Mean, Median & Mode — Practice Set
Bank code: STA-MMM · Section: Statistics · 30 questions — Easy 5 · Medium 8 · Hard 12 · Extreme 5
Sum = mean x count; sort before you take the median; outliers move the mean, not the median.
Every question below is live in the question bank under the ID shown — the sheet and the portal are the same questions. Attempt a level with the clock running, then check Part C.
Part A — Questions
Quantitative Comparison — the four choices are always the same, so they are not reprinted: (A) Column A is greater · (B) Column B is greater · (C) The two quantities are equal · (D) The relationship cannot be determined from the information given
Level 1 · Easy — 5 questions · ~4 min
warm-up — these must be automatic
Q1 · STA-MMM-002 · MCQ · 45s
What is the median of the five measurements 6, 19, 4, 12, 9?
(A) 4
(B) 6
(C) 9
(D) 10
(E) 12
Q2 · STA-MMM-004 · Numeric Entry · 50s
The mean of five numbers is 12. Four of the numbers are 8, 15, 10, and 14. What is the fifth number? Enter your answer as a number.
Numeric entry — write the number.
Q3 · STA-MMM-005 · MCQ · 45s
What is the median of the set {3, 14, 5, 10}?
(A) 5
(B) 7.5
(C) 8
(D) 8.5
(E) 10
Q4 · STA-MMM-007 · QC · 45s
Set S = {10, 20, 30, 40, 50}
Column A: Mean of Set S Column B: Median of Set S
Q5 · STA-MMM-009 · MCQ · 45s
Which of the following correctly describes the mode or modes of the set {3, 3, 5, 7, 7}?
(A) The set has no mode
(B) Exactly one mode: 3
(C) Exactly one mode: 7
(D) Two modes: 3 and 7
(E) Three modes: 3, 5, and 7
Level 2 · Medium — 8 questions · ~10 min
two or three steps, one planted trap each
Q6 · STA-MMM-017 · QC · 80s
A set contains the five values 2, 4, 6, 8, and n, where n > 6.
Column A: Mean of the set Column B: Median of the set
Q7 · STA-MMM-018 · Numeric Entry · 70s
Class A has 10 students with a mean score of 78. Class B has 15 students with a mean score of 88. What is the mean score of all 25 students combined? Enter your answer as a number.
Numeric entry — write the number.
Q8 · STA-MMM-026 · Select all that apply · 85s
The set {p, p, 8, 14, q} has a mean of 10, where p and q are positive integers. Which of the following must be true? Select all that apply.
(A) 2p + q = 28
(B) p is a mode of the set
(C) q > p
(D) The median of the set is 8
(E) p < 14
Q9 · STA-MMM-027 · MCQ · 65s
The average (arithmetic mean) of five consecutive even integers is 18. What is the largest of the five integers?
(A) 18
(B) 20
(C) 22
(D) 26
(E) 28
Q10 · STA-MMM-032 · MCQ · 80s
In the set {2, 5, 8, 11, x}, x is a positive integer and the mean of the set equals the median of the set. What is x?
(A) 6
(B) 6.5
(C) 8
(D) 14
(E) 40
Q11 · STA-MMM-033 · Numeric Entry · 70s
A shop's sales, in units, on seven consecutive days were 12, 20, 12, 25, 20, 12, and 15. What is the sum of the median and the mode of these seven values? Enter your answer as a number.
Numeric entry — write the number.
Q12 · STA-MMM-034 · MCQ · 65s
The mean of four numbers is 25. A fifth number is added to the set, and the mean of the five numbers is 23. What is the fifth number?
(A) 2
(B) 15
(C) 17
(D) 21
(E) 23
Q13 · STA-MMM-038 · Select all that apply · 85s
Set S = {4, 8, 8, 10, 20}. Set T is formed by adding 6 to each value of S. Set U is formed by multiplying each value of S by 6. Which of the following are true? Select all that apply.
(A) The mean of T is 6 greater than the mean of S
(B) The median of T equals the median of S
(C) The range of T equals the range of S
(D) The mode of U is 48
(E) The range of U is 6 greater than the range of S
Level 3 · Hard — 12 questions · ~19 min
where 162+ is won or lost
Q14 · STA-MMM-039 · MCQ · 95s
The average (arithmetic mean) of five numbers is 30. The average of the three largest is 36 and the average of the three smallest is 25. What is the median of the five numbers?
(A) 25
(B) 30
(C) 30.5
(D) 33
(E) 36
Q15 · STA-MMM-040 · QC · 100s
A set of 10 positive integers has a mean of 15 and a median of 12.
Column A: The sum of the 5 largest values Column B: The sum of the 5 smallest values
Q16 · STA-MMM-041 · Numeric Entry · 100s
Team A has 8 players with a mean height of 72 inches. Team B has 12 players with a mean height of 75 inches. The two teams merge into a single 20-player team, and then one player whose height is 70 inches leaves. What is the mean height, in inches, of the remaining players? Enter your answer as a number.
Numeric entry — write the number.
Q17 · STA-MMM-042 · MCQ · 90s
A set of 9 values has a mean of 40 and a median of 35. The largest value in the set is then increased by 30, and nothing else changes. Which of the following describes the new mean and the new median?
