GRE Multi-Step Data Analysis and Graphs 1 — Questions and Answers

Question 1: A company's annual revenue (in millions) over five years was: Year 1: $120M, Year 2: $144M, Year 3: $172.8M, Year 4: $207.4M, Year 5: $248.8M. What was the approximate average percentage growth rate per year? (A) 15% (B) 18% (C) 20% (D) 22%

15%
18%
20% (Correct answer)
22%

Correct answer: 20%

Each year the revenue grows by exactly 20%: 120 × 1.2 = 144, 144 × 1.2 = 172.8, etc. The answer is 20%.

Year 1 to 2: 144/120 = 1.20 (20% growth). Year 2 to 3: 172.8/144 = 1.20. Year 3 to 4: 207.4/172.8 ≈ 1.20. Year 4 to 5: 248.8/207.4 ≈ 1.20. The growth rate is consistently 20% each year.

Question 2: A bar chart shows quarterly sales for two products: Product A: Q1=40, Q2=55, Q3=70, Q4=85 (units) Product B: Q1=80, Q2=70, Q3=60, Q4=50 (units) In which quarter does the combined total first exceed 120 units? (A) Q1 (B) Q2 (C) Q3 (D) Q4

Q1
Q2 (Correct answer)
Q3
Q4

Correct answer: Q2

Q1: 40+80=120 (not exceeding). Q2: 55+70=125 > 120. The combined total first exceeds 120 in Q2.

Q1: 40 + 80 = 120 (equals but does not exceed). Q2: 55 + 70 = 125 > 120. Q3: 70 + 60 = 130. Q4: 85 + 50 = 135. The first quarter where the combined total exceeds 120 is Q2.

Question 3: A pie chart shows the distribution of 360 survey respondents by preferred news source: TV = 30%, Online = 45%, Print = 15%, Radio = 10%. How many more respondents prefer Online compared to the combined total of Print and Radio? (A) 54 (B) 72 (C) 81 (D) 90

54
72 (Correct answer)
81
90

Correct answer: 72

Online: 45% × 360 = 162. Print + Radio: (15% + 10%) × 360 = 25% × 360 = 90. Difference: 162 − 90 = 72.

Online: 0.45 × 360 = 162 respondents. Print: 0.15 × 360 = 54. Radio: 0.10 × 360 = 36. Print + Radio = 54 + 36 = 90. 162 − 90 = 72 more respondents prefer Online over the combined Print and Radio total.

Question 4: A line graph shows a city's population (in thousands) from 2000 to 2020 in 5-year intervals: 2000=200, 2005=230, 2010=272, 2015=306, 2020=340. Between which two 5-year periods was the absolute increase in population the greatest? (A) 2000–2005 (B) 2005–2010 (C) 2010–2015 (D) 2015–2020

2000–2005
2005–2010 (Correct answer)
2010–2015
2015–2020

Correct answer: 2005–2010

Increases: 2000–05: 30K; 2005–10: 42K; 2010–15: 34K; 2015–20: 34K. The 2005–2010 period had the largest absolute increase of 42,000.

2000–2005: 230 − 200 = 30 thousand. 2005–2010: 272 − 230 = 42 thousand. 2010–2015: 306 − 272 = 34 thousand. 2015–2020: 340 − 306 = 34 thousand. The greatest absolute increase was 42,000 during 2005–2010.

Question 5: A table shows test scores for two classes: Class A: Mean = 78, Standard Deviation = 5, n = 30 Class B: Mean = 82, Standard Deviation = 12, n = 20 Which class had greater score variability, and by approximately what factor? (A) Class A; variability 2.4 times greater (B) Class B; variability 2.4 times greater (C) Class A; variability 1.5 times greater (D) Class B; variability 1.5 times greater

Class A; variability 2.4 times greater
Class B; variability 2.4 times greater (Correct answer)
Class A; variability 1.5 times greater
Class B; variability 1.5 times greater

Correct answer: Class B; variability 2.4 times greater

Standard deviation measures variability. Class B: SD=12; Class A: SD=5. 12/5 = 2.4. Class B had 2.4× greater variability.

Standard deviation is the measure of variability in this context. Class B has SD = 12, Class A has SD = 5. The ratio is 12 ÷ 5 = 2.4. Therefore, Class B's scores were 2.4 times more variable than Class A's.

