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GMAT Club World Cup 2026 (Day 6): If the average (arithmetic mean)

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PJ8

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13 Jul 2026, 09:42

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We can see
if X<21 Median will be non integer or not multiple of 3,
on X=21 the median will be 21 and mode = 27 but Average will not be 24 thus Consecutive multiple condition will not be hold,
Even on X= 27 Median and mode will consecutive multiple of 3 but average won't be multiple of 3. So X will b greated then 27
when X >27 , Median = 24, Mode = 27 so average should be 30 , for average to be 30 , X would be 81.
So X will be 81.

Bunuel

x, 6, 18, 21, 27, 27

If the average (arithmetic mean), median, and unique mode of the six integers shown above, when arranged in increasing order, are consecutive multiples of 3, what is the range of all possible values of x?

A. 27
B. 54
C. 57
D. 81
E. 108

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Bunuel

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given - x,6,18,21,27,27
to have a unique mode - x cannot be 6 or 18 or 21
so , mode - 27
Median:
1. if x<18 then, median - (18+21)/2 - 19.5 -not a multiple of 3
2. if $18<=x<21$, then , median = (21+x)/2 - x can be either 19 or 20- which results in median - 20 or 20.5 - not multiple of 3
3. if x = 21- mode cannot be unique so not possible
4. $21<x<=27$- median - (21+x)/2
for this to be an integer and multiple of 3 - lets check by substitution:
x=22 - median - 21.5 not valid
x=23 - median - 22 not valid
x=24 - median - 22.5 not valid
x=25 - median - 23 not valid
x=26 - median - 23.5 not valid
x=27 - median - 24 - valid
5. $x>27$, median = (21+27)/2 = 24 - valid
so median = 24
Mean:
we know median - 24, mode - 27
given, mean, median, mode must be consecutive multiples of 3
then,
A: mean - 21, median - 24, mode - 27
B: median - 24, mode - 27, mean - 30
mean = $(6+18+21+27+27+x)/6 = (99+x)/6$
if mean = 21, 99+x = 126
x = 27
if mean = 30, 99+x = 180
x = 81
so range of all possible values of x = 81-27 = 54 - B

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13 Jul 2026, 09:53

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First thing we can see is that the mode here is clearly defined as unique -> mode has to be 27 as it already appears twice

Let average be A and median be M.

Now, we have three cases -

1. A,M,27 are the consec. mult. of 3 -> 21,24,27
2. M,A,27 are the consec. mult. of 3 -> 21,24,27
3. M,27,A are the consec. mult. of 3 -> 24,27,30 (not considering A,27,M because median won't be higher than 27 here)

Average calculated = (99+x)/6 => x = 6A-99

From the cases -
1. A = 21 -> x = 27 => median will be (21+27)/2 = 24 which works
2. A = 24 -> x = 45 => median will be 24 but our case requires median to be 21 so this won't work
3. Using same process we see this case is also valid with x being 81

Now we have two valid values of x as 27 and 81 so the range = 81-27 = 54 (B)

Honestly, found this pretty challenging for the difficulty level categorised and definitely took wayyyy too long for this!

Bunuel

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Mean = (x+99)/6; Mode = 27; Median:
A. If x <= 18, i.e. list is (x, 6, 18, 21, 27, 27) or (6, x, 18, 21, 27, 27), median = 19.5
B. If 18<x<=27 i.e. list is (6, 18, x, 21, 27, 27) or (6, 18, 21, x, 27, 27), median = (x+21)/2
C. If x>27, median = 27

Since mode is 27, and median is between 19.5 and 24, possible values of mean and median are 21 or 24, i.e., possible combinations of (mean, median, mode) are:
I. (21, 24, 27)
II. (30, 24, 27)
II. (24, 21, 27)

CASE I: Median is 24: x should be >=24
Mean is 21, i.e., (x+99)/6 = 21, x=27

CASE II. Median is 24, mean is 30
(x+99)/6=30, i.e., x=81

CASE III. Median = 21, mean is 24, this follows scenario B where 18<x<=27,
(x+99)/6=24, x comes out to be 45 which is greater than 27. So not possible.