(A) The mean increases by 30; the median is unchanged
(B) The mean increases by 30/9; the median is unchanged
(C) The mean increases by 30/9; the median increases by 30/9
(D) The mean increases by 30/9; the median increases by 30
(E) Both the mean and the median increase by 30
Q18 · STA-MMM-043 · Numeric Entry · 95s
A set of 5 distinct positive integers has a mean of 18 and a median of 16. The smallest value is 8 and the largest is 30. What is the greatest possible value of the second largest number in the set? Enter your answer as a number.
Numeric entry — write the number.
Q19 · STA-MMM-044 · QC · 105s
A set of 6 positive integers has a mean of 10, and its two smallest values are each equal to 1.
Column A: The greatest possible value of the median of the set Column B: 11
Q20 · STA-MMM-045 · Select all that apply · 90s
The mean of a list of 10 numbers is 20. One additional number is added to the list, and the mean of the resulting 11 numbers is an integer. Which of the following could be the number that was added? Select all that apply.
(A) 9
(B) 20
(C) 30
(D) 31
(E) 50
Q21 · STA-MMM-046 · MCQ · 90s
The mean of 12 numbers is 15. The mean of the first 5 of them is 12, and the mean of the last 4 is 18. What is the mean of the remaining 3 numbers?
(A) 12
(B) 15
(C) 16
(D) 18
(E) 48
Q22 · STA-MMM-048 · Numeric Entry · 95s
A set of 8 numbers has a mean of 30. When its largest and its smallest values are both removed, the remaining 6 numbers have a mean of 28. The largest value is 5 times the smallest. What is the largest value? Enter your answer as a number.
Numeric entry — write the number.
Q23 · STA-MMM-049 · MCQ · 95s
Section 1 has 20 students with a mean score of 72. Section 2 has n students with a mean score of 84. When the two sections are combined, the mean score of all the students is 80. What is n?
(A) 10
(B) 20
(C) 30
(D) 40
(E) 50
Q24 · STA-MMM-051 · Select all that apply · 100s
A set of 8 integers has a mean of 25. Which of the following changes to the set would leave the mean exactly 25? Select all that apply.
(A) Adding one new value equal to 25
(B) Multiplying every value in the set by 2
(C) Adding two new values whose sum is 50
(D) Increasing one value by 12 and decreasing another value by 12
(E) Adding 25 to one of the values
Q25 · STA-MMM-052 · MCQ · 90s
In a data set of n numbers the mean is 50. One value of 80 is replaced by a value of 20, and the mean falls by 3. What is n?
(A) 10
(B) 20
(C) 30
(D) 47
(E) 60
Level 4 · Extreme — 5 questions · ~10 min
165+ — expect to need the insight, not the grind
Q26 · STA-MMM-054 · MCQ · 125s
A set of 2n numbers, listed from least to greatest as a1 <= a2 <= ... <= a2n, has mean M and median K, where M > K. Each of the n smallest values is replaced by the number M. Which of the following describes the mean and the median of the new set?
(A) The mean increases and the median increases
(B) The mean is unchanged and the median increases
(C) The mean increases and the median is unchanged
(D) The mean decreases and the median increases
(E) Both the mean and the median are unchanged
Q27 · STA-MMM-055 · Numeric Entry · 120s
A set of 5 positive integers has a mean of 16, a median of 15, and a unique mode of 12 that appears exactly twice. What is the greatest possible value of the largest number in the set? Enter your answer as a number.
Numeric entry — write the number.
Q28 · STA-MMM-056 · QC · 120s
A company has 100 employees. The mean salary is $60,000, the median salary is $50,000, and the highest-paid employee earns $500,000. That employee's salary is then changed to $60,000, and no other salary changes.
Column A: The new mean salary Column B: The new median salary
Q29 · STA-MMM-057 · Select all that apply · 115s
A set of 5 integers, not necessarily distinct, has a mean of 10, a smallest value of 4, and a range of 14. Which of the following could be the median of the set? Select all that apply.
(A) 4
(B) 5
(C) 9
(D) 12
(E) 14
Q30 · STA-MMM-060 · QC · 125s
The numbers a1 <= a2 <= ... <= a2n, where n >= 2, are such that the average of the n smallest equals the average of the n largest, and each of those two averages is M.