Question 6: A scatterplot shows the relationship between hours studied per week (x-axis, range 0–20) and exam score (y-axis, range 40–100). The best-fit line has the equation y = 3x + 50. A student studies 15 hours per week. What score does the model predict, and how does it compare to a student who studies 10 hours? (A) Predicts 95; 15 points higher than the 10-hour student (B) Predicts 95; 10 points higher than the 10-hour student (C) Predicts 100; 20 points higher than the 10-hour student (D) Predicts 90; 20 points higher than the 10-hour student

Predicts 95; 15 points higher than the 10-hour student (Correct answer)
Predicts 95; 10 points higher than the 10-hour student
Predicts 100; 20 points higher than the 10-hour student
Predicts 90; 20 points higher than the 10-hour student

Correct answer: Predicts 95; 15 points higher than the 10-hour student

15 hours: y = 3(15) + 50 = 95. 10 hours: y = 3(10) + 50 = 80. Difference: 95 − 80 = 15 points.

For 15 hours: y = 3(15) + 50 = 45 + 50 = 95. For 10 hours: y = 3(10) + 50 = 30 + 50 = 80. The 15-hour student scores 95, which is 95 − 80 = 15 points higher than the 10-hour student.

Question 7: A table shows monthly electricity costs for an office building: Jan–Mar (avg): $1,200/month Apr–Jun (avg): $900/month Jul–Sep (avg): $1,500/month Oct–Dec (avg): $1,050/month What is the total annual electricity cost? (A) $13,800 (B) $13,950 (C) $14,100 (D) $14,400

$13,800 (Correct answer)
$13,950
$14,100
$14,400

Correct answer: $13,800

Each average applies to 3 months: (1200×3) + (900×3) + (1500×3) + (1050×3) = 3600 + 2700 + 4500 + 3150 = $13,950.

Jan–Mar: $1,200 × 3 = $3,600. Apr–Jun: $900 × 3 = $2,700. Jul–Sep: $1,500 × 3 = $4,500. Oct–Dec: $1,050 × 3 = $3,150. Total: $3,600 + $2,700 + $4,500 + $3,150 = $13,950. The correct answer is (B) $13,950.

Question 8: A double bar chart compares men's and women's marathon finish times (in minutes) across 4 age groups: Age 20–29: Men=210, Women=235 Age 30–39: Men=215, Women=242 Age 40–49: Men=225, Women=255 Age 50–59: Men=240, Women=275 In which age group is the absolute difference between men's and women's times the greatest? (A) 20–29 (B) 30–39 (C) 40–49 (D) 50–59

20–29
30–39
40–49
50–59 (Correct answer)

Correct answer: 50–59

Differences: 20–29: 25min; 30–39: 27min; 40–49: 30min; 50–59: 35min. The 50–59 age group has the greatest difference.

20–29: 235 − 210 = 25 minutes. 30–39: 242 − 215 = 27 minutes. 40–49: 255 − 225 = 30 minutes. 50–59: 275 − 240 = 35 minutes. The greatest absolute difference is 35 minutes in the 50–59 age group.

Question 9: An investment portfolio had the following returns over 4 years: Year 1: +20%, Year 2: −10%, Year 3: +25%, Year 4: −5%. What was the approximate net percentage change over all 4 years? (A) +23% (B) +28% (C) +30% (D) +35%

+23%
+28% (Correct answer)
+30%
+35%

Correct answer: +28%

Net return = 1.20 × 0.90 × 1.25 × 0.95 = 1.08 × 1.25 × 0.95 = 1.35 × 0.95 = 1.2825 ≈ 28.25% gain.

Convert each return to a multiplier: 1.20, 0.90, 1.25, 0.95. Multiply: 1.20 × 0.90 = 1.08; 1.08 × 1.25 = 1.35; 1.35 × 0.95 = 1.2825. This means the portfolio grew to 128.25% of its original value — a net gain of approximately 28.25%, closest to 28%.

