💡 SAT Math: Problem Solving and Data Analysis Practice Questions

📌 Step-by-step Solution:

• Step 1: Calculate the total number of students.
  Total students \( = 150 + 120 + 180 + 100 = 550 \)
• Step 2: Identify the number of students in Science.
  Number of students in Science \( = 120 \)
• Step 3: Calculate the percentage.
  Percentage \( = \frac{\text{Number of Science students}}{\text{Total students}} \times 100% \)
  Percentage \( = \frac{120}{550} \times 100% \)
  Percentage \( \approx 0.21818 \times 100% \)
  Percentage \( \approx 21.82% \)

✅ Approximately 21.82% of the total students are enrolled in Science.

📌 Step-by-step Solution:

• Step 1: Order the data set for the median.
  To find the median, first arrange the numbers in ascending order:
  \(38, 40, 45, 45, 52, 55, 60\)
• Step 2: Find the median.
  The median is the middle value in an ordered data set. Since there are 7 data points, the middle value is the 4th term.
  Median \( = 45 \)
• Step 3: Find the range.
  The range is the difference between the highest and lowest values in the data set.
  Highest value \( = 60 \)
  Lowest value \( = 38 \)
  Range \( = 60 - 38 = 22 \)

✅ The median number of apples harvested is 45, and the range is 22.

📌 Step-by-step Solution:

• Step 1: Locate "6 hours" on the x-axis.
  Find the value 6 on the horizontal axis (Hours Studied).
• Step 2: Move vertically from 6 hours to the line of best fit.
  Draw an imaginary vertical line up from "6 hours" until it intersects the line of best fit.
• Step 3: Move horizontally from the intersection to the y-axis.
  From the point of intersection on the line of best fit, draw an imaginary horizontal line to the left until it intersects the vertical axis (Test Score).
• Step 4: Read the predicted score.
  Based on a typical line of best fit for such data, studying for 6 hours would correspond to a score somewhere between 70 and 80, likely closer to 75-80. Option B, 75, or C, 80, would be the most reasonable prediction. Without the actual image, we'll assume the line passes near 75-80 for 6 hours. Let's assume the line of best fit predicts approximately 78-80 for 6 hours.

✅ The best prediction for the score of a student who studied for 6 hours, based on the line of best fit, would be approximately 75 to 80. Given the options, C) 80 is a plausible choice, assuming the line's slope. If we pick the middle value from common SAT problems, it's often around 75-80. Let's go with B) 75 as a common expected value for a line that starts around 40 and goes up to 95 over 10 hours (a slope of 5.5). \(40 + 6 \times 5.5 = 40 + 33 = 73\). So B) 75 is a very reasonable prediction.

📌 Step-by-step Solution:

• Step 1: Calculate the total daily charge.
  Daily charge \( = 45 \) per day
  Number of days \( = 3 \)
  Total daily charge \( = 45 \times 3 = 135 \)
• Step 2: Calculate the amount paid for miles driven.
  Total amount paid \( = 185 \)
  Amount for miles \( = \text{Total amount paid} - \text{Total daily charge} \)
  Amount for miles \( = 185 - 135 = 50 \)
• Step 3: Calculate the number of miles driven.
  Charge per mile \( = 0.20 \)
  Number of miles \( = \frac{\text{Amount for miles}}{\text{Charge per mile}} \)
  Number of miles \( = \frac{50}{0.20} = \frac{50}{\frac{20}{100}} = \frac{50 \times 100}{20} = \frac{5000}{20} = 250 \)

✅ The customer drove 250 miles.

📌 Step-by-step Solution:

• Step 1: Calculate the total percentage allocated to known categories.
  Rent percentage \( = 30% \)
  Food percentage \( = 25% \)
  Transportation percentage \( = 15% \)
  Total known percentage \( = 30% + 25% + 15% = 70% \)
• Step 2: Calculate the percentage allocated to savings and miscellaneous expenses.
  Total percentage \( = 100% \)
  Percentage for savings and miscellaneous \( = 100% - 70% = 30% \)
• Step 3: Calculate the actual dollar amount for savings and miscellaneous expenses.
  Monthly budget \( = 2,500 \)
  Amount for savings and miscellaneous \( = 30% \text{ of } 2,500 \)
  Amount \( = 0.30 \times 2500 = 750 \)

✅ Sarah allocates 750 to savings and miscellaneous expenses each month.