Column A: The mean of all 2n numbers Column B: The median of all 2n numbers
Part B — Answer Key
| Q | ID | Level | Type | Answer |
|---|---|---|---|---|
| 1 | STA-MMM-002 |
easy | MCQ | C |
| 2 | STA-MMM-004 |
easy | Numeric Entry | 13 |
| 3 | STA-MMM-005 |
easy | MCQ | B |
| 4 | STA-MMM-007 |
easy | QC | C |
| 5 | STA-MMM-009 |
easy | MCQ | D |
| 6 | STA-MMM-017 |
medium | QC | D |
| 7 | STA-MMM-018 |
medium | Numeric Entry | 84 |
| 8 | STA-MMM-026 |
medium | Select all that apply | A, B, E |
| 9 | STA-MMM-027 |
medium | MCQ | C |
| 10 | STA-MMM-032 |
medium | MCQ | D |
| 11 | STA-MMM-033 |
medium | Numeric Entry | 27 |
| 12 | STA-MMM-034 |
medium | MCQ | B |
| 13 | STA-MMM-038 |
medium | Select all that apply | A, C, D |
| 14 | STA-MMM-039 |
hard | MCQ | D |
| 15 | STA-MMM-040 |
hard | QC | A |
| 16 | STA-MMM-041 |
hard | Numeric Entry | 74 |
| 17 | STA-MMM-042 |
hard | MCQ | B |
| 18 | STA-MMM-043 |
hard | Numeric Entry | 27 |
| 19 | STA-MMM-044 |
hard | QC | A |
| 20 | STA-MMM-045 |
hard | Select all that apply | A, B, D |
| 21 | STA-MMM-046 |
hard | MCQ | C |
| 22 | STA-MMM-048 |
hard | Numeric Entry | 60 |
| 23 | STA-MMM-049 |
hard | MCQ | D |
| 24 | STA-MMM-051 |
hard | Select all that apply | A, C, D |
| 25 | STA-MMM-052 |
hard | MCQ | B |
| 26 | STA-MMM-054 |
extreme hard | MCQ | A |
| 27 | STA-MMM-055 |
extreme hard | Numeric Entry | 25 |
| 28 | STA-MMM-056 |
extreme hard | QC | A |
| 29 | STA-MMM-057 |
extreme hard | Select all that apply | B, C, D |
| 30 | STA-MMM-060 |
extreme hard | QC | C |
Part C — Worked Solutions
Q1 · STA-MMM-002 — Answer: C
Step 1 — sort first. The median is defined by POSITION, and the list as printed is out of order. Sorted: 4, 6, 9, 12, 19. Step 2 — the count is 5, an odd number, so the median is the single middle value: the 3rd one. The 3rd value is 9. Check: two values (4 and 6) lie below 9 and two values (12 and 19) lie above it, so 9 really is the middle. Correct. Answer: 9.
Trap. Choice A, 4, is the middle entry of the list as printed - the error of taking the median without sorting. Choice D, 10, is the mean, (6 + 19 + 4 + 12 + 9)/5 = 50/5 = 10, which is a different measure of centre. Choices B and E are the values one position either side of the middle after sorting.
Q2 · STA-MMM-004 — Answer: 13
Step 1 — convert the average into a total. Sum = mean x count = 12 x 5 = 60. Step 2 — add the four known values: 8 + 15 + 10 + 14 = 47. Step 3 — the fifth number is whatever closes the gap in the TOTAL: 60 - 47 = 13. Check: (8 + 15 + 10 + 14 + 13)/5 = 60/5 = 12. Correct. Answer: 13.
Trap. Working with averages instead of sums. The four known numbers average 47/4 = 11.75, which is only 0.25 below 12, so the fifth number 'feels' like about 12. The shortfall must be made up in the total, not in the average: the set is 60 - 47 = 13 short, and all of it sits in the fifth number.
Q3 · STA-MMM-005 — Answer: B
Step 1 — sort: 3, 5, 10, 14. Step 2 — the count is 4, an even number, so there is no single middle entry. The median is the average of the two middle values, the 2nd and the 3rd. Step 3 — (5 + 10)/2 = 15/2 = 7.5. Notice that 7.5 is not a member of the set at all. For an even-sized set the median usually is not. Answer: 7.5.
Trap. Choices A and E are the two middle ELEMENTS, 5 and 10, picked instead of averaged - the standard even-count error. Choice C, 8, is the mean, (3 + 14 + 5 + 10)/4 = 32/4 = 8. Choice D, 8.5, is the midrange, (3 + 14)/2, which averages the extremes rather than the middle pair.
Q4 · STA-MMM-007 — Answer: C
Column B: the set is already sorted and has 5 values, so the median is the 3rd value, 30. Column A: sum = 10 + 20 + 30 + 40 + 50 = 150, and 150/5 = 30. Both are 30, so the two quantities are equal. Faster route: the values are evenly spaced (a constant gap of 10), and for an evenly spaced set mean = median = (first + last)/2 = (10 + 50)/2 = 30. No addition needed.
Trap. Assuming the two must differ because they are different words, and guessing A. The shortcut only fails when the spacing is uneven: change the 50 to a 90 and the mean jumps to 38 while the median stays at 30. Even spacing is exactly what makes them agree.
Q5 · STA-MMM-009 — Answer: D
Step 1 — count how often each value appears: 3 appears twice, 5 appears once, 7 appears twice. Step 2 — the mode is the value with the highest frequency. The highest frequency here is 2, and TWO values reach it. Step 3 — the set is therefore bimodal, with modes 3 and 7. Answer: two modes, 3 and 7.
Trap. Choice C picks 7 because it is the larger of the two repeated values, and choice B picks 3 because it appears first. Nothing in the definition of the mode breaks ties - a set can have two modes, or three, or none. Choice A would be right only if every value appeared the same number of times, which is not the case here since 5 appears just once.