Question 10: A histogram shows the distribution of test scores for 100 students: 50–59: 5 students 60–69: 15 students 70–79: 35 students 80–89: 30 students 90–99: 15 students What percentage of students scored below 80? (A) 45% (B) 50% (C) 55% (D) 60%

45%
50%
55% (Correct answer)
60%

Correct answer: 55%

Students below 80: 5 + 15 + 35 = 55. 55/100 = 55%.

Score ranges below 80 are 50–59 (5), 60–69 (15), and 70–79 (35). Total below 80: 5 + 15 + 35 = 55 students. Percentage: 55/100 × 100% = 55%.

Question 11: A two-way table shows survey results on coffee preference by age group: Under 30: Coffee=40, Tea=60, Total=100 Over 30: Coffee=90, Tea=60, Total=150 Total: Coffee=130, Tea=120, Grand Total=250 What percentage of coffee drinkers are over 30? (A) 60% (B) 65% (C) 69% (D) 72%

60%
65%
69% (Correct answer)
72%

Correct answer: 69%

Coffee drinkers over 30: 90. Total coffee drinkers: 130. 90/130 ≈ 0.692 = 69.2% ≈ 69%.

Total coffee drinkers = 130. Coffee drinkers over 30 = 90. Proportion: 90 ÷ 130 = 0.6923... ≈ 69%. This is a conditional probability: P(over 30 | coffee drinker) = 90/130.

Question 12: A stacked bar chart shows the composition of a city's workforce (1,000 workers total): Government 25%, Private sector 55%, Self-employed 15%, Other 5%. If the workforce grows by 20% next year with all proportions staying the same, how many more private sector workers will there be? (A) 100 (B) 110 (C) 120 (D) 130

100
110 (Correct answer)
120
130

Correct answer: 110

Current private sector: 55% × 1000 = 550. New total: 1200. New private sector: 55% × 1200 = 660. Increase: 660 − 550 = 110.

Current workforce = 1,000. Private sector = 55% × 1,000 = 550 workers. New workforce = 1,000 × 1.20 = 1,200. New private sector = 55% × 1,200 = 660. Increase = 660 − 550 = 110 additional private sector workers.

Question 13: A line graph shows two companies' quarterly profits (in $M): Company X: Q1=10, Q2=14, Q3=19, Q4=25 Company Y: Q1=22, Q2=20, Q3=18, Q4=15 In which quarter do the lines intersect (i.e., Company X's profit first equals or exceeds Company Y's)? (A) Q2 (B) Q3 (C) Q4 (D) They never intersect within the given data

Q2
Q3
Q4 (Correct answer)
They never intersect within the given data

Correct answer: Q4

Q1: X=10 < Y=22. Q2: X=14 < Y=20. Q3: X=19 > Y=18. Actually X first exceeds Y at Q3. But wait: we need when X first equals OR exceeds Y. At Q3, X=19 > Y=18, so Q3 is correct.

Q1: X=10, Y=22 → X < Y. Q2: X=14, Y=20 → X < Y. Q3: X=19, Y=18 → X > Y. Company X first exceeds Company Y in Q3. The lines cross between Q2 and Q3. Answer (B) Q3 is correct.

Question 14: A table shows the number of defective items per 1,000 produced at three factories over two years: Factory A: Year 1=25, Year 2=18 Factory B: Year 1=40, Year 2=28 Factory C: Year 1=15, Year 2=12 Which factory achieved the greatest percentage reduction in defects? (A) Factory A (B) Factory B (C) Factory C (D) Factories A and B tied

Factory A
Factory B (Correct answer)
Factory C
Factories A and B tied

Correct answer: Factory B

A: (25−18)/25 = 28% reduction. B: (40−28)/40 = 30% reduction. C: (15−12)/15 = 20% reduction. Factory B had the greatest percentage reduction at 30%.

Factory A: (25 − 18)/25 = 7/25 = 28% reduction. Factory B: (40 − 28)/40 = 12/40 = 30% reduction. Factory C: (15 − 12)/15 = 3/15 = 20% reduction. Despite reducing defects by the largest absolute number (12), Factory B also leads in percentage reduction at 30%.

Question 15: A graph shows that a city's carbon emissions (in million tons) follow the equation E = −0.4t + 12, where t = years after 2000. According to this model, in what year will emissions reach zero? (A) 2025 (B) 2028 (C) 2030 (D) 2032

2025
2028
2030 (Correct answer)
2032

Correct answer: 2030

Set E = 0: 0 = −0.4t + 12 → 0.4t = 12 → t = 30. Year 2000 + 30 = 2030.

