Bluebook SAT Test Problem Solving and Data Analysis 1 — Questions and Answers
Question 1: A scatter plot shows hours studied (x) vs. test score (y). The line of best fit is y = 8x + 42. What score does the model predict for 5 hours of study?
- 82 (Correct answer)
- 90
- 72
- 62
Correct answer: 82
y = 8(5) + 42 = 40 + 42 = 82.
The line of best fit y = 8x + 42 models the relationship between study hours (x) and test scores (y). For x = 5: y = 8(5) + 42 = 40 + 42 = 82. The slope of 8 means each additional hour of study is associated with 8 more points on the test. The y-intercept of 42 represents the predicted score with zero hours of study.
Question 2: In a class of 30 students, 18 passed math, 20 passed English, and 12 passed both. How many students passed neither subject?
- 4 (Correct answer)
- 6
- 8
- 2
Correct answer: 4
By inclusion-exclusion: passed at least one = 18+20-12 = 26. Passed neither = 30-26 = 4.
The inclusion-exclusion principle states |A∪B| = |A|+|B|-|A∩B|. Students who passed at least one subject: 18+20-12 = 26. Students who passed neither: 30-26 = 4. Drawing a Venn diagram helps: 12 in both, 6 only in math (18-12), 8 only in English (20-12), total accounted for: 12+6+8=26. Remaining: 30-26=4.
Question 3: The table below shows monthly sales data. What is the mean monthly sales? Month: Jan, Feb, Mar, Apr, May Sales: $3,200, $4,800, $2,600, $5,400, $4,000
- $4,000 (Correct answer)
- $4,200
- $3,800
- $3,600
Correct answer: $4,000
Mean = (3200+4800+2600+5400+4000)/5 = 20000/5 = $4,000.
The mean (average) is calculated by summing all values and dividing by the count. Sum = 3200+4800+2600+5400+4000 = 20,000. Number of months = 5. Mean = 20,000/5 = $4,000. Note that the mean ($4,000) happens to equal one of the data values (May), which is a coincidence — the mean need not be one of the actual data values.
Question 4: A survey of 200 people found that 35% prefer coffee and 45% prefer tea. The rest prefer neither. How many people prefer neither?
- 40 (Correct answer)
- 80
- 60
- 70
Correct answer: 40
35% + 45% = 80% prefer coffee or tea. 20% prefer neither. 0.20 × 200 = 40.
Percentage preferring coffee or tea: 35% + 45% = 80%. Percentage preferring neither: 100% - 80% = 20%. Number of people: 20% × 200 = 0.20 × 200 = 40. This is a straightforward percentage application, but be careful to check whether the categories are mutually exclusive (which they are here as stated).
Question 5: A data set has values: 12, 15, 11, 18, 14, 13, 20, 9, 16, 12. What is the median?
- 13.5 (Correct answer)
- 14
- 12
- 13
Correct answer: 13.5
Sorted: 9,11,12,12,13,14,15,16,18,20. Median = average of 5th and 6th: (13+14)/2 = 13.5.
Sorting the 10 values: 9, 11, 12, 12, 13, 14, 15, 16, 18, 20. With 10 values (even count), the median is the average of the 5th and 6th values. The 5th value is 13, the 6th is 14. Median = (13+14)/2 = 27/2 = 13.5. The median is not affected by extreme values (unlike the mean), making it a robust measure of center.
Question 6: If a bag contains 4 red, 6 blue, and 2 green marbles, what is the probability of randomly selecting a blue marble?
- 1/2 (Correct answer)
- 1/3
- 2/3
- 1/4
Correct answer: 1/2
Total marbles = 12. Blue marbles = 6. P(blue) = 6/12 = 1/2.
Total marbles: 4+6+2 = 12. Favorable outcomes (blue): 6. Probability = 6/12 = 1/2. This means if you randomly select one marble, there's a 50% chance it's blue. Probability is always between 0 and 1, and a probability of 1/2 means the event is equally likely to happen as not to happen.
Question 7: A two-way table shows that 60 students play sports, of whom 25 also play music. Of 40 students who don't play sports, 15 play music. What percentage of all students play music?
- 40% (Correct answer)
- 25%
- 37.5%
- 50%
Correct answer: 40%
Total students = 100. Music players = 25+15 = 40. Percentage = 40/100 = 40%.
Total students: 60+40 = 100. Students playing music: 25 (sports + music) + 15 (no sports + music) = 40. Percentage playing music: 40/100 = 40%. This two-way table problem tests the ability to read category intersections. The data could be organized: sports/music=25, sports/no music=35, no sports/music=15, no sports/no music=25.
Question 8: A company's profits grew from $50,000 in Year 1 to $80,000 in Year 4. What is the average annual increase?
- $10,000 (Correct answer)
- $15,000
- $7,500
- $30,000
Correct answer: $10,000
Total increase = $30,000 over 3 years (from year 1 to year 4). Average = $30,000/3 = $10,000/year.
From Year 1 to Year 4 is a span of 3 years (not 4). Total change: $80,000-$50,000 = $30,000. Average annual increase: $30,000 ÷ 3 = $10,000 per year. A common mistake is dividing by 4 instead of 3 (the number of intervals, not endpoints). Linear growth rate = total change / number of periods.
Bluebook SAT Test Problem Solving and Data Analysis 1
A scatter plot shows hours studied (x) vs. test score (y).
The line of best fit is y = 8x + 42.
What score does the model predict for 5 hours of study?