Q6 · STA-MMM-017 — Answer: D
Step 1 — pin down the median. Since n > 6, when the set is sorted n lands in the 4th or the 5th position, so the middle (3rd) value is always 6. Column B = 6, whatever n is. Step 2 — the mean is (2 + 4 + 6 + 8 + n)/5 = (20 + n)/5. Step 3 — the mean beats 6 exactly when 20 + n > 30, that is when n > 10. Step 4 — test both sides of that threshold, staying inside n > 6: n = 7 gives mean = 27/5 = 5.4, which is LESS than the median 6, so Column B is greater. n = 20 gives mean = 40/5 = 8, which is MORE than 6, so Column A is greater. Two legal values of n give opposite answers, so the relationship cannot be determined.
Trap. Choosing A on the reasoning that n could be enormous and would drag the mean up. All we are told is n > 6, and every n from just above 6 up to 10 leaves the mean BELOW the fixed median of 6. On any mean-versus-median comparison with a free variable, test one value just inside the constraint and one far outside it before committing.
Q7 · STA-MMM-018 — Answer: 84
Step 1 — turn each average into a total. Class A: 10 x 78 = 780. Class B: 15 x 88 = 1320. Step 2 — combine: total score = 780 + 1320 = 2100 over 25 students. Step 3 — 2100/25 = 84. Faster route (alligation): Class B holds 15 of the 25 students, or 3/5 of them, so the combined mean sits 3/5 of the way along the 10-point gap from 78 to 88: 78 + 0.6 x 10 = 84. Check: 25 x 84 = 2100. Correct. Answer: 84.
Trap. Averaging the two averages: (78 + 88)/2 = 83. That is legal only when the two groups are the same size. Class B is larger, so the combined mean must lie closer to 88 than to 78 - and 84 is indeed 6 above 78 but only 4 below 88.
Q8 · STA-MMM-026 — Answer: A, B, E
The count is 5 and the mean is 10, so the sum is 5 x 10 = 50: p + p + 8 + 14 + q = 50. A - that simplifies to 2p + q = 28. TRUE. B - p occupies two of the five slots. The only way another value could match that is if q equals 8 or 14, and then it too appears exactly twice - never more. So no value ever appears more often than p, which makes p a mode (sometimes one of two). TRUE. C - take p = 10, forcing q = 28 - 20 = 8. Then q < p. FALSE. D - with p = 10 and q = 8 the sorted set is {8, 8, 10, 10, 14} and the median is 10, not 8. FALSE. E - q is a positive integer, so q >= 1 and 2p = 28 - q <= 27, giving p <= 13.5, so p <= 13. TRUE. Answer: A, B, E.
Trap. Rejecting B by reading it as 'p is the ONLY mode' - with p = 7 the set is {7, 7, 8, 14, 14}, which is bimodal, yet p is still a mode. C and D both look safe because the set is written in what appears to be increasing order; it is not sorted, and p can be larger than q and larger than 8. E is the one most people skip: the fact that q must be at least 1 is what caps p.
Q9 · STA-MMM-027 — Answer: C
Step 1 — consecutive even integers are evenly spaced, and for an evenly spaced set mean = median. So 18 is the MIDDLE (3rd) term. Step 2 — build the list outward from 18 in steps of 2: 14, 16, 18, 20, 22. Step 3 — the largest is two steps above the middle: 18 + 2 x 2 = 22. Check: (14 + 16 + 18 + 20 + 22)/5 = 90/5 = 18. Correct. Answer: 22.
Trap. Choice E adds five steps to the mean, 18 + 2 x 5 = 28, and choice D adds four, 18 + 8 = 26, as if 18 were the first term. The mean of an evenly spaced set sits in the middle, so the largest term is only two steps - 4 - above it. Choice A is the mean itself.
Q10 · STA-MMM-032 — Answer: D
Step 1 — the four known values sum to 2 + 5 + 8 + 11 = 26, so the mean is (26 + x)/5. Step 2 — the median depends on where x lands. Try the case x >= 11: then the sorted set is {2, 5, 8, 11, x} and the median is the 3rd value, 8. Step 3 — set mean = median: (26 + x)/5 = 8, so 26 + x = 40 and x = 14. Step 4 — check the case holds together: 14 >= 11, so the sorted set really is {2, 5, 8, 11, 14}. Mean = 40/5 = 8 and median = 8. Correct. (If x were allowed to be a non-integer, the case 5 <= x <= 8 would make x itself the median and give (26 + x)/5 = x, so 4x = 26 and x = 6.5. That is why the stem restricts x to integers.) Answer: 14.
Trap. Choice E, 40, is the total the five numbers must reach, not the missing value - subtract the 26 already present. Choice A comes from dividing by 4 instead of 5, and choice B is the average of the four known values. Choice C sets x equal to the median without checking the mean: with x = 8 the set is {2, 5, 8, 8, 11}, whose mean is 34/5 = 6.8, not 8.
Q11 · STA-MMM-033 — Answer: 27
Step 1 — sort before touching the median: 12, 12, 12, 15, 20, 20, 25. Step 2 — with 7 values the median is the 4th one: 15. Step 3 — count frequencies: 12 appears three times, 20 appears twice, 15 and 25 appear once each. The mode is 12. Step 4 — median + mode = 15 + 12 = 27. Answer: 27.