−0.4t + 12 = 0. Subtract 12 from both sides: −0.4t = −12. Divide by −0.4: t = 30. Since t represents years after 2000, the year is 2000 + 30 = 2030.

Question 16: A pie chart shows that a school's budget is divided as follows: Instruction 48%, Administration 18%, Facilities 14%, Technology 12%, Other 8%. The total budget is $5 million. How much more is spent on Instruction than on Administration and Technology combined? (A) $800,000 (B) $900,000 (C) $1,000,000 (D) $1,200,000

$800,000
$900,000
$1,000,000 (Correct answer)
$1,200,000

Correct answer: $1,000,000

Instruction: 48% × $5M = $2.4M. Admin + Tech: (18%+12%) × $5M = 30% × $5M = $1.5M. Difference: $2.4M − $1.5M = $0.9M = $900,000.

Instruction = 0.48 × $5,000,000 = $2,400,000. Administration = 0.18 × $5,000,000 = $900,000. Technology = 0.12 × $5,000,000 = $600,000. Admin + Tech = $900,000 + $600,000 = $1,500,000. Difference = $2,400,000 − $1,500,000 = $900,000.

Question 17: A scatter plot shows data points that suggest a strong positive correlation between advertising spend and sales revenue. The correlation coefficient is r = 0.92. What does this tell us? (A) Advertising spending causes higher sales (B) About 85% of the variance in sales is explained by advertising spend (C) For every $1 increase in advertising, sales increase by exactly $0.92 (D) The relationship is linear and perfectly predictable

Advertising spending causes higher sales
About 85% of the variance in sales is explained by advertising spend (Correct answer)
For every $1 increase in advertising, sales increase by exactly $0.92
The relationship is linear and perfectly predictable

Correct answer: About 85% of the variance in sales is explained by advertising spend

r² = 0.92² ≈ 0.846. About 84.6% ≈ 85% of the variance in sales is explained by advertising spend. Correlation does not imply causation.

r = 0.92 means strong positive correlation. r² = 0.92² = 0.8464, meaning about 84.6% of variance in sales is explained by advertising spend. Option A is wrong — correlation does not prove causation. Option C confuses r with regression slope. Option D is wrong — r=1 would be perfect, not 0.92.

Question 18: A table shows the number of applications and acceptances at a university over 3 years: Year 1: Applications=8,000, Acceptances=1,200 Year 2: Applications=9,500, Acceptances=1,330 Year 3: Applications=11,000, Acceptances=1,430 In which year was the acceptance rate highest? (A) Year 1 (B) Year 2 (C) Year 3 (D) All years had the same acceptance rate

Year 1 (Correct answer)
Year 2
Year 3
All years had the same acceptance rate

Correct answer: Year 1

Year 1: 1200/8000 = 15%. Year 2: 1330/9500 ≈ 14%. Year 3: 1430/11000 = 13%. Year 1 had the highest acceptance rate.

Year 1: 1,200 ÷ 8,000 = 0.15 = 15%. Year 2: 1,330 ÷ 9,500 ≈ 0.14 = 14%. Year 3: 1,430 ÷ 11,000 ≈ 0.13 = 13%. Although absolute acceptances increased each year, the rate decreased because applications grew faster than acceptances.

Question 19: A graph shows exponential population growth modeled by P = 500 × 2^(t/10), where t = years. What is the population at t = 30? (A) 2,000 (B) 3,000 (C) 4,000 (D) 6,000

2,000
3,000
4,000 (Correct answer)
6,000

Correct answer: 4,000

P = 500 × 2^(30/10) = 500 × 2^3 = 500 × 8 = 4,000.

P = 500 × 2^(t/10). At t = 30: P = 500 × 2^(30/10) = 500 × 2^3 = 500 × 8 = 4,000. The population doubles every 10 years (t=0: 500; t=10: 1,000; t=20: 2,000; t=30: 4,000).

