25 hard SAT One-variable data: center & spread questions
Real questions from the SAT Climb bank, all at the hard difficulty tier. Pick an answer before you open the explanation. Every question tells you why the answer is right and why each wrong choice is tempting.
Math · Problem-Solving and Data Analysis~2 per testHard tier
Ignores frequencies and averages the distinct values.
Confuses spread (SD) with center (mean).
Question 1Hard
Data set X consists of 9 values: 12, 15, 18, 18, 20, 22, 24, 30, 45. How much greater is the mean than the median for data set X?
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Why B is right
The mean is (12+15+18+18+20+22+24+30+45)/9 = 204/9 ≈ 22.67. The median of the 9 values in ascending order is the 5th value, which is 20. The difference is 22.67 − 20 = 2.67, or approximately 2.7.
Why the others are wrong
AThis incorrectly computes the difference as 22 − 20, using the rounded mean instead of the precise value.
CThis incorrectly uses 23 as the median (perhaps averaging the 4th and 5th values as if there were an even number of values), giving 23 − 20 = 3.
DThis incorrectly identifies the median as the 4th value (18) due to an off-by-one error in counting to the middle, giving 22.67 − 18 ≈ 4.7.
Question 2Hard
A data set consists of the 8 values shown: 34, 38, 38, 42, 46, 46, 50, 54. A 9th value, w, is added to the data set. If w is less than the original median, by how much does the median decrease?
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Why B is right
The original median of the 8 values is the average of the 4th and 5th values: (42 + 46)/2 = 44. When a 9th value w (where w < 44) is added, the new median is the 5th value of the ordered 9-value set. Since w is less than 44, it will be inserted before the original 4th or 5th value, making the original 4th value (42) the new 5th value. The new median is 42, so the decrease is 44 - 42 = 2.
Why the others are wrong
AThis incorrectly assumes that adding a value below the median does not change the median, failing to account for how median position shifts with odd versus even counts.
CThis uses the difference between the mean and median or miscalculates the shift by considering the wrong values as the new median.
DThis results from finding the difference between the original median and a value two positions away, rather than properly identifying the new median position.
Question 3Hard
Data set R consists of 7 integers: 12, 14, 16, 18, 20, 22, and 48. Data set S consists of the same 7 integers as data set R. Which of the following is greater for data set R than for data set S?
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Why D is right
Since data sets R and S consist of the exact same 7 integers, their mean, median, and range are all identical. Therefore, none of these measures is greater for data set R than for data set S.
Why the others are wrong
AThis incorrectly assumes the mean differs between identical data sets.
BThis incorrectly assumes the median differs between identical data sets.
CThis incorrectly assumes the range differs between identical data sets.
Question 4Hard
Data set A: 10, 12, 14, 16, 18
Data set B: 8, 13, 14, 15, 20
Which data set has the greater standard deviation?
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Why B is right
Data set A has values evenly spaced around the mean of 14, all within 4 units. Data set B has the same mean of 14 but contains values 8 and 20, which are 6 units from the mean, showing greater spread from the center.
Why the others are wrong
AThis incorrectly focuses on the range between consecutive values in data set A, which is constant, but standard deviation measures distance from the mean, not uniformity of spacing.
CThis results from noting both sets have the same mean and sample size, but standard deviation depends on how far individual values deviate from that mean, which differs between the sets.
DAll necessary information is provided; standard deviation can be compared qualitatively by examining how clustered the values are around their respective means.
Question 5Hard
A data set consists of 7 values with a mean of 16 and a range of 18. If the maximum value is 26, what is the minimum value?
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Why C is right
The range is the difference between the maximum and minimum values. Given that the maximum is 26 and the range is 18, the minimum is 26 - 18 = 8. The mean is not needed to answer this question, though it confirms the data set is consistent.
Why the others are wrong
AThis results from subtracting 19 instead of 18, an off-by-one error.
BThis incorrectly uses the mean 16 in the calculation instead of the maximum value.
DThis incorrectly uses the range value as the answer without performing the subtraction.
Question 6Hard
Data set P consists of 9 values with a mean of 24 and a median of 20. Data set Q consists of all the values in data set P as well as the value 24. Which of the following must be true for data set Q? I. The mean is greater than the median. II. The range is greater than the range of data set P.