Trap. Reading the median off the unsorted list, whose 4th entry is 25, and answering 25 + 12 = 37. The second slip is calling 20 the mode because it is the largest repeated value - frequency alone decides the mode, and 12 appears three times to 20's two, which would give 15 + 20 = 35.
Q12 · STA-MMM-034 — Answer: B
Step 1 — old total = 25 x 4 = 100. Step 2 — new total = 23 x 5 = 115. Step 3 — the fifth number is the difference between the two totals: 115 - 100 = 15. Check: (100 + 15)/5 = 115/5 = 23. Correct. Faster route: the new number has to drag all five values down by 2 each, a total pull of 5 x 2 = 10, so it must sit 10 below the old mean: 25 - 10 = 15. Answer: 15.
Trap. Choice A reports the drop in the mean (2) as the number itself, and choice D is the new mean minus that drop. The subtle error is choice C: spreading the 2-point drop over only the four OLD numbers, 25 - 2 x 4 = 17. The new mean applies to all five values, so the deficit to cover is 2 x 5 = 10.
Q13 · STA-MMM-038 — Answer: A, C, D
First describe S: it is already sorted, so mean = 50/5 = 10, median = 8 (3rd of 5), mode = 8, range = 20 - 4 = 16. T = {10, 14, 14, 16, 26} and U = {24, 48, 48, 60, 120}. A - mean of T = 80/5 = 16 = 10 + 6. Adding a constant shifts the mean by that constant. TRUE. B - median of T = 14, while the median of S is 8. A shift moves the median too. FALSE. C - range of T = 26 - 10 = 16, the same as the range of S. Every value moved by the same amount, so the gap between the largest and the smallest is untouched. TRUE. D - in U the value 48 appears twice and everything else once, so the mode is 48 = 6 x 8. Scaling multiplies the mode. TRUE. E - range of U = 120 - 24 = 96, which is 6 TIMES the range of S, not 6 more than it. FALSE. Answer: A, C, D.
Trap. B and E swap the two rules. Adding a constant moves the centre (mean, median, mode) but leaves the spread (range, standard deviation) alone; multiplying by a constant scales the centre AND the spread. Picking B treats the median as if it were a measure of spread; picking E treats multiplication as if it were addition.
Q14 · STA-MMM-039 — Answer: D
Step 1 — write the sorted values as a <= b <= c <= d <= e. The median is c. Step 2 — turn every average into a sum: all five: a + b + c + d + e = 5 x 30 = 150 three smallest: a + b + c = 3 x 25 = 75 three largest: c + d + e = 3 x 36 = 108 Step 3 — add the two group sums: 75 + 108 = 183. That total counts each of a, b, d, e exactly once, but counts c TWICE, because the middle value belongs to both groups. Step 4 — so 183 = (a + b + c + d + e) + c = 150 + c, giving c = 33. Check: a + b = 75 - 33 = 42 and d + e = 108 - 33 = 75, so the set could be 20, 22, 33, 37, 38 - sum 150, three smallest averaging 25, three largest averaging 36. Correct. Answer: 33.
Trap. Choice B, 30, assumes the median must equal the mean. It is the overlap that pins the answer down: the two groups share the middle value, so 75 + 108 - 150 = 33. Choice C averages the two group means, (25 + 36)/2 = 30.5, and choices A and E simply report one of the group averages.
Q15 · STA-MMM-040 — Answer: A
Step 1 — the whole set: sum = 10 x 15 = 150. The two columns split that total, so Column A + Column B = 150. Step 2 — use the median. For 10 sorted values the median is the average of the 5th and 6th, so the 5th value + the 6th value = 24. Since the 5th is no larger than the 6th, the 5th value is at most 12. Step 3 — every one of the 5 smallest values is at most the 5th value, hence at most 12. So Column B <= 5 x 12 = 60. Step 4 — therefore Column A = 150 - Column B >= 150 - 60 = 90. Column A is at least 90 while Column B is at most 60, so Column A is greater no matter what the actual numbers are.
Trap. Choosing D because no individual values are given. Two facts settle it: the median caps each of the bottom five at 12 (at most 60 altogether), while the mean fixes the whole set at 150. Whenever a set's mean sits above its median, the weight of the set is in its upper half.
Q16 · STA-MMM-041 — Answer: 74
Step 1 — convert each average into a total. Team A: 8 x 72 = 576 inches. Team B: 12 x 75 = 900 inches. Step 2 — merged team: sum = 576 + 900 = 1476 over 20 players (a mean of 73.8). Step 3 — the departure changes BOTH the sum and the count: new sum = 1476 - 70 = 1406, new count = 20 - 1 = 19. Step 4 — 1406/19 = 74. Check: 19 x 74 = 1406. Correct. Answer: 74.
Trap. Subtracting the 70 from the total but still dividing by 20, which gives 1406/20 = 70.3. When a member leaves, the count must fall with the sum. A second error is averaging the two team means, (72 + 75)/2 = 73.5, which ignores that Team B has more players and so pulls the merged mean toward 75.
