PSAT Math Quiz: Graphs

What this quiz covers

This quiz focuses on Graphs, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT 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.

Question 1

1. $\dfrac{30}{80}$
2. $\dfrac{50}{80}$ (correct answer)
3. $\dfrac{50}{220}$
4. $\dfrac{80}{500}$

Explanation: Undecided total: 30 (18–44) + 50 (45+) = 80. Age 45+ undecided: 50. Fraction = 50/80. (A) uses the 18–44 undecided numerator. (C) divides 45+ undecided by total 45+ respondents (wrong conditional). (D) divides total undecided by grand total.

Question 2

1. $60$
2. $63$
3. $72.5$ (correct answer)
4. $75$

Explanation: The best-fit line is labeled $y=2.5x+10$, where $x$ is years after 2000. For 2025, $x=25$ so $y=2.5(25)+10=72.5$ thousand.
A and B underestimate by using $x=20$ or rounding. D uses $x=26$ or rounds up indiscriminately.

Question 3

1. The mean decreases by approximately 0.32. (correct answer)
2. The mean decreases by approximately 0.42.
3. The mean decreases by approximately 0.50.
4. The mean stays the same.

Explanation: Original sum: (0)(3) + (1)(5) + (2)(6) + (3)(3) + (4)(2) + (8)(1) = 0+5+12+9+8+8 = 42. Mean = 42/20 = 2.10. Remove the 8: new sum = 34, 19 values. New mean = 34/19 ≈ 1.789. Decrease ≈ 2.10 − 1.79 = 0.31, ≈ 0.32. (B) miscounts data. (C) divides by 20 again. (D) ignores the outlier effect.

Question 4

1. 2
2. 3 (correct answer)
3. 4
4. 5

Explanation: Using horizontal line $y = k$: k = −3: 1 intersection; k = −2: 2 intersections; k = −1: 3 intersections; k = 0: 2 intersections; k = 1: 3 intersections; k = 2: 2 intersections; k = 3: 1 intersection. Values with exactly two solutions: k = −2, 0, 2 → three values.

Question 5

1. The student who studied for 6 hours scored higher than the line of best fit predicts.
2. For every additional hour studied, a student's actual exam score increased by exactly 4.2 points.
3. The predicted score for a student who studied 0 hours is greater than the actual score of any student in the data set.
4. The residual for the student who studied 10 hours is approximately 8 points. (correct answer)

Explanation: The student who studied 10 hours scored 108 on the graph. The predicted score is $4.2(10) + 58 = 100$. The residual is $108 - 100 = 8$. (A) is false because the 6-hour student scored 78, below the predicted $4.2(6)+58 = 83.2$. (B) misinterprets slope — the slope describes predicted change, not actual change for each individual. (C) is false because several plotted scores exceed 58.

Question 6

1. $-1$
2. $2$
3. $5$ (correct answer)
4. $7$

Explanation: $h(4) = g(4-2) + 3 = g(2) + 3$. From the graph, $g(2) = 2$. So $h(4) = 2 + 3 = 5$. (A) computes $g(4) - 3$. (B) forgets the +3 shift. (D) computes $g(2+2) + 3 = g(4)+3 = 4+3$.

Question 7

1. Club A
2. Club B
3. Club C (correct answer)
4. Club D

Explanation: Compute percent change: Club A: $(36-30)/30 = 20\%$. Club B: $(45-36)/36 = 25\%$. Club C: $(33-20)/20 = 65\%$. Club D: $(60-42)/42 \approx 42.9\%$. Only Club C exceeds 50%. Students who compute absolute differences (Club D has largest absolute gain of 18) would incorrectly choose D.

Question 8

1. 20%
2. 30% (correct answer)
3. 35%
4. 40%

Explanation: At 50 cm the graph shows 28 plants or 70 % at or below that height. Plants taller than 50 cm: $40-28=12$, which is $12/40=30\%$.
Other choices misread the cumulative counts.

Question 9

1. $0.75$
2. $1.2$
3. $1.5$ (correct answer)
4. $3.0$

Explanation: At 8 a.m. the graph shows $61^{\circ}\text{F}$, and at 2 p.m. (14:00) it shows $70^{\circ}\text{F}$. The change is $70-61=9^{\circ}\text{F}$ over $14-8=6$ hours, giving $9/6=1.5^{\circ}\text{F per hour}$.
A: $0.75$ misuses the total 12-hour span. B: $1.2$ divides the 6-hour change by 7.5 hours in error.
D: $3.0$ confuses the change with the rate for a 3-hour interval.

Question 10

1. 0–2 minutes
2. 2–6 minutes (correct answer)
3. 6–8 minutes
4. 8–10 minutes

Explanation: The graph is horizontal from 2 to 6 minutes, showing no increase in distance. A, C, and D correspond to positive-slope segments indicating motion.