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Why A is right
Adding 24 to data set P creates data set Q with 10 values. The new mean is (9×24+24)/10=240/10=24. The new median is the average of the 5th and 6th values. Since the original median was 20, and we're adding 24, the 5th value is at most 20 and the 6th value is at least 20. If the 6th value was originally ≥24, then the new median could be (20+24)/2=22 or higher. But we know mean = 24>20, so there must be values above 24. The new median will be between 20 and 24, so mean > median. Statement I is true. Statement II is false because adding a value that's not an extreme doesn't necessarily change the range.
Why the others are wrong
BThis assumes the range must increase, but adding 24 (which is the mean) doesn't affect the range if 24 is between the existing minimum and maximum.
CThis incorrectly assumes both statements are true, but the range doesn't necessarily increase.
DThis incorrectly concludes that neither statement is true, missing that the mean remains greater than the median in data set Q.
Question 7Hard
Data set X consists of 9 values: 12, 15, 15, 18, 20, 22, 24, 27, and 45. How much greater is the mean of data set X than the median of data set X?
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Why B is right
The median of the 9 values in ascending order is the 5th value, which is 20. The sum of the values is 198, so the mean is 198/9 = 22. Therefore, the mean is 22 - 20 = 2 greater than the median.
Why the others are wrong
AThis incorrectly assumes the mean and median are equal, which would only be true if the distribution were perfectly symmetric.
CThis incorrectly calculates the median as 19 instead of 20, likely by averaging the 4th and 5th values instead of taking the middle value of the odd-sized dataset.
DThis results from an off-by-one error in computing either the mean (as 21) or the median (as 21).
Question 8Hard
Data set R consists of 15 integers with a median of 35 and a mean of 38. Data set S is created by adding a single integer to data set R. The median of data set S is 36. Which of the following could be the mean of data set S?
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Why D is right
The sum of data set R is 15 × 38 = 570. The median of 15 integers is the 8th value, which is 35. When one integer x is added, data set S has 16 integers with median 36, meaning the average of the 8th and 9th values is 36. Since the original 8th value was 35 and the new median is 36, the added value must be positioned such that it shifts the middle upward. For the median to increase to 36, the added value must be at least 37 (to push the 8th and 9th values to average 36). If x=106, the new sum is 570 + 106 = 676, and the mean is 676/16 = 42.25. Rounding or choosing an appropriate value gives a mean around 42.
Why the others are wrong
AThis incorrectly assumes the mean of S equals the new median.
BThis assumes the mean equals the median, ignoring that the original mean was higher than the median.
CThis assumes the mean remains unchanged despite adding a new value, which is only possible if the added value exactly equals the original mean.
Question 9Hard
Data set J: 40, 44, 46, 48, 52
Data set K: 42, 44, 46, 48, 50
Which data set has the greater standard deviation?
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Why C is right
Both data sets have a mean of 46. Data set J has extreme values of 40 and 52, each 6 units from the mean. Data set K has extreme values of 42 and 50, each 4 units from the mean, indicating less variability.
Why the others are wrong
AAlthough both sets share the same three middle values and have the same total range of 12, the extreme values are positioned differently relative to the mean, yielding different standard deviations.
BThis reverses the comparison, possibly from observing that data set K's values are more evenly distributed but misinterpreting even distribution as higher variability.
DThe context of the data is irrelevant for comparing standard deviations; the numerical values alone determine which set has greater spread around its mean.
Question 10Hard
Data set R consists of the 13 positive integers from 40 to 52. Data set S is created by removing the value 52 from data set R. How much greater is the range of data set R than the range of data set S?
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Why B is right
Data set R has values from 40 to 52, so the range is 52-40=12. When 52 is removed, data set S has values from 40 to 51, giving a range of 51-40=11. The difference in ranges is 12-11=1.
Why the others are wrong
AThis assumes that removing the maximum value does not affect the range, failing to recognize that the new maximum becomes 51.
CThis double-counts the effect of removing 52, possibly subtracting 2 from the original range incorrectly.