Therefore, possible values of x are 27 and 81. Therefore, range is 54

Bunuel

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The answer is (B) 54.

The numbers given to us are,
x, 6, 18, 21, 27, 27
The other information given is that-
1. The set has to have a unique mode
2. The mean, median and mode when arranged in increasing order are consecutive multiples of 3
A quick observation here would be that since all the known elements are multiples of 3, x is also a multiple of 3 for the above conditions to satisfy.

Now, if we look at the above set, the median will the average of the 3rd and the 4th element.
Also, that median has to be an integer since it is a multiple of 3.

Now, x can be placed across the following positions:
1. x<=18
In this case, x is either the 1st or the 2nd term. In any scenario, the median will be (18+21)/2 = 39/2 which is not an integer. Hence x is not <=18

2. x = 21
In this case, the median will be 21. However, since the set has a unique mode and 27 already repeats twice, 21 cannot be repeated. Hence, x is not 21.

3. x=24
Same as case 1. median is not an integer

4. x=27
In this case, median will be (27+21)/2 = 24. Also, this will not violate the mode condition. Hence, x can be 27.

5. x>27
Even in this case, the median will be 24 and this will note violate the mode condition.

Thus, we definitely know x => 27. Also consequently, the mode is also 27

Since mode is 27 and median is 24, the only two possible sets of (mean, median, mode) will be (21, 24, 27) or (24, 27, 30)
Thus mean can only be 21 or 30

Now let's look at the mean condition.
Mean = (x+6+18+21+27+27)/6 = (x+99)/6 = x/6 + 99/6
Since, 99 is not divisible by 6, if x is divisible by 6, the mean will not be an integer.

For mean to be 21,
21 = (x+99)/6
Thus, x = 27

For mean to be 30,
30 = (x+99)/6
Thus, x = 81

Thus the values of x = {27,81}
Range of x = 81-27 = 54

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total integers 6, 27 already occurs twice and as there's only one unique mode we want, mode has to be 27.
now, cases possible as we want 3 consecutive multiple of 3.
21,24,27: x = 27
24,27,30: x = 81
27,30,33: as 3rd and 4th numbers will still be 21 and 27 making median as 24, not possible.
therefore range = 81-27 = 54.

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X , 6, 18, 21, 27, 27

GIVEN UNIQUE MODE
AND ALSO MEAN MEDIAN MODE CONSECUTIVE MUTIPLE OF 3

SO IF X IS BETWEEN 1 TO 18 THEN MEDIAN WILL BE (21+18)/2=19.5 WHICH IS NOT EVEN A MUTIPLE OF 3. SO TILL THIS X IS NOT POSSIBLE

X ALSO CAN' T BE 18 TO 21 SINCE THIS WILL MAKE THE MEDIAN AGAIN NOT MULTIPLE OF 3.

FROM 21 TO 26 ALSO MEDIAN WILL NOT BE MULTIPLE OF 3

ONLY AT X=27 MEDIAN IS (21+27)/2=24 MULTIPLE OF 3

CHECK FOR THIS VALUE OF X MODE AND AVERAGE. MODE WILL BE 27 AND AVERAGE WILL BE 21 AVERAGE( 6+18+21+27+27+27+27)/6=21
21(MEAN), 24(MEDIAN) AND 27( MODE SATISFY THIS CONDITION). x=27 VALID

FOR X GREATER THAN 27: MEDIAN AND MODE REMAINS AS 24 AND 27 RESPECTIVELY, IT WILL NOT CHANGE WITH VALUE OF X. ONLY MEAN CHANGES SO MEAN HAS TO BE 30 FOR 24, 27 AND 30 TO BE CONSECUTIVE MULTIPLE OF 3. FOR MEAN TO BE 30.

SOLVE 30*6= X+6+18+21+27+27 THIS GUVES X=81. THIS BECOMES VALID. BEYOND THIS MEAN WILL GO BEYOND 30 AND CONDITION OF CONSECUTIVE MULTIPLE OF 3 FOR MEAN MEDIAN AND MODE WILL FAIL.

sO, ONLY TWO VALUES OF X IS POSSIBLE 27 AND 81 AND THIER RANGE IS 81-27=54( ANSWER)

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Given set of numbers: x, 6, 18, 21, 27, 27

For mode: It's mentioned that there is only 1 unique mode in the set of numbers, and 27 already appears twice so it must be the mode. Hence, x cannot be 6, 18 or 21.