Q17 · STA-MMM-042 — Answer: B
Step 1 — the mean. Old sum = 9 x 40 = 360. Raising one value by 30 makes the new sum 390. Step 2 — new mean = 390/9 = 43.33..., which is 40 + 30/9 = 40 + 3.33. The extra 30 is shared out over all 9 values, so the mean rises by 30/9, not by 30. Step 3 — the median. With 9 sorted values the median is the 5th. The value that changed was already the largest and it only got larger, so no value changes rank and the 5th value is still 35. The mean rises by 30/9; the median does not move. Answer: B.
Trap. Choice A adds the whole 30 to the mean, forgetting that a mean divides by the count. Choices C, D and E move the median: the median depends only on POSITION, and pushing the top value further up changes no ranks at all. Had the largest value been DECREASED by 30 it could have fallen past the middle, and then the median would have moved.
Q18 · STA-MMM-043 — Answer: 27
Step 1 — sum = 5 x 18 = 90. Sorted, the set looks like {8, b, 16, d, 30}, where 8 < b < 16 < d < 30. Step 2 — the sum equation: 8 + b + 16 + d + 30 = 90, so b + d = 90 - 54 = 36. Step 3 — d is largest when b is smallest. The integers are distinct and b must sit strictly between 8 and 16, so the smallest b can be is 9. Step 4 — d = 36 - 9 = 27. Check: {8, 9, 16, 27, 30} - five distinct positive integers, sum 90, mean 18, median 16, and 27 is still below the largest value 30. Correct. Answer: 27.
Trap. Answering 36, which is the SUM b + d rather than either value on its own. The other slip is allowing b = 8 (which would give d = 28): the integers are distinct, so the second value must be at least 9. And d = 27 is only legal because it stays under the stated largest value, 30.
Q19 · STA-MMM-044 — Answer: A
Step 1 — sum = 6 x 10 = 60. Two of the values are 1, so the other four sum to 58. Call them v3 <= v4 <= v5 <= v6. Step 2 — with 6 sorted values the median is (v3 + v4)/2, so the job is to make v3 + v4 as large as possible. Step 3 — the ceiling comes from the top pair: v5 and v6 are each at least v4, so v5 + v6 >= 2 x v4 >= v3 + v4. Since (v3 + v4) + (v5 + v6) = 58, that forces v3 + v4 <= 29. Step 4 — 29 is not reachable with integers: it would need v4 >= 15, leaving only 29 for v5 + v6 while both must be at least 15. So try 28: v3 = v4 = 14 and v5 = v6 = 15 gives {1, 1, 14, 14, 15, 15}, which sums to 1 + 1 + 14 + 14 + 15 + 15 = 60. Valid. Step 5 — the greatest possible median is 28/2 = 14, and 14 > 11. Column A is greater.
Trap. Assuming the median cannot exceed the mean of 10 and choosing B. Two values are pinned at 1, which frees a large share of the total for the middle of the set. The real ceiling comes from the requirement that the two values ABOVE the median be at least as large as the median's own pair - that is what caps v3 + v4 at 28.
Q20 · STA-MMM-045 — Answer: A, B, D
Step 1 — the original list sums to 10 x 20 = 200. Step 2 — after adding x the list has 11 numbers, so the new mean is (200 + x)/11. That is an integer exactly when 200 + x is a multiple of 11. A - 200 + 9 = 209 = 11 x 19, so the new mean is 19. TRUE. B - 200 + 20 = 220 = 11 x 20, so the new mean is 20; adding a value equal to the mean leaves the mean alone. TRUE. C - 200 + 30 = 230, and 230/11 = 20.909..., not an integer. FALSE. D - 200 + 31 = 231 = 11 x 21, so the new mean is 21. TRUE. E - 200 + 50 = 250, and 250/11 = 22.727..., not an integer. FALSE. Answer: A, B, D.
Trap. Choices C and E are exactly the values that make the new SUM a multiple of 10 - the OLD count. Once the eleventh number joins, the divisor is 11. Notice the workable additions sit 11 apart: 9, 20, 31, 42, 53, and so on.
Q21 · STA-MMM-046 — Answer: C
Step 1 — total sum = 12 x 15 = 180. Step 2 — first group: 5 x 12 = 60. Last group: 4 x 18 = 72. Step 3 — the remaining 3 numbers must supply 180 - 60 - 72 = 48. Step 4 — their mean = 48/3 = 16. Check: 60 + 72 + 48 = 180 across 5 + 4 + 3 = 12 values. Correct. Answer: 16.
Trap. Choice E is the leftover sum, 48, reported before dividing by 3; choice A divides that 48 by 4 instead of 3. The conceptual trap is choice B: assuming a leftover group must average the overall mean. It would only do so if the two known groups already averaged 15 between them - in fact they average (60 + 72)/9 = 14.67, so the final three have to come in above 15 to compensate.
Q22 · STA-MMM-048 — Answer: 60
Step 1 — total sum = 8 x 30 = 240. Step 2 — the 6 survivors sum to 6 x 28 = 168. Step 3 — so the two removed values together account for 240 - 168 = 72. Step 4 — with L = 5S, the pair is 5S + S = 6S = 72, so S = 12 and L = 5 x 12 = 60. Check: 168 + 12 + 60 = 240, and 240/8 = 30. Also 60 = 5 x 12, and 60 and 12 are indeed the largest and smallest. Correct. Answer: 60.