DThis confuses the range with another measure, possibly the original range itself or the difference between means.
Question 11Hard
A data set consists of 9 values: 6, 9, 12, 15, 18, 21, 24, 27, and m, where m > 27. If the mean of the data set is 21, by how much is the mean greater than the median?
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Why B is right
The mean is 21, so the sum of the 9 values is 9 × 21 = 189. The sum of the known values is 6+9+12+15+18+21+24+27 = 132, so m=189−132=57. The data set in order is 6, 9, 12, 15, 18, 21, 24, 27, 57. The median is the 5th value, which is 18. The difference between the mean and median is 21 - 18 = 3.
Why the others are wrong
AThis results from computing the median as 19 instead of 18, possibly by averaging adjacent values incorrectly.
CThis results from computing the median as 17 or the mean as 22, possibly from a computational error in finding m.
DThis results from computing the median as 16 by selecting the 4th value instead of the 5th value.
Question 12Hard
A data set contains 9 values: 12, 15, 18, 18, 20, 22, 24, 30, and 45. By how much is the mean greater than the median for this data set?
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Why B is right
The median is the middle value when the 9 values are in order: 20. The mean is the sum divided by 9: (12+15+18+18+20+22+24+30+45)/9 = 204/9 ≈ 22.7. The difference is 22.7 - 20 = 2.7.
Why the others are wrong
AThis incorrectly uses the difference between the mean and a value near the mean rather than computing the actual median.
CThis results from an arithmetic error in computing the mean or from treating the median as the average of adjacent values incorrectly.
DThis results from computing the median as the average of the 4th and 6th values (off-by-one error in identifying the middle position).
Question 13Hard
Data set M consists of 11 integers. The mean of data set M is 45, and the median is 42. If the smallest integer in data set M is increased by 15, which of the following statements must be true about the resulting data set?
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Why B is right
The median of 11 integers is the 6th value when ordered, which is 42. Increasing the smallest integer by 15 does not change the 6th value unless the smallest integer becomes larger than the original 6th value. Since the smallest value is being increased and was originally the minimum, it is highly unlikely to exceed 42 (the median) after adding only 15, so the median remains 42. However, the sum of all 11 integers increases by 15, so the mean increases by 15/11.
Why the others are wrong
AThis incorrectly assumes that changing any value in the dataset affects the median, but the median depends only on the middle position.
CThis reverses the effects, incorrectly thinking the mean is unaffected while the median changes.
DThis ignores that increasing any value in the dataset increases the sum and therefore the mean.
Question 14Hard
Data set P: 22, 24, 25, 26, 28
Data set Q: 20, 24, 25, 26, 30
Which data set has the greater standard deviation?
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Why B is right
Both data sets have a mean of 25. In data set P, the extreme values are 22 and 28, each 3 units from the mean. In data set Q, the extreme values are 20 and 30, each 5 units from the mean, indicating greater spread.
Why the others are wrong
AThis incorrectly assumes that because data set P has a smaller maximum gap between consecutive values (4 versus 6 in Q), it must have greater variability overall.
CWhile both sets have identical middle three values, the extreme values in data set Q are farther from the mean, creating unequal standard deviations.
DThe qualitative comparison of spread does not depend on whether the data represent samples or populations; data set Q shows objectively greater dispersion in either case.
Question 15Hard
A data set consists of 5 values: 20, 24, 28, 32, and k, where k > 32. The mean of the data set is 30. By how much is the mean greater than the median?
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Why C is right
The mean is 30, so the sum of the 5 values is 5 × 30 = 150. The sum of the known values is 20+24+28+32 = 104, so k=150−104=46. The data set in order is 20, 24, 28, 32, 46. The median is the 3rd value, which is 28. The difference between the mean and median is 30 - 28 = 2.
Why the others are wrong
AThis incorrectly assumes the mean and median are equal, ignoring the effect of the outlier k = 46.
BThis results from computing the median as 29 by averaging 28 and 30, or from a computational error in finding k.
DThis results from computing the median as 27 by selecting the wrong middle value or averaging incorrectly.
Question 16Hard
A data set contains 9 test scores: 72, 75, 78, 81, 84, 84, 87, 90, and 96. A 10th score is added to the data set. If the median of the 10 scores is 84, what is the smallest possible value for the 10th score?