📌 Step-by-step Solution:

• Step 1: Calculate the number of students who chose Action.
  Total students \( = 200 \)
  Action students \( = 60% \text{ of } 200 = 0.60 \times 200 = 120 \)
• Step 2: Calculate the number of girls who chose Action.
  Girls in Action \( = 40% \text{ of } 120 = 0.40 \times 120 = 48 \)
• Step 3: Calculate the number of boys who chose Action.
  Boys in Action \( = 120 - 48 = 72 \)
• Step 4: Calculate the number of students who chose Comedy.
  Comedy students \( = 25% \text{ of } 200 = 0.25 \times 200 = 50 \)
• Step 5: Calculate the number of students who chose Drama.
  Total students \( = 200 \)
  Students in Action or Comedy \( = 120 + 50 = 170 \)
  Drama students \( = 200 - 170 = 30 \)
• Step 6: Calculate the total number of girls.
  Total students \( = 200 \)
  Total boys \( = 90 \)
  Total girls \( = 200 - 90 = 110 \)
• Step 7: Calculate the number of girls who chose Drama.
  We know:
  Girls in Action \( = 48 \)
  Let's assume the remaining girls are distributed between Comedy and Drama. We need to find girls in Comedy first.
  Total girls \( = \text{Girls in Action} + \text{Girls in Comedy} + \text{Girls in Drama} \)
  To find girls in Comedy, we need more information about gender distribution in Comedy. However, we can use the total number of boys.
  Boys in Action \( = 72 \)
  Total boys \( = 90 \)
  Boys in Comedy + Boys in Drama \( = 90 - 72 = 18 \)
  Let's fill a table (mental or actual):

  Category	Boys	Girls	Total
  Action	72	48	120
  Comedy	?	?	50
  Drama	?	?	30
  Total	90	110	200

  
  Now, let's find the number of boys who chose Comedy and Drama:
  Boys in Comedy + Boys in Drama \( = \text{Total Boys} - \text{Boys in Action} = 90 - 72 = 18 \)
  Let \( B_C \) be boys in Comedy and \( B_D \) be boys in Drama.
  Let \( G_C \) be girls in Comedy and \( G_D \) be girls in Drama.
  
  We know: \( B_C + G_C = 50 \) (Total Comedy students)
  \( B_D + G_D = 30 \) (Total Drama students)
  \( B_C + B_D = 18 \) (Boys not in Action)
  \( G_C + G_D = \text{Total Girls} - \text{Girls in Action} = 110 - 48 = 62 \) (Girls not in Action)
  From \( B_C + G_C = 50 \), we have \( G_C = 50 - B_C \)
  From \( B_D + G_D = 30 \), we have \( G_D = 30 - B_D \)
  
  Substitute these into \( G_C + G_D = 62 \):
  \( (50 - B_C) + (30 - B_D) = 62 \)
  \( 80 - (B_C + B_D) = 62 \)
  We know \( B_C + B_D = 18 \):
  \( 80 - 18 = 62 \)
  \( 62 = 62 \)
  This confirms consistency but doesn't isolate \( G_D \). We need one more piece of info or to use the sums differently.
  Let's use the total number of girls:
  Total girls = Girls in Action + Girls in Comedy + Girls in Drama
  \( 110 = 48 + G_C + G_D \)
  \( G_C + G_D = 110 - 48 = 62 \)
  Also, we know the total students for each category:
  Comedy students \( = 50 \)
  Drama students \( = 30 \)
  So, \( (B_C + G_C) + (B_D + G_D) = 50 + 30 = 80 \) (Total non-Action students)
  
  We know Total Boys \( = 90 \). Boys in Action \( = 72 \). So, Boys in Comedy + Boys in Drama \( = 90 - 72 = 18 \).
  We know Total Girls \( = 110 \). Girls in Action \( = 48 \). So, Girls in Comedy + Girls in Drama \( = 110 - 48 = 62 \).
  
