9.1 Mean, Median & Mode

Key Takeaways

  • The GMAT identity is Sum = n × Mean. Almost every 'the average is' sentence is a sum in disguise, including items that add, drop, or replace one value.
  • A weighted average uses counts (or other stated weights), not a halfway point between subgroup means. The combined mean sits closer to the larger group and always between the two subgroup means.
  • For an ordered list of n values, the median is at position (n+1)/2 when n is odd, and the average of the values at positions n/2 and n/2+1 when n is even. Always sort first.
  • The median resists an outlier; the mean does not. Adding a number equal to the current mean leaves the mean unchanged; adding a number above the mean raises the mean, but only partway toward that new number.
  • If every value is replaced by kx + c, the mean, median, and mode each become k times the old measure, plus c. Adding c slides all three; multiplying by k scales all three.
Last updated: August 2026

What GMAC Actually Tests

Official Guide Math Review 3.4 puts statistics on the Quantitative Reasoning map. On the current GMAT (Focus Edition) that does not mean a statistics course. You will not run a regression, look up a z-table, or test a hypothesis. You will work with a short list, a frequency table, or a word problem that hides an average. The section is 21 Problem Solving questions in 45 minutes with no calculator, so GMAC writes items that reward identities rather than keystrokes.

Three location numbers describe a list: the mean (arithmetic average), the median (middle of the ordered list), and the mode (most frequent value). GMAT items mix these on purpose. A tester who treats average as a synonym for middle will miss every item that inserts one extreme value.

The Mean and the Sum Identity

For a list of n numbers:

Mean = (sum of the values) / n

The exam-useful rewrite is the sum identity:

Sum = n × Mean

If five numbers have mean 18, the sum is 90. Write that before you invent algebra. Almost every sentence that starts with "the average is" is a sum in disguise.

Signed deviations add to zero

The mean is the balance point of the list. Subtract the mean from every value and those signed deviations add to exactly 0:

(x1 − mean) + (x2 − mean) + ... + (xn − mean) = 0

Worked example. The list 12, 15, 18, 22, x has mean 17.

Deviations from 17: (12 − 17) + (15 − 17) + (18 − 17) + (22 − 17) + (x − 17) = 0

That is −5 + −2 + 1 + 5 + (x − 17) = 0, so −1 + (x − 17) = 0, and x = 18.

The same arithmetic as "sum = 5 × 17 = 85," but with smaller integers. On a no-calculator section, smaller integers are the point.

Adding, dropping, or replacing one number

If a list of n values has mean M, and you add a new number y:

  • New sum = nM + y
  • New mean = (nM + y) / (n + 1)

Consequences you can use without rebuilding the whole list:

  • If y = M, the mean does not change.
  • If y > M, the mean increases, but not all the way to y. The new mean lies strictly between M and y.
  • If y < M, the mean decreases, and again lands strictly between y and M.
  • Replacing a value a with a value b changes the sum by (b − a) and changes the mean by (b − a) / n.
  • Dropping a value a changes the sum to nM − a and the mean to (nM − a) / (n − 1).

GMAT wording to translate immediately: "a sixth number is included," "one score is dropped," "the largest value is replaced." Each of those is a sum update.

Example. Five scores have mean 20. A sixth score of 32 is added. The old sum is 100; the new sum is 132; the new mean is 132 / 6 = 22. The mean moved 2 points toward 32, not 12 points. The extra value is one of six, so it receives only 1/6 of the gap: (32 − 20) / 6 = 2.

Example. Seven numbers average 12, so the sum is 84. Replace the largest number, 25, with 11. The new sum is 84 − 25 + 11 = 70, and the new mean is 70 / 7 = 10. You never needed the other five numbers.

Weighted Averages

When subgroups have different sizes, do not average the subgroup means. Weight by counts — or by whatever the problem names as a weight (headcount, dollars, minutes, credits).

Combined mean = (total combined sum) / (total combined count)

= (w1 × m1 + w2 × m2 + ...) / (w1 + w2 + ...)

Example. 18 students average 84; 12 students average 74; the combined class has 30 students.

  • Student-group sum = 18 × 84 = 1,512
  • Other-group sum = 12 × 74 = 888
  • Combined sum = 2,400
  • Combined mean = 2,400 / 30 = 80

Lever rule for two groups

The combined mean sits closer to the larger group. Distances from the two subgroup means are inversely proportional to the counts:

(count of group 1) / (count of group 2) = (distance from m2 to the mix) / (distance from m1 to the mix)

Here 18 : 12 = 3 : 2. The gap from 74 to 84 is 10. The mix is 3/5 of the way from 74 toward 84 because the 18-student group carries 3 of the 5 parts: 74 + 6 = 80. Same answer, less arithmetic.

If the two groups have equal size, the weighted mean collapses to the ordinary midpoint of the two subgroup means. If an answer lies outside the interval between the two subgroup means, it is impossible for a mix of those two groups.

