Practice Math — Problem Solving & Data Analysis questions for the Scholastic Assessment Test. Every question includes a full explanation of why the correct answer is right and why the tempting distractors are wrong.
24 questions available · medium difficulty · SAT · Free, no registration required
10 example questions with full explanations. Use the interactive practice above to work through the complete set.
Question 1hard
A researcher wants to study the effect of a new study technique on student test scores. She recruits 200 volunteers from an online forum dedicated to academic improvement and randomly assigns 100 to use the new technique and 100 to use their current method. After 8 weeks, the new-technique group scores an average of 12 points higher. Select all that apply: Which of the following conclusions are statistically and methodologically justified based on this study?
- A.The random assignment of participants to groups allows for a causal inference that the new technique caused the score improvement within this sample.✓
- B.The results can be generalized to all high school students because the sample size of 200 is sufficiently large.
- C.The use of volunteers from an academic forum introduces a sampling bias that limits generalizability to the broader student population.✓
- D.The 12-point average difference is meaningful only if the variability (e.g., standard deviation) of scores within each group is also considered.✓
- E.Because participants were randomly assigned, the study is free from all confounding variables and bias.
Correct answer: A, C, D
A is correct because random assignment within an experiment does support causal inference for the sample studied. C is correct because recruiting from an academic forum overrepresents motivated students, creating a biased sample that cannot be generalized to all students. D is correct because a 12-point mean difference could be trivial if the standard deviation is very large (e.g., 80 points), so variability is essential context. B is wrong because sample size alone does not fix a biased sampling method — a large biased sample is still biased. E is wrong because random assignment controls for confounding variables in the assignment process, but it does not eliminate all bias; the volunteer/self-selection bias in recruitment remains.
Question 2hard
The table below shows data from two classes on a 100-point exam.
Class A: scores are 55, 70, 70, 75, 80, 85, 85, 85, 90, 95 (n = 10)
Class B: scores are 60, 62, 65, 70, 72, 75, 78, 80, 83, 85 (n = 10)
Select all that apply: Which of the following statements about the two data sets are correct?
- A.The median of Class A is greater than the median of Class B.✓
- B.The mean of Class A is equal to the mean of Class B.
- C.Class A has a greater range than Class B.✓
- D.Class B has less variability (spread) than Class A.✓
Correct answer: A, C, D
Class A sorted: 55,70,70,75,80,85,85,85,90,95 → mean = 790/10 = 79; median = (80+85)/2 = 82.5. Class B sorted: 60,62,65,70,72,75,78,80,83,85 → mean = 730/10 = 73; median = (72+75)/2 = 73.5. A is correct: 82.5 > 73.5. B is incorrect: 79 ≠ 73. C is correct: Class A range = 95−55 = 40; Class B range = 85−60 = 25, so Class A has greater range. D is correct: Class B's scores span only 25 points versus 40, indicating less overall spread/variability. The key distractor is B — students may assume equal means without calculating.
Question 3hard
A two-way table summarizes survey data about 300 students' preferred exercise type and grade level.
| | Running | Swimming | Cycling | Total |
|---------------|---------|----------|---------|-------|
| 10th Grade | 45 | 30 | 25 | 100 |
| 11th Grade | 50 | 40 | 10 | 100 |
| 12th Grade | 25 | 50 | 25 | 100 |
| Total | 120 | 120 | 60 | 300 |
A student is selected at random. Select all that apply: Which of the following probability statements are correct?
- A.The probability that a randomly selected student prefers Swimming, given that the student is in 12th grade, is 1/2.✓
- B.The probability that a randomly selected student is in 11th grade and prefers Cycling is 1/30.✓
- C.The probability that a randomly selected student prefers Running is 2/5.✓
- D.The events 'prefers Cycling' and 'is in 11th grade' are independent.
Correct answer: A, B, C
A: P(Swimming | 12th grade) = 50/100 = 1/2. ✓ B: P(11th grade AND Cycling) = 10/300 = 1/30. ✓ C: P(Running) = 120/300 = 2/5. ✓ D: To test independence, compare P(Cycling | 11th grade) = 10/100 = 1/10 with P(Cycling) = 60/300 = 1/5. Since 1/10 ≠ 1/5, the events are NOT independent, making D false. Students commonly confuse joint probability with conditional probability (distractor B) and may incorrectly assume independence without checking the condition.
Question 4hard
A scatterplot shows the relationship between hours of weekly exercise (x) and resting heart rate in beats per minute (y) for 50 adults. The line of best fit is given by the equation ŷ = −2.4x + 85, with a correlation coefficient of r = −0.78. Select all that apply: Which of the following interpretations of the model are valid?
- A.For each additional hour of weekly exercise, the model predicts a decrease of 2.4 beats per minute in resting heart rate, on average.✓
- B.The correlation coefficient of −0.78 indicates that exercise causes resting heart rate to decrease.