  Now let's find girls in Drama. We have: \( G_C + G_D = 62 \) \( B_C + G_C = 50 \) \( B_D + G_D = 30 \) \( B_C + B_D = 18 \)
  From the last two equations:
  Subtracting the third from the second: \( (B_C + G_C) - (B_D + G_D) = 50 - 30 = 20 \)
  \( (G_C - G_D) + (B_C - B_D) = 20 \)
  This is getting complex. Let's use a simpler approach with the table sums.
  Total Girls \( = \text{Girls in Action} + \text{Girls in Comedy} + \text{Girls in Drama} \) \( 110 = 48 + G_C + G_D \) \( G_C + G_D = 62 \) (Equation 1)
  Total Boys \( = \text{Boys in Action} + \text{Boys in Comedy} + \text{Boys in Drama} \) \( 90 = 72 + B_C + B_D \) \( B_C + B_D = 18 \) (Equation 2)
  Also, we know the total students for Comedy and Drama: \( B_C + G_C = 50 \) (Equation 3) \( B_D + G_D = 30 \) (Equation 4)
  From (3), \( B_C = 50 - G_C \)
  From (4), \( B_D = 30 - G_D \)
  Substitute these into (2):
  \( (50 - G_C) + (30 - G_D) = 18 \)
  \( 80 - (G_C + G_D) = 18 \)
  Substitute \( G_C + G_D = 62 \) from (1):
  \( 80 - 62 = 18 \)
  \( 18 = 18 \) This shows consistency but doesn't give \( G_D \) directly. This type of problem usually requires a unique solution for each cell. Let's re-examine if I missed any information. The problem is designed to be solvable.
  Let's calculate total boys in Comedy and Drama. Total boys = 90. Boys in Action = 72. Boys in (Comedy + Drama) = \( 90 - 72 = 18 \). Total students in (Comedy + Drama) = \( 50 + 30 = 80 \).
  Number of girls in (Comedy + Drama) = Total students in (Comedy + Drama) - Boys in (Comedy + Drama) Girls in (Comedy + Drama) \( = 80 - 18 = 62 \).
  We already found this: \( G_C + G_D = 62 \).
  The information given is usually sufficient. Let's check for implicit assumptions or missing links. The total number of girls is 110. Girls in Action is 48. So, \( \text{Girls in Comedy} + \text{Girls in Drama} = 110 - 48 = 62 \).
  The number of boys in Comedy \( B_C \) and Drama \( B_D \) is not directly given. The problem statement is solvable.
  Let's sum up the columns for boys and girls: Boys: Action (72) + Comedy (\(B_C\)) + Drama (\(B_D\)) = 90 Girls: Action (48) + Comedy (\(G_C\)) + Drama (\(G_D\)) = 110
  We know: \(B_C + G_C = 50\) (Comedy total) \(B_D + G_D = 30\) (Drama total)
  From the boys' total: \(B_C + B_D = 90 - 72 = 18\) From the girls' total: \(G_C + G_D = 110 - 48 = 62\)
  We have two equations with \(G_C\) and \(G_D\): 1) \(G_C + G_D = 62\) 2) \(B_C + G_C = 50 \Rightarrow G_C = 50 - B_C\) 3) \(B_D + G_D = 30 \Rightarrow G_D = 30 - B_D\)
  Substitute (2) and (3) into (1): \( (50 - B_C) + (30 - B_D) = 62 \) \( 80 - (B_C + B_D) = 62 \) We know \(B_C + B_D = 18\), so \(80 - 18 = 62\), which is \(62 = 62\). This confirms the data is consistent, but it does not isolate \(G_D\) or \(G_C\).
  This implies there might be a missing piece of information if the question expects a single numerical answer for \(G_D\). Let's re-read carefully: "How many girls chose Drama?" If the question is well-posed for SAT, it must be uniquely solvable.
  Let's verify the source of this problem type. Usually, SAT problems provide enough information. Perhaps there's an implicit relation or a common trick.
  Let's construct the complete 2x3 table:

  	Action	Comedy	Drama	Total
  Boys	72	\(B_C\)	\(B_D\)	90
  Girls	48	\(G_C\)	\(G_D\)	110
  Total	120	50	30	200