SituationCombined mean
Equal-size groupsMidpoint of the two subgroup means
Unequal-size groupsCloser to the larger group's mean
Add a value equal to the current meanMean unchanged
Add a value above the current meanMean rises, stays below the new value
Add a value below the current meanMean falls, stays above the new value
Replace a with bMean changes by (b − a) / n

Median: Position, Not Average

The median is the middle of the ordered list. Always sort first. GMAT lists are often unsorted on purpose.

  • Odd n: one middle value, at position (n + 1) / 2. For {3, 7, 9, 14, 21}, n = 5, position 3, median = 9.
  • Even n: no single middle value. Average the two central values, at positions n / 2 and n / 2 + 1. For {4, 8, 11, 15, 18, 24}, n = 6, positions 3 and 4 are 11 and 15, median = 13.
nMedian positionWhat you report
53rd ordered valueThat value
63rd and 4thTheir average
74th ordered valueThat value
84th and 5thTheir average
95th ordered valueThat value

Adding a number changes n, so it can change which positions are middle. Adding a value far in the right tail of an odd-length list often leaves the median unchanged, because the old middle value is still the middle. Adding a value to an even-length list creates an odd-length list, so the new median is a single ordered entry rather than an average of a pair.

Example. Ordered six-number list: 7, 9, 12, 18, 21, 30. Median = (12 + 18) / 2 = 15. Insert 16. The new ordered list is 7, 9, 12, 16, 18, 21, 30, n = 7, position 4. New median = 16. The mean also changes, but by a different amount: old sum 97, new sum 113, new mean 113 / 7, which is not 16. Mean and median do not have to move together.

Outlier resistance

The median ignores how far the tails go. The mean does not.

  • Set A: {10, 20, 30, 40, 50}. Mean = 30, median = 30.
  • Set B: {10, 20, 30, 40, 1000}. Median is still 30 (3rd of 5). Mean = 1,100 / 5 = 220.

GMAC uses this contrast constantly: a single extreme salary, a single long commute, a single high test score. If the question asks which measure is least affected by that extreme, the median (or the mode) is the candidate; the mean moves. Dropping or replacing the extreme value moves the mean by (change in sum) / n and typically leaves the median untouched if the ordered middle is still the same entry.

Mode

The mode is the value with the highest frequency.

  • Unimodal: one winner. {2, 3, 3, 5, 8} has mode 3.
  • Bimodal: two values tie for highest frequency. {1, 2, 2, 4, 5, 5, 9} has modes 2 and 5.
  • No mode: every value appears equally often, including a list of distinct numbers. {4, 7, 9, 12, 15} has no mode.

The mode can sit far from the mean. It is a frequency fact, not a center-of-mass fact. On a frequency table, the mode is the value with the largest count — you do not need the mean to name it. A list can have a mode of 12, a median of 15, and a mean of 16 at the same time; those three numbers answer three different questions.

Linear Transformations

Apply the same operation to every element.

Add (or subtract) a constant c. The whole list slides. Spacing is unchanged.

  • New mean = old mean + c
  • New median = old median + c
  • New mode = old mode + c

Multiply (or divide) by a constant k. Every location measure scales by k.

  • New mean = k × old mean
  • New median = k × old median
  • New mode = k × old mode

A combined rule y = kx + c (multiply, then add) sends each measure to k × (old measure) + c. Track the order: adding 4 and then multiplying by 2 is y = 2(x + 4) = 2x + 8, not 2x + 4.

Example. Mean 40, median 42. Transform y = 3x + 5. New mean = 3 × 40 + 5 = 125. New median = 3 × 42 + 5 = 131. You never need the original nine numbers. The sum of the new mean and the new median is 256.

Subtracting the mean from every value produces a new mean of 0. The new median is (old median − old mean). That recentering is the first step of a standard-deviation calculation in the next section; it does not change how spread out the list is.

Transformation on every xiMeanMedianMode
OriginalMMedMo
xi + cM + cMed + cMo + c
xi − cM − cMed − cMo − c
k · xikMk · Medk · Mo
k · xi + ckM + ck · Med + ck · Mo + c

Traps on Quant

  • Averaging two group means without weights.
  • Forgetting to sort before naming a median.
  • Using position n/2 on an odd list, or reporting only one of the two central values on an even list.
  • Assuming a mode exists, or assuming the mode equals the mean.
  • Applying a transformation to the mean but not to the median when the question asks for both.
  • Thinking an added outlier must move the median.
  • Thinking an added number equal to the mean must change the mean.
Loading diagram...
Weighted mean sits closer to the larger group
Test Your Knowledge

A department has 20 analysts with an average (arithmetic mean) score of 72 and 30 managers with an average score of 84. What is the average score for all 50 people?

A
B
C
D
Test Your Knowledge

A set of seven integers has a median of 10 and a mean of 12. The largest integer in the set is 25. If 25 is replaced by 11, which statement is true?

A
B
C
D
Test Your Knowledge

A list of numbers has mean 16 and median 15. Every number in the list is first increased by 4 and then multiplied by 2. What is the sum of the new mean and the new median?

A
B
C
D