}\); likely added \(4 \times 3 + 240 - \text{something}\)
    1. \(400\): used rate of \(\frac{240}{4.2} \approx 57\) or added \(240 + 160 = 400\) incorrectly
    1. \(480\): multiplied \(240 \times 2\) (doubled the distance for double the time, incorrectly treating 7 as double 4)

Answer: C

Question 3

The list below shows the scores of 7 students on a test.

\(62,\ 75,\ 81,\ 81,\ 88,\ 91,\ 94\)

What is the median score?




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The data is already in order. With 7 values, the median is the 4th value:

\[62,\ 75,\ 81,\ \mathbf{81},\ 88,\ 91,\ 94\]

Median \(= 81\).

    1. \(82\): averaged the two middle values as if there were an even number of data points: \(\frac{81+83}{2}\)… or averaged 81 and 83
    1. \(84\): computed the mean of all 7 values: \((62+75+81+81+88+91+94)/7 = 572/7 \approx 81.7\); not quite 84 — likely summed incorrectly
    1. \(88\): identified the 5th value instead of the 4th

Answer: A

Question 4

A store sells a jacket for \(\$120\). If the price is reduced by \(25\%\), what is the sale price of the jacket, in dollars?

Enter your answer:

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A \(25\%\) reduction means the customer pays \(75\%\) of the original price:

\[120 \times 0.75 = 90\]

Alternatively: discount amount \(= 120 \times 0.25 = 30\). Sale price \(= 120 - 30 = 90\).

Answer: 90

Question 5

A bag contains 4 red marbles, 5 blue marbles, and 3 green marbles. If one marble is drawn at random, what is the probability that it is blue?




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Total marbles \(= 4 + 5 + 3 = 12\).

\[P(\text{blue}) = \frac{5}{12}\]

    1. \(\dfrac{1}{4}\): divided 5 by 20 or used \(\frac{3}{12}\) (the green proportion)
    1. \(\dfrac{5}{7}\): used only the non-red marbles (\(5 + 3 - 1 = 7\)) as the denominator
    1. \(\dfrac{7}{12}\): used the combined count of red and blue (\(4 + 3 = 7\)) as the numerator instead of blue only

Answer: B

Question 6

A 5-liter jug of water weighs 5.4 kilograms. The empty jug weighs 0.4 kilograms. What is the weight of water per liter, in kilograms per liter?




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Weight of water only \(= 5.4 - 0.4 = 5.0\) kilograms.

Weight per liter \(= \dfrac{5.0}{5} = 1.0\) kg/L.

    1. \(0.9\): divided total weight \(5.4\) by \(6\) (added jug weight to liter count)
    1. \(1.08\): divided total weight \(5.4\) by \(5\) without subtracting the jug’s weight
    1. \(1.18\): divided \((5.4 + 0.4) / 5\), adding jug weight instead of subtracting

Answer: B

Question 7

The scatterplot below shows the relationship between study time (hours) and exam score (out of 100) for 10 students.

1 2 3 4 5 6 7 Study Time (hours) 40 50 60 70 80 90 100 Score

Which of the following best describes the relationship shown in the scatterplot?




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As study time increases, exam scores consistently increase — the points follow closely along an upward-sloping line. This is a strong positive association.

    1. Weak positive: the points cluster tightly around the line, indicating strength, not weakness
    1. Strong negative: the trend is positive (scores rise with hours), not negative
    1. No association: there is a clear directional trend

Answer: C

Question 8

A class of 30 students has an average (mean) score of 78 on a test. When the teacher includes a student who was absent, the average drops to 77. What was the absent student’s score?

Enter your answer:

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Total score of 30 students \(= 30 \times 78 = 2{,}340\).

Total score of 31 students \(= 31 \times 77 = 2{,}387\).

Absent student’s score \(= 2{,}387 - 2{,}340 = 47\).

Answer: 47

Question 9

A two-way table shows the results of a survey of 200 people about their preferred beverage and whether they exercise regularly.