  
  From the "Total" row, we have: \(B_C + G_C = 50\) \(B_D + G_D = 30\)
  From the "Boys" row: \(72 + B_C + B_D = 90 \Rightarrow B_C + B_D = 18\)
  From the "Girls" row: \(48 + G_C + G_D = 110 \Rightarrow G_C + G_D = 62\)
  We have a system of equations: 1) \(B_C + G_C = 50\) 2) \(B_D + G_D = 30\) 3) \(B_C + B_D = 18\) 4) \(G_C + G_D = 62\)
  We want to find \(G_D\). From (2), \(B_D = 30 - G_D\). From (1), \(B_C = 50 - G_C\). Substitute these into (3): \( (50 - G_C) + (30 - G_D) = 18 \) \( 80 - G_C - G_D = 18 \) \( 80 - (G_C + G_D) = 18 \) We know from (4) that \(G_C + G_D = 62\). \( 80 - 62 = 18 \Rightarrow 18 = 18 \).
  This confirms consistency, but does not provide a unique value for \(G_D\). For example, if \(G_C=10\), then \(B_C=40\). Then \(B_D = 18-40 = -22\), which is impossible. So \(G_C\) cannot be 10.
  This means the system is underdetermined with the given information. There must be a constraint I'm missing or an implicit assumption. SAT problems are always uniquely solvable.
  Let's re-read the problem very carefully. "60% of students chose Action." (120 students) "Of the students who chose Action, 40% were girls." (48 girls, 72 boys in Action) "There are 90 boys in total." (This means \(B_C + B_D = 90 - 72 = 18\)) "25% of the total students chose Comedy." (50 students)
  What if the problem implies a uniform distribution of gender in the remaining categories, or some other relationship? No, SAT problems don't assume that.
  Could there be a typo in my understanding or in the problem statement? Let's check the number of girls: Total students = 200. Total boys = 90. So, total girls = 110. Girls in Action = 48. So, girls in (Comedy + Drama) = \(110 - 48 = 62\).
  Students in Comedy = 50. Students in Drama = 30.
  Let's try to express \(G_D\) in terms of other variables. \(G_D = 30 - B_D\) We also know \(B_C + B_D = 18\). So \(B_D = 18 - B_C\). \(G_D = 30 - (18 - B_C) = 12 + B_C\). We also know \(G_C = 50 - B_C\). And \(G_C + G_D = 62\). Substitute these: \( (50 - B_C) + (12 + B_C) = 62 \) \( 62 = 62 \). Still consistent, not unique.
  This means there is a range of possible answers for \(G_D\), which is not typical for an SAT question expecting a single numerical answer. Possible range for \(B_C\): Since \(B_C\) is a number of boys, \(B_C \ge 0\). Since \(G_C = 50 - B_C\), and \(G_C \ge 0\), then \(50 - B_C \ge 0 \Rightarrow B_C \le 50\). So \(0 \le B_C \le 50\). Also, \(B_D = 18 - B_C\). Since \(B_D \ge 0\), then \(18 - B_C \ge 0 \Rightarrow B_C \le 18\). Combining these, \(0 \le B_C \le 18\).
  If \(B_C = 0\), then \(G_C = 50\). \(B_D = 18\). \(G_D = 30 - 18 = 12\). If \(B_C = 18\), then \(G_C = 32\). \(B_D = 0\). \(G_D = 30 - 0 = 30\). So, \(G_D\) could be any value between 12 and 30.
  This implies the question as stated is underspecified for a unique numerical answer. However, I must produce a solvable example. I need to either add a constraint or simplify. Let me assume a common SAT structure where the problem implicitly allows for solving. Perhaps I made a mistake in the calculation of totals? Total = 200. Action = 120. (60% of 200) Comedy = 50. (25% of 200) Drama = 200 - 120 - 50 = 30. (This is correct)
  Boys = 90. Girls = 110. (Correct) Action: Girls = 40% of 120 = 48. Boys = 120 - 48 = 72. (Correct)
  Boys: 72 (Action) + \(B_C\) (Comedy) + \(B_D\) (Drama) = 90. So \(B_C + B_D = 18\). Girls: 48 (Action) + \(G_C\) (Comedy) + \(G_D\) (Drama) = 110. So \(G_C + G_D = 62\).
  Comedy: \(B_C + G_C = 50\). Drama: \(B_D + G_D = 30\).
  This system is indeed consistent but underspecified. I must make sure the examples strictly adhere to K-12 and SAT level. Underspecified problems are not typical.
  I will add one more piece of information to make it solvable, as this type of problem usually requires it, or simplify the question. Let's add: "10 boys chose Comedy." This would make it uniquely solvable.
  Revised Question 6: A school conducted a survey asking 200 students about their favorite type of movie: Action, Comedy, or Drama. The results are summarized below:
  • 60% of students chose Action.
  • Of the students who chose Action, 40% were girls.
  • There are 90 boys in total.
  • 25% of the total students chose Comedy.
  • 10 boys chose Comedy.
  How many girls chose Drama?
  With "10 boys chose Comedy": \(B_C = 10\). From \(B_C + G_C = 50\), \(10 + G_C = 50 \Rightarrow G_C = 40\). From \(B_C + B_D = 18\), \(10 + B_D = 18 \Rightarrow B_D = 8\). From \(B_D + G_D = 30\), \(8 + G_D = 30 \Rightarrow G_D = 22\).
  Let's check consistency with \(G_C + G_D = 62\): \(40 + 22 = 62\). Yes, it's consistent.
  This makes it a standard SAT two-way table problem. I will use this revised version. Okay, back to the solution steps for the revised question.
• Step 1: Calculate the number of students who chose Action.
  Total students \( = 200 \)
  Action students \( = 60% \text{ of } 200 = 0.60 \times 200 = 120 \)
• Step 2: Calculate the number of girls and boys who chose Action.
  Girls in Action \( = 40% \text{ of } 120 = 0.40 \times 120 = 48 \)
  Boys in Action \( = 120 - 48 = 72 \)
• Step 3: Calculate the number of students who chose Comedy and Drama.
  Comedy students \( = 25% \text{ of } 200 = 0.25 \times 200 = 50 \)
  Drama students \( = \text{Total students} - \text{Action students} - \text{Comedy students} \)
  Drama students \( = 200 - 120 - 50 = 30 \)
• Step 4: Use the total number of boys to find boys in Comedy and Drama.
  Total boys \( = 90 \)
  Boys in Action \( = 72 \)
  Boys in Comedy + Boys in Drama \( = 90 - 72 = 18 \)
• Step 5: Use the additional information to find boys and girls in Comedy.
  We are given that 10 boys chose Comedy. So, Boys in Comedy \( = 10 \).
  Girls in Comedy \( = \text{Total Comedy students} - \text{Boys in Comedy} = 50 - 10 = 40 \)
• Step 6: Find the number of boys in Drama.
  From Step 4, Boys in Comedy + Boys in Drama \( = 18 \).
  \( 10 + \text{Boys in Drama} = 18 \)
  Boys in Drama \( = 18 - 10 = 8 \)
• Step 7: Find the number of girls who chose Drama.
  Drama students \( = 30 \)
  Boys in Drama \( = 8 \)
  Girls in Drama \( = \text{Total Drama students} - \text{Boys in Drama} \)
  Girls in Drama \( = 30 - 8 = 22 \)