Question 20: A frequency table shows the grades of 40 students: A=8, B=14, C=12, D=4, F=2. What is the probability that a randomly selected student earned either a B or C? (A) 0.55 (B) 0.60 (C) 0.65 (D) 0.70

0.55
0.60
0.65 (Correct answer)
0.70

Correct answer: 0.65

B + C = 14 + 12 = 26. P = 26/40 = 0.65.

Students with B or C: 14 + 12 = 26. Total students: 40. P(B or C) = 26/40 = 0.65. This uses the addition rule for mutually exclusive events: P(B or C) = P(B) + P(C) = 14/40 + 12/40 = 26/40.

Question 21: A stacked line graph shows three regions' share of global renewable energy production over time. In 2015, Region A=40%, Region B=35%, Region C=25%. By 2020, Region A=35%, Region B=40%, Region C=25%. If total production was 2,000 TWh in 2015 and 3,000 TWh in 2020, how much more energy did Region B produce in 2020 vs. 2015? (A) 500 TWh (B) 520 TWh (C) 600 TWh (D) 700 TWh

500 TWh (Correct answer)
520 TWh
600 TWh
700 TWh

Correct answer: 500 TWh

Region B in 2015: 35% × 2000 = 700 TWh. Region B in 2020: 40% × 3000 = 1200 TWh. Increase: 1200 − 700 = 500 TWh.

2015: Region B = 35% × 2,000 TWh = 700 TWh. 2020: Region B = 40% × 3,000 TWh = 1,200 TWh. Increase = 1,200 − 700 = 500 TWh. Despite Region A losing share, Region B gained both in percentage and in total volume.

Question 22: A box-and-whisker plot for a dataset shows: minimum=20, Q1=35, median=50, Q3=70, maximum=95. What is the interquartile range (IQR), and what does it represent? (A) IQR=45; the range of the entire dataset (B) IQR=35; the spread of the middle 50% of data (C) IQR=30; the average distance from the mean (D) IQR=75; the distance from min to max

IQR=45; the range of the entire dataset
IQR=35; the spread of the middle 50% of data (Correct answer)
IQR=30; the average distance from the mean
IQR=75; the distance from min to max

Correct answer: IQR=35; the spread of the middle 50% of data

IQR = Q3 − Q1 = 70 − 35 = 35. The IQR represents the spread of the middle 50% of the data.

IQR = Q3 − Q1 = 70 − 35 = 35. The IQR is the range of the middle 50% of values (from the 25th percentile to the 75th percentile). The full range is max − min = 95 − 20 = 75. The IQR is more robust than the full range because it is not affected by extreme outliers.

Question 23: A comparative table shows the fuel efficiency (mpg) of 5 car models in city and highway driving: Model A: City=22, Highway=30 Model B: City=18, Highway=25 Model C: City=35, Highway=42 Model D: City=28, Highway=35 Model E: City=15, Highway=20 If a driver spends 60% of miles on the highway and 40% in the city, which model gives the best weighted average mpg? (A) Model A (B) Model C (C) Model D (D) Model E

Model A
Model C (Correct answer)
Model D
Model E

Correct answer: Model C

Weighted avg = 0.4(city) + 0.6(highway). C: 0.4(35)+0.6(42) = 14+25.2 = 39.2. All others are lower. Model C wins.

Model A: 0.4(22) + 0.6(30) = 8.8 + 18 = 26.8. Model B: 0.4(18) + 0.6(25) = 7.2 + 15 = 22.2. Model C: 0.4(35) + 0.6(42) = 14 + 25.2 = 39.2. Model D: 0.4(28) + 0.6(35) = 11.2 + 21 = 32.2. Model E: 0.4(15) + 0.6(20) = 6 + 12 = 18. Model C has the highest weighted average at 39.2 mpg.

Question 24: A bar chart shows annual revenue growth rates for a company: Year 1: 5%, Year 2: 8%, Year 3: −3%, Year 4: 12%, Year 5: 6%. If Year 1 starting revenue was $100M, what was the revenue at the end of Year 5? (A) $128M (B) $131M (C) $133M (D) $136M

$128M
$131M
$133M (Correct answer)
$136M

Correct answer: $133M

100 × 1.05 × 1.08 × 0.97 × 1.12 × 1.06 = 100 × 1.3326 ≈ $133M.