Coffee Tea Total
Exercises 72 48 120
Does not exercise 44 36 80
Total 116 84 200

What is the probability that a randomly selected person who prefers tea also exercises regularly?




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This is a conditional probability: \(P(\text{exercises} \mid \text{prefers tea})\).

\[P(\text{exercises} \mid \text{tea}) = \frac{\text{exercises AND tea}}{\text{total tea}} = \frac{48}{84} = \frac{4}{7}\]

    1. \(\dfrac{48}{200}\): uses total survey population as denominator — gives joint probability, not conditional
    1. \(\dfrac{48}{120}\): uses total exercisers as denominator instead of total tea drinkers
    1. \(\dfrac{84}{200}\): gives the overall proportion who prefer tea, not the conditional probability

Answer: C

Question 10

A student received scores of 82, 90, 76, and 88 on four tests. Each test is worth the same amount. What score does the student need on a fifth test to have a mean score of exactly 85?




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Required total for mean of 85 over 5 tests \(= 85 \times 5 = 425\).

Current total \(= 82 + 90 + 76 + 88 = 336\).

Fifth score needed \(= 425 - 336 = 89\).

    1. \(85\): assumed the fifth score equals the target mean — only works if all four current scores also equal 85
    1. \(87\): arithmetic error in summing — used \(338\) instead of \(336\)
    1. \(92\): used a target mean of 86 or summed incorrectly

Answer: C

Question 11

Two fair six-sided dice are rolled. What is the probability that the sum of the two dice equals 8?




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Total outcomes \(= 6 \times 6 = 36\).

Outcomes that sum to 8: \((2,6),(3,5),(4,4),(5,3),(6,2)\) — that’s 5 outcomes.

\[P(\text{sum} = 8) = \frac{5}{36}\]

    1. \(\dfrac{1}{9} = \dfrac{4}{36}\): counted only 4 outcomes (missed one pair, perhaps \((4,4)\))
    1. \(\dfrac{1}{6} = \dfrac{6}{36}\): counted 6 outcomes (added an extra invalid pair)
    1. \(\dfrac{7}{36}\): confused with sum \(= 7\) which has 6 outcomes, or counted 7 outcomes in error

Answer: B

Question 12

The boxplot below summarizes the distribution of heights (in inches) of players on a basketball team.

64 68 72 76 80 84 88 Height (inches)

Based on the boxplot, which of the following statements is true?




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From the boxplot: min \(= 68\), Q1 \(= 72\), median \(= 76\), Q3 \(= 80\), max \(= 84\).

  • Range \(= 84 - 68 = 16\) inches, not 12 → A is false
  • IQR \(= Q3 - Q1 = 80 - 72 = 8\) inches, not 12 → B is false
  • Middle 50% of the data lies between Q1 and Q3 (72 to 80 inches) → C is true ✓
  • Median \(= 76\) inches, not 80 → D is false

Answer: C

Question 13

A class of 25 students has a mean quiz score of 84. A second class of 15 students has a mean quiz score of 76. What is the mean quiz score for all 40 students combined?

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Total score for class 1 \(= 25 \times 84 = 2{,}100\).

Total score for class 2 \(= 15 \times 76 = 1{,}140\).

Combined total \(= 2{,}100 + 1{,}140 = 3{,}240\).

Combined mean \(= \dfrac{3{,}240}{40} = 81\).

Answer: 81

Question 14

A survey finds that students who sleep more than 8 hours per night have higher GPA scores on average. Which of the following conclusions is best supported by this finding?




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An observational survey can establish association (correlation) but not causation. The finding shows that more sleep and higher GPA tend to occur together, but it does not prove that sleep causes the higher GPA (other factors, such as study habits or stress levels, may be involved).

    1. Incorrectly concludes causation from correlation — a survey cannot establish a causal relationship
    1. Overstates the conclusion — the survey says “more than 8 hours,” not “exactly 8,” and correlation doesn’t prescribe an exact amount
    1. Draws a much stronger conclusion than the data support; the survey shows averages, not guaranteed outcomes for individuals

Answer: B

Question 15

Two datasets, A and B, each have the same mean. Dataset A has values \(\{10, 10, 10, 10, 10\}\). Dataset B has values \(\{2, 6, 10, 14, 18\}\). Which of the following correctly compares their standard deviations?