Question 11

1. Coal (correct answer)
2. Natural gas
3. Renewable
4. Nuclear

Explanation: Coal drops from 45 % to 25 %, a 20-percentage-point change. Renewable rises 15 → 30 (15 points), natural gas 30 → 35 (5), nuclear stays at 10 (0). The largest change is coal. Others change less.

Question 12

1. 12%
2. 19% (correct answer)
3. 25%
4. 32%

Explanation: Total students $48+56+38+34=176$. Seniors: $34/176\approx0.193=19\%$.
A: 12 % uses 22 seniors. C: 25 % rounds 34/136. D: 32 % treats 56 seniors instead of 34.

Question 13

1. $\dfrac16$
2. $\dfrac13$
3. $\dfrac12$ (correct answer)
4. $\dfrac23$

Explanation: Bars for 80–89 and 90–99 contain $9+6=15$ students. Total students: $2+5+8+9+6=30$. Probability $=15/30=1/2$.
A and B undervalue by using only one bar. D assumes 20 students scored 80 or higher.

Question 14

1. 25%
2. 50% (correct answer)
3. 75%
4. Cannot be determined

Explanation: The interval from the first quartile (6 h) to the third quartile (12 h) contains the middle 50 % of the data.
A and C confuse quartile spacing.
D ignores the definition of a box plot.

Question 15

1. $\dfrac{85}{200}$
2. $\dfrac{85}{400}$
3. $\dfrac{125}{400}$
4. $\dfrac{125}{200}$ (correct answer)

Explanation: Juniors + Seniors total 100 + 100 = 200. Drivers among juniors/seniors: 40 + 85 = 125. Probability = 125/200. (A) uses only seniors numerator with juniors+seniors denominator. (B) uses seniors who drive over entire sample. (C) uses all drivers over entire sample, ignoring the 'junior or senior' condition.

Question 16

1. Enrollment increased by approximately the same percentage each year. (correct answer)
2. Enrollment increased by approximately the same number of students each year.
3. Enrollment growth slowed each year, approaching a constant total.
4. Enrollment grew exponentially and then declined.

Explanation: Ratios year-to-year: 2640/2400 = 1.10, 2904/2640 = 1.10, 3194/2904 ≈ 1.10, 3514/3194 ≈ 1.10. Each ratio is about 1.10, indicating roughly 10% growth per year — not constant absolute increase (differences are 240, 264, 290, 320, increasing). (B) matches a linear pattern, not this data. (C) contradicts the accelerating absolute growth. (D) has no decline.

Question 17

1. $\$150{,}000$
2. $\$300{,}000$
3. $\$450{,}000$ (correct answer)
4. $\$600{,}000$

Explanation: Education = 30% of 8{,}000{,}000 = \2{,}400{,}000$. Increase of 25$$600{,}000$. Split 3:1 between Parks and Other means Parks contributes$\frac{3}{4}(600{,}000) = $450{,}000$. (A) takes 25% of Parks alone. (B) splits evenly. (D) takes entire increase from Parks.

Question 18

1. The slope will increase substantially because (10, 20) lies far above the current line.
2. The slope will decrease because (10, 20) lies below the predicted y-value for x = 10. (correct answer)
3. The slope will remain essentially unchanged because the point lies near the line of best fit.
4. The slope will decrease because the point has the largest x-value in the data set.

Explanation: From the graph, the line of best fit is approximately $y = 2.5x + 3$, so at $x = 10$, the predicted y is 28. The new point (10, 20) lies 8 units below the line. Since this point is at the far-right (high x), it exerts strong leverage, pulling the right end of the regression line down, which decreases the slope. (A) incorrectly claims the point is above the line. (C) incorrect — far-right low point has high leverage. (D) misattributes the cause to x-position alone; the y-value matters.

Question 19

1. February
2. March (correct answer)
3. April
4. May

Explanation: Percent difference = |A−B|/min(A,B). Feb: |30−20|/20 = 50%. Mar: |40−15|/15 ≈ 167%. Apr: |25−45|/25 = 80%. May: |50−30|/30 ≈ 67%. Greatest is March. (A), (C), (D) correspond to months with large absolute differences but smaller percent differences relative to the smaller value.

Question 20

1. The median is in the interval $60\leq x < 70$.
2. The median is in the interval $70\leq x < 80$. (correct answer)
3. The median is in the interval $80\leq x < 90$.
4. The median cannot be determined from the histogram.

Explanation: Cumulative counts: 50–60: 4; 60–70: 4+10 = 14; 70–80: 14+18 = 32; 80–90: 32+12 = 44; 90–100: 44+6 = 50. The median is the average of the 25th and 26th values. Both fall in 70–80 since cumulative count reaches 14 before and 32 through this interval. (A) gives only 14 values. (C) forgets to count earlier intervals. (D) is incorrect — the interval can be determined.