Trap. Answering 72, which is the combined value of the two removed numbers rather than the larger one. The other slip is splitting 72 by dividing by 5 (giving 14.4): the two pieces are 5 parts and 1 part, so there are 6 parts in total and each part is 12.
Q23 · STA-MMM-049 — Answer: D
Step 1 — write the combined mean as one equation in sums: (20 x 72 + 84n)/(20 + n) = 80 Step 2 — clear the denominator: 1440 + 84n = 1600 + 80n. Step 3 — 4n = 160, so n = 40. Faster route (alligation): the combined mean of 80 sits 8 above Section 1's mean and 4 below Section 2's. The group sizes go in the INVERSE ratio of those gaps, so Section 1 : Section 2 = 4 : 8 = 1 : 2. With 20 students in Section 1, Section 2 has 40. Check: (1440 + 40 x 84)/60 = (1440 + 3360)/60 = 4800/60 = 80. Correct. Answer: 40.
Trap. Choice B, 20, assumes the sections must be the same size. They cannot be: equal sizes would put the combined mean at (72 + 84)/2 = 78, not 80. The combined mean sits twice as far from 72 as it does from 84, so the group averaging 84 must be twice as large.
Q24 · STA-MMM-051 — Answer: A, C, D
The set starts with sum = 8 x 25 = 200 across 8 values. A - new sum 225 across 9 values: 225/9 = 25. Adding a value equal to the mean never moves the mean. TRUE. B - new sum 400 across 8 values: 400/8 = 50. Scaling every value scales the mean by the same factor. FALSE. C - new sum 250 across 10 values: 250/10 = 25. The two new values average 25 between them, so they behave like two copies of the mean. TRUE. D - the sum stays at 200 (the +12 and the -12 cancel) and the count stays at 8: 200/8 = 25. TRUE. E - new sum 225 across 8 values: 225/8 = 28.125. FALSE. Answer: A, C, D.
Trap. A and E look like the same move and are not: adding 25 AS a new member raises the sum and the count in step, while adding 25 TO an existing member raises the sum with no extra member to absorb it. B is the other bait - doubling every value doubles the mean rather than preserving it.
Q25 · STA-MMM-052 — Answer: B
Step 1 — find the change in the sum. Swapping 80 for 20 removes 80 - 20 = 60 from the total, and the count does not change. Step 2 — a drop of 60 in the sum lowers the mean by 60/n. Step 3 — set that equal to the stated drop: 60/n = 3, so n = 20. Check: with n = 20 the old sum is 20 x 50 = 1000; the new sum is 940, and 940/20 = 47 = 50 - 3. Correct. Answer: 20.
Trap. Choice E is the drop in the SUM and choice D is the new mean - neither answers 'how many numbers'. Choice A uses the wrong gap: 80 - 50 = 30 measures the removed value's distance from the mean, but what actually leaves the total is 80 - 20 = 60, the distance between the old and new values. Choice C halves that 60 because two values were 'involved', though only one slot changed.
Q26 · STA-MMM-054 — Answer: A
Two facts to set up: the whole set sums to 2nM, and K = (a_n + a_{n+1})/2, so a_n <= K <= a_{n+1}. Step 1 — the bottom half is small. Each of a1 ... a_n is at most a_n, and a_n <= K, so the bottom half sums to at most nK, which is strictly less than nM (because K < M). Step 2 — the mean. Replacing that bottom half with n copies of M swaps a total of at most nK for a total of exactly nM, so the sum strictly rises while the count stays at 2n. The mean INCREASES above M. Step 3 — the median. In the new set, n values equal M and every survivor is at least a_{n+1} >= K. So at most n of the 2n values can be K or below, which means the (n+1)th smallest value is strictly above K while the nth smallest is at least K. Their average - the new median - is therefore strictly greater than K. The median INCREASES. Concrete check: {1, 1, 10, 12} has M = 24/4 = 6 and K = (1 + 10)/2 = 5.5, so M > K. Replacing the two smallest with 6 gives {6, 6, 10, 12}: mean = 34/4 = 8.5 > 6 and median = (6 + 10)/2 = 8 > 5.5. Both rose. Answer: A.
Trap. Choice B is the magnet: replacing values with the mean 'should' leave the mean alone. It only would if the replaced values already averaged M, and here they are the bottom half of a set whose mean sits above its median, so they averaged strictly less - swapping them for M lifts the total. Choice C fails for the sister reason: the median is built from the top of the bottom half, so rewriting the bottom half does move it.
Q27 · STA-MMM-055 — Answer: 25
Step 1 — sum = 5 x 16 = 80. Write the sorted set as v1 <= v2 <= v3 <= v4 <= v5, with v3 = 15 since the median of 5 values is the 3rd. Step 2 — locate the two 12s. Everything from position 3 onward is at least 15, so neither 12 can sit at position 3, 4 or 5. Both must sit at the bottom: v1 = v2 = 12. Step 3 — the sum: 12 + 12 + 15 + v4 + v5 = 80, so v4 + v5 = 41. Step 4 — to maximise v5, minimise v4. We know v4 >= 15, but v4 = 15 would make 15 a second value appearing twice, so the set would be bimodal and 12 would no longer be the UNIQUE mode. Hence v4 >= 16. Step 5 — v4 = 16 gives v5 = 41 - 16 = 25. Check: {12, 12, 15, 16, 25} has sum 80 (mean 16), median 15, and 12 is the only value appearing more than once. Correct. Answer: 25.