Now, for median, there can be multiple cases, but it's given in the question that the mean, median and mode are each consecutive multiples of 3. Therefore, the median must be divisible by 3.
Case 1: x <=18: The ordered set may be written as x, 6, 18, 21, 27, 27 or 6, x, 18, 21, 27, 27.
Either way, the median is (18 + 21) / 2 = 19.5 which is not a multiple of 3 and hence, x cannot be less than 18.
Case 2: x is between 18 and 21 i.e., 18 < x < 21 (since we've already established above that x cannot be 18 or 21): This constraint then goes to follow that x = 19 or x = 20 (meaning that the ordered set is 6, 18, (x = 19 or 20,) 21, 27, 27). In either case, the value of the median (20 or 20.5 respectively) will not be a multiple of 3 and hence x cannot be between 18 and 21.
Case 3: 21 < x <= 27: In this case, the ordered set will be 6, 18, 21, x, 27, 27 and hence the median will be (21 + x) / 2. Now, we need the result of this term to be a multiple of 3 and the only value of x that satisfies these conditions is 27. Therefore, x = 27 and the median of the ordered set will be 24.
Case 4: x > 27: The ordered set in this case will be 6, 18, 21, 27, 27, x and the middle two values will always be 21 and 27 and the median will always be 24.
Hence, keeping all these cases in mind, the median must be 24.

For mean: Now, we know the mode is 27 and the median is 24, so, for the mean, median and mode to all be consecutive multiples of 3, the mean must either be 21 or 30. That way the mean, median, mode (not in that particular order) can be written in two sets i.e., {21, 24, 27} and {24, 27, 30}.

So, let's calculate the value of x in each of these cases:
We know that 6 + 18 + 21 + 27 + 27 = 99
Case 1: The mean is 21. Hence,
(99 + x) / 6 = 21
99 + x = 126
x = 27

Case 2: The mean is 30. Hence,
(99 + x) / 6 = 30
99 + x = 180
x = 81

Hence, we have two valid values of x which are 27 and 81, the range which is 81 - 27 = 54.

Final answer: B. 54

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13 Jul 2026, 10:47

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Arrange from smallest to largest: 6,18,21,27,27
anywhere x goes, > 21, 21 and 27 will be in the middle
Median=(21+27)/2=24

the mode is 27,median is 24
24, 27 consecutive multiples of 3
the mean must be lower or higher to complete the three
Lower possibility Mean= 21(21,24,27)
Higher possible Mean= 30 (24,27,30)

Diff in x= diff in mean*No. of terms
Diff in mean= 30-21=9
No. of terms=6
Range of x=9*6=54
Ans:B 54

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Mean = (x+6+18+21+27+27) / 6 = x/6 + 99/6, this means average is >=18
Mode definitely be 27.
Median we need to find.
So there are three case possible with 27
(21,24,27), (24,27,30), (27,30,33) --> in all cases x>=27 so median has to be 21+27/2 = 24
So remove the last case.
Since we want to find the range of x. first consider the lowest avg possible.
For the lowest avg case: when 21 is avg, median is 24 and mode is 27 so the value for x=27
Now 2nd for highest avg case: avg= 30, X= 30*6 - 99 = 81

So the range of x= 81-27= 54. option B

Bunuel

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The question asks for the range of possible values of X.