- C.The model predicts a resting heart rate of 85 bpm for an adult who exercises 0 hours per week.✓
- D.Approximately 60.8% of the variation in resting heart rate can be explained by the linear relationship with weekly exercise hours.✓
- E.The negative value of r indicates a negative association between exercise hours and resting heart rate, but does not by itself establish causation.✓
Correct answer: A, C, D, E
A is correct: the slope −2.4 means each additional hour predicts a 2.4 bpm decrease on average. C is correct: the y-intercept 85 is the predicted heart rate when x = 0. D is correct: r² = (−0.78)² = 0.6084 ≈ 60.8%, which is the coefficient of determination representing explained variation. E is correct: correlation describes association strength and direction but never implies causation. B is incorrect and is the key trap — correlation (even strong correlation) does not establish causation; there may be confounding variables such as overall health, diet, or genetics that explain the relationship.
Question 5medium
A store sells a jacket for $84 after marking it down 30% from its original price. A few weeks later, the store increases the sale price by 30%. What is the final price of the jacket after this increase, to the nearest cent?
- A.$84.00
- B.$112.00
- C.$120.00
- D.$109.20✓
Correct answer: D
After the 30% discount, the sale price is $84.Asubsequent 30% increase is applied to this new price: $84 × 1.30 = $109.20.Acommon error (choice A) is assuming the two 30% changes cancel out and return to the original price — they do not, because the percentages apply to different base values. Choice C ($120) is the original price before any discount, and choice B ($112) incorrectly applies the increase to a miscalculated base.
Question 6medium
The table below shows the results of a survey in which 200 students were asked whether they prefer reading fiction or nonfiction and whether they prefer print or digital books.
| | Print | Digital | Total |
|---------------|-------|---------|-------|
| Fiction | 54 | 66 | 120 |
| Nonfiction | 48 | 32 | 80 |
| Total | 102 | 98 | 200 |
A student is selected at random from those who prefer digital books. What is the probability that this student prefers fiction?
- A.33/49✓
- B.33/100
- C.11/20
- D.3/5
Correct answer: A
Since we are told the student prefers digital books, we restrict our sample space to the 98 digital-book students. Of those, 66 prefer fiction. The probability is 66/98, which simplifies to 33/49. Choice C(33/100) incorrectly uses the total of 200 students as the denominator. Choice D (3/5) is the overall proportion of fiction readers (120/200). Choice A(11/20) confuses the fiction-print cell with the fiction-digital cell.
Question 7medium
A recipe that serves 6 people requires 2.25 cups of flour and 1.5 teaspoons of baking powder. A baker wants to scale the recipe to serve 10 people. How many teaspoons of baking powder will the baker need, rounded to the nearest tenth?
Correct answer: B
The scaling factor is 10/6 ≈ 1.6667. Multiply the original amount of baking powder by this factor: 1.5 × (10/6) = 15/6 = 2.5 teaspoons. Choice A (2.0) results from incorrectly using a scaling factor of 4/3. Choice C (3.0) doubles the baking powder, as if scaling to 12 servings. Choice D (3.5) may result from adding the difference in servings (4) multiplied by an incorrect unit rate.
Question 8medium
The dot plots below show the distribution of quiz scores (out of 20) for two classes of students.
Class A scores: 10, 11, 12, 12, 13, 14, 14, 14, 15, 15
Class B scores: 5, 8, 11, 12, 14, 14, 16, 17, 19, 20
Which statement best compares the variability of the two distributions?
- A.Class A has greater variability because its mean is closer to the center of the score range.
- B.The two classes have equal variability because they have the same number of students.
- C.Class B has greater variability because its scores are more spread out from the mean.✓
- D.Class A has greater variability because it has more repeated values (a mode), indicating more data points near the mean.
Correct answer: C
Variability measures how spread out data values are from the center. Class A's scores range from 10 to 15 (range = 5), and the values cluster tightly around the mean of 13. Class B's scores range from 5 to 20 (range = 15), and the values are spread much farther from the mean of 13.6, indicating much greater variability. Choice B is incorrect because sample size does not determine variability. Choice A confuses mean position with spread. Choice D incorrectly interprets having a mode as evidence of greater variability — in fact, repeated values near the center suggest less spread.
Question 9medium
A social media post received 240 likes in January. By March, the number of likes had increased to 348. What was the approximate percent increase in likes from January to March?
Correct answer: D
Percent increase = (new − original) / original × 100 = (348 − 240) / 240 × 100 = 108 / 240 × 100 = 45%. Option A (31%) results from incorrectly dividing 108 by 348 (the new value) instead of the original. Option B (69%) might come from a misreading or arithmetic error. Option C (108%) confuses the raw difference (108) with the percent change, forgetting to divide by the original value.
Question 10medium
A scatterplot shows the relationship between the number of hours spent practicing piano per week (x) and the score on a music proficiency test (y) for 20 students. The line of best fit is given by the equation y = 4.2x + 51. According to this model, how much higher is the predicted test score for a student who practices 10 hours per week compared to a student who practices 4 hours per week?
- A.25.2 points✓
- B.6 points
- C.51 points
- D.93 points
Correct answer: A
The predicted score for 10 hours: y = 4.2(10) + 51 = 42 + 51 = 93. The predicted score for 4 hours: y = 4.2(4) + 51 = 16.8 + 51 = 67.8. The difference is 93 − 67.8 = 25.2 points. Alternatively, the slope of 4.2 means each additional hour adds 4.2 points, so 6 extra hours adds 4.2 × 6 = 25.2 points. OptionA(6 points) is just the difference in hours, not test score points. Option C (51) is the y-intercept, not a difference. Option D (93) is the score at 10 hours, not the difference between the two scores.