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Standard deviation measures how spread out the values are from the mean.

Dataset A \(= \{10, 10, 10, 10, 10\}\): all values are identical — zero spread, so standard deviation \(= 0\).

Dataset B \(= \{2, 6, 10, 14, 18\}\): values range from 2 to 18 around the mean of 10 — there is substantial spread, so standard deviation \(> 0\).

Dataset B has the greater standard deviation.

    1. Dataset A has no variation — its standard deviation is 0, the smallest possible
    1. They cannot be equal: Dataset A has SD \(= 0\) while Dataset B clearly has \(\text{SD} > 0\)
    1. Standard deviation is defined by spread around the mean, not the median

Answer: B

Question 16

A restaurant menu has 4 appetizers, 6 entrées, and 3 desserts. If a customer chooses one item from each course, how many different meal combinations are possible?




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By the Multiplication (Counting) Principle, the total number of combinations is the product of the number of choices at each course:

\[4 \times 6 \times 3 = 72\]

    1. \(13\): added the choices instead of multiplying (\(4 + 6 + 3 = 13\))
    1. \(36\): multiplied only two of the three courses (\(6 \times 3 \times 2 = 36\)… or \(4 \times 9\))
    1. \(216\): computed \(6^3 = 216\), using only the entrée count three times

Answer: C

Question 17

A bag contains 4 red chips and 2 blue chips. Two chips are drawn at random without replacement. What is the probability that both chips are red?

Enter your answer:

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Without replacement, the draws are dependent events.

\[P(\text{both red}) = P(\text{1st red}) \times P(\text{2nd red} \mid \text{1st red}) = \frac{4}{6} \times \frac{3}{5} = \frac{12}{30} = \frac{2}{5} = 0.4\]

Answer: 0.4

Question 18

The scatterplot below shows the number of hours of TV watched per day and the reading score for 10 students. A line of best fit is shown.

1 2 3 4 5 TV Hours / Day 40 50 60 70 80 90 100 Reading Score

Based on the line of best fit, what is the approximate predicted reading score for a student who watches 2.5 hours of TV per day?




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From the line of best fit equation \(\text{score} \approx 97 - 9.4t\) where \(t\) is TV hours:

\[\text{score} \approx 97 - 9.4(2.5) = 97 - 23.5 = 73.5 \approx 74\]

Alternatively, reading directly from the graph at \(x = 2.5\): the line passes through approximately \(y = 74\).

    1. \(62\): reads the line at \(x = 3.7\) hours instead of \(2.5\)
    1. \(70\): reads the line at \(x = 3.0\) hours
    1. \(78\): reads the line at \(x = 2.0\) hours

Answer: C

Question 19

A survey of 400 voters found that 240 supported Candidate A. The margin of error is \(\pm 4\) percentage points. Which of the following is a reasonable conclusion?




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Sample proportion \(= \dfrac{240}{400} = 0.60 = 60\%\).

With a margin of error of \(\pm 4\) percentage points, the confidence interval for the true population proportion is \(60\% - 4\% = 56\%\) to \(60\% + 4\% = 64\%\).

    1. “Exactly 60%” is overly precise — the margin of error acknowledges uncertainty around this estimate
    1. The survey shows support exceeding 50%, but margin-of-error intervals and factors like voter turnout mean the outcome is not guaranteed
    1. A properly conducted random sample is used to make inferences about the broader population, not just the sample itself

Answer: B

Question 20

In a school of 800 students, a random sample of 40 students was surveyed. Of those surveyed, 14 said they preferred online learning. Based on the sample, what is the best estimate of the number of students in the entire school who prefer online learning?

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Sample proportion \(= \dfrac{14}{40} = 0.35 = 35\%\).

Estimated number in the full school \(= 0.35 \times 800 = 280\).

Answer: 280