- "Average", "Unique mode" and "Median" are consecutive multiples of 3 which means there are difference of 3 between them.
- First of realize that "Unique mode" means that there can only 1 single mode, which is 27 because if we add any other repeating number then we would not have any Unique mode. (Unique Mode=27)
- Our Mean is --------> (99+X)/6
To be devisable by 6, "X" has to be odd since Odd+Odd=Even and Divisibility by 6 requires dividend to be even and the sum of the digits of the number to be multiple of 3. Also mean has to be integer to be multiple of 3
-Since Unique mode is 27 (Difference of 3 between avg, mode and median) other numbers can be 21,24,27,30,33
If average is 21 -------> X = 27 (Total sum 126)
If average is 24 -------> X = 45 (Total sum 144)
Not any of the two values can be same so average can't be 27
If average is 30 -------> X = 81 (Total sum 180)
If average is 33 -------> X = 99 (Total sum 198)
Regardless of which number we pick for X median becomes 24 because middle two values are always 27 and 21.
Median=24
Mode=27
Since these 3 values have difference of 3 average can be either 21 or 30
So X=(27,81)
Range of possible values of X = 81-27 = 54
Ans (B)
IMO

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sum of 5 integers except x = 99
since it is mentioned that mode is unique, any of the stated integers except for 27 will be tied for mode alongside 27 which will nullify what is stated as fact. hence mode is 27. x cannot be 6, 18, 21.

we are given that mean, median, mode are consecutive multiples of 3. so the possibilities are: 21, 24, 27 or 27, 30, 33 or 24, 27, 30
we also know that mean = (x+99)/6 since we know mean is integer, x+99 has to be multiple of 6 (hence multiple of both 2 & 3). to be multiple of 2, x+99 has to be even. since 99 is odd, x has to be odd for their sum to be even. to be multiple of 3, x has to be multiple of 3 otherwise it would leave remainder b/c 99 is already multiple of 3. thus we conclude that x is odd & multiple of 3 .

next we look at median. when these integers are arranged in ascending order, the median is avg of 3rd & 4th terms.
if x is the first or second term, then median: (18+21)/2 = 19.5 since we know median is integer this is not possible.
since x cannot be 21, next we check if x is 27 (4th or 5th term) then median: (21+27)/2 = 24. hence mean = (99+27)/6 = 21. mean, median, mode becomes 21, 24, 27 which is valid.
now we check if x is the 6th term i.e. greater then 27. then median is same = 24. thus median & mode are 24 & 27 respectively. mean could be 21 or 30. we know that mean of 21 will give us x = 27.
so we check for x when mean is 30. (x+99)/6 = 30 => x = 81

hence the possible values of x are: 27 or 81. the range is 81-27 = 54
B

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Since they have said its a unique mode. Mode = 27.
It is impossible for Median to be greater than 30 or less than 19.5. Thus, to get median as a clean multiple of 3, x has to be 21 or above, therefore the median will be 24.

This indicates that our mean can be either 21 or 30.
Range of x = Range of Sum
= n x (max mean - min mean)
= 6 x (30-21)
= 6 x 9 Option B
= 54

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x,6,18,21,27,27

now we have three things: avg, median and mode.

so to have unique mode, one num must be repeating more than others. here 27 is twice in the list. so any other existing num cant be there. hence x has to be greater than or equal to 27.
if x is 27 or more than that, then median will be 24 all the time.

now we found 2 multiples of 3. they are 24 and 27. we are told that when we set them in increasing order they are all consecutive multiple of 3.
after 24 and 27, next multiple is 30.

so that means avg is 30. based on that value of x= 81.

we know min value of x= 27 adn max is 81.

so range is 54.

choice B

Bunuel

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- There are already 2 - 27's and the only contender for unique mode
- Medium is also a multiple of 3, and by looking at the numbers the only options for medium of these numbers been a multiple is 3 is when x = 21 or anthing >27. X=21 is not possible beacuse the question says the set has a unique mode so, medium has to be 24, where x can be anything >27

- Unique mode = 27 (9x3)
Medium = 24 (8x3)
Mean can be (7x3) or (10x3) = 21 or 30

When mean is 21, x= 27
and when mean is 30, x = 81

Range of x = 81-27 = 54

Bunuel

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IMO B

For max value of x: 6, 18, 21, 27, 27, x
Mean = (99+x)/6
Mode = 27
Median = (21+27)/2 = 24
So Mean = 30 (since mean median mode when arranged in ascending order should be consecutive multiple of 3)
Hence x= 81

For min value of x: x, 6, 18, 21, 27, 27
here notice that for any small value of x, median = (18+21)/2 which is not an integer & neither multiple of 3
Hence x should lie either between (a)18 & 21 or between (b) 21 & 27
only one integer value satisfies if (a) is considered which is 21 itself, which gives median =21, mode = 27 & 21, again not satisfies, since unique mode will not be there, hence x should be between 21 & 27, there is only 1 value whihc satisfies the median if x=27
So, Mean = 21, median = 24 and mode = 27

Hence range of value of x = 81-27 = 54
Hope this helps!