Year 1: 100 × 1.05 = 105. Year 2: 105 × 1.08 = 113.4. Year 3: 113.4 × 0.97 = 110.0. Year 4: 110.0 × 1.12 = 123.2. Year 5: 123.2 × 1.06 = 130.6. Wait: let me recalculate precisely: 100 × 1.05 = 105; ×1.08 = 113.4; ×0.97 = 110.0; ×1.12 = 123.2; ×1.06 = 130.6 ≈ $131M. Answer is (B) $131M.

Question 25: A graph shows that a city's water usage (gallons per person per day) decreased from 180 to 135 over 10 years. What was the percentage decrease? (A) 20% (B) 25% (C) 30% (D) 33%

20%
25% (Correct answer)
30%
33%

Correct answer: 25%

% decrease = (180 − 135)/180 × 100 = 45/180 × 100 = 25%.

(180 − 135) / 180 = 45 / 180 = 0.25 = 25%. The city reduced per-person water usage by 25% over the decade.

Question 26: A data table shows the results of a clinical trial: Treatment Group: 200 patients, 160 improved Control Group: 200 patients, 120 improved What is the relative risk reduction (RRR) of the treatment compared to control? (A) 20% (B) 25% (C) 33% (D) 40%

20%
25%
33% (Correct answer)
40%

Correct answer: 33%

Control improvement rate: 120/200 = 60%. Treatment improvement rate: 160/200 = 80%. RRR = (treatment rate − control rate)/control rate = (80%−60%)/60% = 20/60 ≈ 33%.

Control rate = 120/200 = 0.60 = 60%. Treatment rate = 160/200 = 0.80 = 80%. Absolute risk reduction = 80% − 60% = 20%. Relative risk reduction = 20% / 60% = 1/3 ≈ 33.3%.

Question 27: A frequency histogram has 5 bins of width 10: [10,20): 4, [20,30): 8, [30,40): 12, [40,50): 6, [50,60): 2. What is the approximate median of this data set (32 total observations)? (A) 28 (B) 32 (C) 35 (D) 38

28
32
35 (Correct answer)
38

Correct answer: 35

Cumulative: by 30 we have 4+8=12 values (37.5%). By 40 we have 12+12=24 (75%). The 16th and 17th values (median of 32) fall in [30,40). Estimate: 30 + ((16−12)/12)×10 = 30 + 3.3 ≈ 33. Closest answer is 35.

32 observations → median is average of 16th and 17th values. Cumulative counts: up to 20: 4; up to 30: 12; up to 40: 24. Both the 16th and 17th values fall in [30, 40). Interpolating: 30 + ((16−12)/12) × 10 = 30 + 3.33 ≈ 33. Of the choices given, 35 is the closest reasonable estimate.

Question 28: A table shows the average monthly temperature (°C) and ice cream sales (units) for 6 months: Temp: 10, 15, 20, 25, 30, 35 Sales: 200, 350, 500, 650, 800, 950 Which equation best models the relationship? (A) Sales = 30 × Temperature − 100 (B) Sales = 25 × Temperature + 200 (C) Sales = 30 × Temperature − 100 (D) Sales = 20 × Temperature + 200

Sales = 30 × Temperature − 100 (Correct answer)
Sales = 25 × Temperature + 200
Sales = 30 × Temperature − 100
Sales = 20 × Temperature + 200

Correct answer: Sales = 30 × Temperature − 100

From 10→15 (ΔT=5, ΔS=150, slope=30). Check: 30(10)−100=200 ✓, 30(15)−100=350 ✓, 30(20)−100=500 ✓. Equation A works.

Slope = (350 − 200)/(15 − 10) = 150/5 = 30. Using point (10, 200): 200 = 30(10) + b → b = 200 − 300 = −100. Equation: Sales = 30 × Temperature − 100. Verify with T=35: 30(35) − 100 = 1050 − 100 = 950 ✓.

Question 29: A comparative analysis shows two countries' GDP growth: Country A: 2018=$1T, 2023=$1.4T Country B: 2018=$500B, 2023=$800B Which country had a higher percentage GDP growth from 2018 to 2023? (A) Country A with 40% growth (B) Country B with 60% growth (C) Both had equal growth rates (D) Country A with 50% growth

Country A with 40% growth
Country B with 60% growth (Correct answer)
Both had equal growth rates
Country A with 50% growth

Correct answer: Country B with 60% growth

Country A: (1.4−1)/1 = 40%. Country B: (800−500)/500 = 60%. Country B had higher percentage growth.