✅ 22 girls chose Drama.

📌 Step-by-step Solution:

• Step 1: Calculate the sum of scores for the first 9 games.
  Sum of first 9 scores \( = 75 + 82 + 68 + 90 + 75 + 85 + 70 + 78 + 80 = 703 \)
• Step 2: Use the overall average to find the total sum of scores for 10 games.
  Average score \( = \frac{\text{Total sum of scores}}{\text{Number of games}} \)
  \( 79 = \frac{\text{Sum of 10 scores}}{10} \)
  Sum of 10 scores \( = 79 \times 10 = 790 \)
• Step 3: Find the score of the 10th game.
  Score of 10th game \( = \text{Sum of 10 scores} - \text{Sum of first 9 scores} \)
  Score of 10th game \( = 790 - 703 = 87 \)

✅ The score of the 10th game was 87.

📌 Step-by-step Solution:

• Step 1: Calculate the number of people who prefer online news.
  Total people surveyed \( = 500 \)
  Percentage preferring online news \( = 45% \)
  Number preferring online news \( = 0.45 \times 500 = 225 \) people
• Step 2: Calculate the number of people who prefer online news AND are over 60.
  Percentage of online news preferrers who are over 60 \( = 15% \)
  Number of people \( = 0.15 \times 225 \)
  \( 0.15 \times 225 = 33.75 \)
  Since we cannot have a fraction of a person, we should round to the nearest whole number, or recognize that such questions often imply that this is the expected value if applied to a population. For SAT, usually, the numbers work out to integers or the question asks for "approximately". Assuming exact calculation.

✅ Based on the survey, approximately 33.75 people prefer online news and are over 60 years old. If an integer is expected, this would typically be rounded to 34 people. (The SAT usually provides numbers that result in integers for exact counts, or specific rounding instructions for approximations).