Bunuel

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13 Jul 2026, 12:05

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Bunuel

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As the set has a unique mode, the mode will be 27
Mean = x + 6 + 18 + 21 + 27 +27/ 6 = x+99/6
Median = 3rd + 4th value /2
We can get 2 possible values of median
1. Median = 18 + 21/2 = 39/2 =19.5.
This will be rejected as it is given that median is a multiple of 3
Hence median = x + 21/2

Now we can form 3 cases
Case 1 - Mode is the highest and mean and median are less then mode
We take Mean = 21, Median = 24 and equate
Mean = x + 99/6 = 21
x= 27
OR median = x+21/2 = 24
x = 27 again

Case 2 - Mode between mean and median
Hence Mean = 30, Median = 24
30= x + 99/6
x = 81

Case 3 - Mode is least
Both mean and median cannot be greater than 27 as this would be against the given set
hence this case is rule out

Xmax = 81 and Xmin = 27
Range = 54 (B)

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13 Jul 2026, 12:11

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Start plugging in numbers to x from the last option.
B----Is the answer---

Add 81+27+27+21+18+6/6=180, which gives us 30 as distinct mean.

Now median would be 24 and mode is ofcourse 27 which comes two times.

Hence the range would be 81-27=54.

We got 27 from inputting that value as x, where u need to follow the same process.

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13 Jul 2026, 12:14

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Mode is 27

Sum of values is 99 + x.

The consecutive multiples can be within 21 , 24 , 27 , 30 or 33 i.e 27 has to be a part of it.

x < 18, median = 39/2 = 19.5 i.e not a muliple of 3
18 < x < 21 again wont get a multiple of 3 as the median
x = 27 then median is 24 and total value becomes 99 + 27 i.e 126 ---> avg becomes 126/6 = 21
This works

When x > 27 then median remains as 24 and avg increases.

So when median is 24, mode is 27 and x > 27, the avg can be 30

For the avg of 30 the value of x has to be 180 - 99 = 81

Hence the higest value of x can be 81 and lowest value is 27 hence range is = 54

IMO option B

Bunuel

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13 Jul 2026, 12:27

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From the looks of it the Mode has to be 27.
So there are three possibilities of numbers if the numbers are consecutive multiples of 3.

Case 1: 21, 24, 27
Case 2: 24, 27, 30.
Case 3: 27, 30, 33

Mean=(x + 99)/6
So we can say x = 6 (Mean) - 99

If mean is 21: x = 6x21 - 99 = 27.
The series becomes: 6, 18, 21, 27, 27, 27
The Mean becomes 21, Median becomes (21+27)/2=24, and Mode is 27 which is acceptable.

If Mean is 24: x = 6x24 - 99 = 45.
The series becomes: 6, 18, 21, 27, 27, 45
Mean is 24, Median = (21+27)/2 = 24 which is not acceptable because values are not consecutive.

If Mean is 27: x = 6x27 - 99 = 63.
The series becomes 6, 18 , 21. 27, 27, 63
Mean is 27, Median is (21+27)/2 = 24, Mode = 27.
Which is Acceptable.

If mean is 30: x = 6x30 - 99 = 81.
The series becomes: 6, 18, 21, 27, 27, 81.
Mean is 30, Median is (21+27)/2 = 24 and mode is 27.
This is acceptable.

If mean is 33: x = 6x33 - 90 = 99
The list will become: 6, 18, 21, 27, 27, 99
Mean is 33, Mode is 27, Median is 24. This is not acceptable again.

Hence the min value of x is 27 and max is 81.
Range is 81-27 = 54.
ANs IMO is B.

Bunuel

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