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Why C is right
With 10 values, the median is the average of the 5th and 6th values when ordered. Currently, the 5th value is 84 and the 6th is 84. For the median to be 84, the average of the 5th and 6th must be 84. If the new score is 84 or greater, it won't displace the current 5th and 6th values. If it's less than 84, it shifts positions. Testing: adding 84 keeps the 5th and 6th as 84 and 84, giving median 84. Any value from 84 through 96 works, so the smallest is 84.
Why the others are wrong
AThis incorrectly assumes a lower value can maintain the required median without considering position shifts.
BThis results from an off-by-one error in determining which positions define the median for 10 values.
DThis ignores that a value of 84 already satisfies the condition and incorrectly requires a higher value.
Question 17Hard
Data set T: 50, 54, 57, 60, 64
Data set U: 52, 55, 57, 59, 62
Which data set has the greater standard deviation?
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Why C is right
Data set T has a mean of 57 with extreme values at 50 and 64, which are 7 units from the mean. Data set U has a mean of 57 with extreme values at 52 and 62, which are 5 units from the mean, showing less spread.
Why the others are wrong
AHaving the same number of values does not guarantee equal standard deviations; variability depends on how the values are distributed around the mean.
BThis reverses the comparison, possibly from noting that data set U has a smaller maximum value but failing to account for the distances from the mean.
DAlthough both data sets have the same mean and range of 14 and 10 respectively, the distribution of individual values around the mean differs, resulting in unequal standard deviations.
Question 18Hard
A data set consists of 5 values. Four of the values are 10, 14, 18, and 22. The mean of the data set is 18. By how much is the mean greater than the median?
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Why A is right
The mean is 18, so the sum of the 5 values is 5 × 18 = 90. The sum of the known values is 10+14+18+22 = 64, so the fifth value is 90 - 64 = 26. The data set in order is 10, 14, 18, 22, 26. The median is the 3rd value, which is 18. The difference between the mean and median is 18 - 18 = 0.
Why the others are wrong
BThis results from computing the median as 17 or 19, possibly by averaging adjacent values incorrectly.
CThis results from computing the median as 16 by selecting the 2nd value or the mean as 20 through a computational error.
DThis results from computing the median as 14 by selecting the 2nd value instead of the 3rd value.
Question 19Hard
The 7 values in data set S have a mean of 30 and a median of 28. Data set W is created by multiplying each value in data set S by 2 and then subtracting 10 from each result. What is the median of data set W?
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Why A is right
The median of data set S is 28 (the 4th value when 7 values are arranged in order). Each value in S is transformed by the rule: new value = 2(old value) - 10. Applying this to the median: 2(28) - 10 = 56 - 10 = 46. Since the transformation is applied to every value, the ordering is preserved, so the median of W is the transformed median of S, which is 46.
Why the others are wrong
BThis ignores the subtraction step in the transformation and incorrectly uses only 2(28) = 56, then somehow arrives at 50, representing a misunderstanding of the outlier effect of the -10 adjustment.
CThis uses the mean of S instead of the median, computing 2(30) - 10 = 50, then mistakenly adding instead of properly executing the transformation, arriving at 56.
DThis is an off-by-one error, likely computing 2(30) = 60 using the mean instead of median and omitting the subtraction entirely.
Question 20Hard
Data set X: 5, 9, 10, 11, 15
Data set Y: 7, 9, 10, 11, 13
Which data set has the greater standard deviation?
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Why C is right
Both data sets have a mean of 10. Data set X has extreme values of 5 and 15, each 5 units from the mean. Data set Y has extreme values of 7 and 13, each only 3 units from the mean, showing less spread.
Why the others are wrong
AThis reverses the comparison, possibly by focusing on the fact that data set Y has a smaller range of values, but a smaller range indicates less variability, not more.
BAlthough both sets share the same three middle values, their extreme values differ in distance from the mean, resulting in different standard deviations.
DSince both data sets are presented in the same (unspecified) units, the comparison of standard deviations is valid and does not depend on knowing what those units are.