Country A: ($1.4T − $1T) / $1T = $0.4T / $1T = 40% growth. Country B: ($800B − $500B) / $500B = $300B / $500B = 60% growth. Although Country A added $400B in absolute terms vs. Country B's $300B, Country B's smaller base means its percentage growth (60%) was higher.

Question 30: A normal distribution has mean = 100 and standard deviation = 15. Approximately what percentage of values fall between 85 and 115? (A) 34% (B) 50% (C) 68% (D) 95%

34%
50%
68% (Correct answer)
95%

Correct answer: 68%

85 and 115 are one standard deviation below and above the mean (100 ± 15). By the empirical rule, ~68% of data falls within ±1 SD.

85 = 100 − 15 = mean − 1σ. 115 = 100 + 15 = mean + 1σ. By the empirical rule (68-95-99.7 rule), approximately 68% of data in a normal distribution falls within one standard deviation of the mean.

Question 31: A graph shows that a town's population, which was 50,000 in 2010, grows at 3% per year. A second town starts with 80,000 people but shrinks at 2% per year. In approximately how many years will the first town surpass the second? (A) 8 years (B) 10 years (C) 12 years (D) 15 years

8 years
10 years
12 years
15 years (Correct answer)

Correct answer: 15 years

Town 1: 50000×1.03^t. Town 2: 80000×0.98^t. Set equal: 1.03^t/0.98^t = 80000/50000 = 1.6. (1.03/0.98)^t = (1.0510)^t = 1.6. t = ln(1.6)/ln(1.051) ≈ 0.470/0.0497 ≈ 9.5 ≈ 10 years.

50,000 × 1.03^t = 80,000 × 0.98^t. Dividing: (1.03/0.98)^t = 80,000/50,000 = 1.6. 1.03/0.98 ≈ 1.0510. So 1.051^t = 1.6. Taking logarithms: t × ln(1.051) = ln(1.6). t = ln(1.6)/ln(1.051) ≈ 0.4700/0.0497 ≈ 9.5 years, so about 10 years. (B) is correct.

Question 32: A survey of 500 employees found the following job satisfaction ratings (scale 1–5): Rating 1: 20 employees Rating 2: 60 employees Rating 3: 150 employees Rating 4: 200 employees Rating 5: 70 employees What is the weighted mean satisfaction rating? (A) 3.4 (B) 3.5 (C) 3.6 (D) 3.7

3.4 (Correct answer)
3.5
3.6
3.7

Correct answer: 3.4

Mean = (1×20 + 2×60 + 3×150 + 4×200 + 5×70)/500 = (20+120+450+800+350)/500 = 1740/500 = 3.48 ≈ 3.5.

Sum of (rating × frequency): 1×20=20, 2×60=120, 3×150=450, 4×200=800, 5×70=350. Total = 20+120+450+800+350 = 1,740. Mean = 1,740 ÷ 500 = 3.48. This is closest to 3.5, so (B) is correct.

Question 33: A graph shows a company's monthly user growth (in thousands) following a pattern: Month 1=2, Month 2=6, Month 3=18, Month 4=54. If this pattern continues, how many users (in thousands) does the company have in Month 6? (A) 162 (B) 486 (C) 1,458 (D) 4,374

162
486 (Correct answer)
1,458
4,374

Correct answer: 486

Pattern: each month multiplies by 3. Month 5: 54×3=162. Month 6: 162×3=486.

2 → 6 (×3), 6 → 18 (×3), 18 → 54 (×3). Common ratio = 3. Month 5: 54 × 3 = 162. Month 6: 162 × 3 = 486 thousand users.

GRE Multi-Step Data Analysis and Graphs 1

A company's annual revenue (in millions) over five years was: Year 1: $120M, Year 2: $144M, Year 3: $172.8M, Year 4: $207.4M, Year 5: $248.8M.
What was the approximate average percentage growth rate per year?
(A) 15%
(B) 18%
(C) 20%
(D) 22%

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