Trap. Answering 26 by taking v4 = 15. That set, {12, 12, 15, 15, 26}, has two values appearing twice, so 12 is not the unique mode. The other error is letting a 12 sit at position 3 or 4: with a median of 15, every value from the third slot on is at least 15, which pins both 12s to the bottom two slots.
Q28 · STA-MMM-056 — Answer: A
Step 1 — Column A is pinned exactly. Old total = 100 x 60,000 = 6,000,000. Replacing 500,000 with 60,000 removes 440,000, so the new total is 5,560,000 and the new mean is 5,560,000/100 = 55,600. Step 2 — cap Column B. Sort the old salaries v1 <= v2 <= ... <= v100. The median says v50 + v51 = 100,000, and since v50 <= v51 the 50th salary is at most 50,000. Step 3 — the new list is v1 ... v99 with 60,000 in place of v100. Because 60,000 is larger than v50 (which is at most 50,000), the bottom 50 positions are still v1 ... v50, and the 51st position is whichever is smaller, v51 or 60,000. Step 4 — so the new median = (v50 + the smaller of v51 and 60,000)/2, which is at most (v50 + v51)/2 = 50,000. Column A is exactly 55,600 and Column B is at most 50,000, so Column A is greater.
Trap. Choosing D because the individual salaries are unknown. The mean is fixed exactly and the median is capped by the old median - no arrangement of the other 99 salaries can push it past 50,000, so the comparison resolves without knowing a single value. Choosing C assumes the mean collapses onto the median once the outlier is gone; it does not, because the other 99 salaries have not moved.
Q29 · STA-MMM-057 — Answer: B, C, D
Step 1 — range = largest - smallest, so the largest value is 4 + 14 = 18. Sum = 5 x 10 = 50. Step 2 — sorted, the set is {4, a, m, b, 18} with 4 <= a <= m <= b <= 18, and 4 + a + m + b + 18 = 50, so a + m + b = 28. Step 3 — how small can the median m be? Since a <= m and b <= 18, we get 28 = a + m + b <= m + m + 18, so 2m >= 10 and m >= 5. Step 4 — how large can m be? Since a >= 4 and b >= m, we get 28 = a + m + b >= 4 + m + m, so 2m <= 24 and m <= 12. Step 5 — so the median can be any integer from 5 to 12, and both ends are reachable. A - 4 is below the floor. Concretely, a median of 4 forces the three smallest to be 4, 4, 4, so the fourth value would be 50 - 4 - 4 - 4 - 18 = 20, which exceeds the largest value 18. FALSE. B - 5 works: {4, 5, 5, 18, 18} has sum 50, median 5, smallest 4, largest 18. TRUE. C - 9 works: {4, 6, 9, 13, 18} has sum 50 and median 9. TRUE. D - 12 works: {4, 4, 12, 12, 18} has sum 50 and median 12. TRUE. E - 14 is above the ceiling of 12. With m = 14 the remaining two values satisfy a + b = 28 - 14 = 14, but a >= 4 and b >= m = 14, so a + b >= 18. Impossible. FALSE. Answer: B, C, D.
Trap. Both wrong choices come from forgetting that the median is squeezed from both sides. Choice A ignores that the two values ABOVE the median plus the 18 must still fit inside a total of 50; choice E ignores that the two values BELOW the median cannot drop under 4. Fixing the range at 14 also fixes the maximum at exactly 18 - it is an equation, not a lower bound.
Q30 · STA-MMM-060 — Answer: C
Step 1 — Column A. The bottom half sums to nM and the top half sums to nM, so all 2n numbers sum to 2nM and the mean is 2nM/(2n) = M. Step 2 — Column B looks unpinned, but the sorted order hides a very strong consequence. Pair each bottom value with the top value the same distance along the list: a1 with a_{n+1}, a2 with a_{n+2}, and so on up to a_n with a_2n. Because the list is sorted, a_i <= a_{n+i} for every i from 1 to n. Step 3 — add those n inequalities: a1 + ... + a_n <= a_{n+1} + ... + a_2n. We are told both sides equal nM, so the sum of the inequalities is actually an equality - which can only happen if EVERY one of them is an equality. Hence a_i = a_{n+i} for all i. Step 4 — combine that with the sorted order: a1 = a_{n+1} forces a1 = a2 = ... = a_{n+1}, and the same chain carries through to a_2n. Every value in the set is the same number, and since the mean is M, every value equals M. Step 5 — the median of a list whose values are all M is M. Column A = Column B = M. The two quantities are equal.
Trap. Choosing D on the grounds that no actual numbers are given. The condition is far stronger than it looks: for a SORTED list, the bottom half can match the top half's total only if the two halves agree value by value, which collapses the whole set to one repeated number. Drop the sorting and the conclusion dies - the six numbers 0, 1, 29 (mean 10) and 9, 9, 12 (mean 10) also have overall mean 10, but their median is 9.