Question 21Hard
A data set consists of 7 values with a sum of 168 and a median of 22. If the smallest value is 18 and the largest value is 32, what is the mean of the data set?
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Why B is right
The mean of a data set is the sum divided by the number of values. The sum is 168 and there are 7 values, so the mean is 168/7 = 24. The median and the smallest/largest values are provided as additional information to verify consistency, but the mean is directly computed from the sum.
Why the others are wrong
AThis incorrectly uses the median value as the answer instead of computing the mean from the sum.
CThis results from adding the smallest and largest values and dividing by 2, which gives (18+32)/2 = 25, confusing this with the mean.
DThis results from an arithmetic error, possibly computing 168/7 incorrectly as 26 instead of 24.
Question 22Hard
A data set of 15 values has a mean of 42 and a median of 38. If a value of 90 is added to the data set, which of the following statements must be true?
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Why A is right
The original sum is 15 × 42 = 630. Adding 90 gives a new sum of 720, and the new mean is 720/16 = 45. The original median (8th value of 15) is 38. With 16 values, the new median is the average of the 8th and 9th values. Since 90 is larger than all or most values, the 8th and 9th values remain near or at 38, making the new median approximately 38 or slightly higher (but certainly less than 45). Thus, the new mean (45) is greater than the new median.
Why the others are wrong
BThis incorrectly assumes adding a large outlier increases the median more than the mean.
CThis ignores the outlier effect, incorrectly assuming mean and median change proportionally.
DThis results from an off-by-one error in identifying the new median position, confusing it with the old mean.
Question 23Hard
The ages in years of 9 committee members are 34, 38, 41, 43, 47, 50, 53, 58, and 71. A 10th member joins the committee. If the median age of the 10 members is 48, what is the age in years of the 10th member?
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Why C is right
With 10 members, the median is the average of the 5th and 6th values when ordered. Currently, the 5th value is 47 and the 6th is 50. For the median to be 48, we need (5th+6th)/2=48, so 5th+6th=96. If the new member's age is between 47 and 50, say 49, the ordered list becomes: 34,38,41,43,47,49,50,53,58,71. The 5th is 47 and 6th is 49, giving median (47+49)/2 = 48. So the 10th member is 49 years old.
Why the others are wrong
AThis incorrectly assumes the new member's age equals the current 5th value without considering the averaging requirement.
BThis results from assuming the new member's age equals the target median directly.
DThis ignores the need to balance the 5th and 6th positions to achieve the required average.
Question 24Hard
Data set S consists of 15 measurements with a mean of 8.4 and a median of 7.2. If three additional measurements of 12, 14, and 16 are added to create data set T, which of the following correctly describes the relationship between the mean and median of data set T?
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Why A is right
The original sum is 15×8.4=126. Adding 12, 14, and 16 gives a new sum of 126+42=168. The new mean is 168/18=9.33. For 18 values, the median is the average of the 9th and 10th values. Since we added three values (12, 14, 16) all above the original median of 7.2, the new median will be higher than 7.2 but likely below 9.33 because the distribution was already right-skewed. Therefore, the mean remains greater than the median.
Why the others are wrong
BThis reverses the relationship, incorrectly assuming the added values shift the median above the mean.
CThis ignores the effect of adding high outliers, which increases the mean more than the median.
DThis is an off-by-one calculation error, using the original difference without proper recalculation.
Question 25Hard
A data set consists of 9 values: 12, 15, 15, 18, 20, 22, 24, 24, and 30. A 10th value, x, is added to this data set, and the resulting mean is equal to the original median. What is the value of x?
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Why B is right
The original median of the 9 values is the 5th value when ordered, which is 20. The sum of the original 9 values is 180, so the mean of the 10 values is (180+x)/10. Setting this equal to 20 gives (180+x)/10=20, so 180+x=200, and x=20.
Why the others are wrong
AThis incorrectly uses the original mean (180/9 = 20) and solves for when the new median equals 18, misunderstanding which statistic should equal which.
CThis uses the average of the original mean and median without properly accounting for how adding a value affects the mean calculation.
DThis results from finding the median position in the new 10-value set (between 5th and 6th values) but failing to compute which x makes the mean equal to the original median.
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