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

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AthaniX

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13 Jul 2026, 23:47

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We can have two possible ranges for mean, median, mode
21,24,27
or
24,27,30

In our case x has to be >= 27

Mode = 27
median = 21+27=24

Average - (x+99)/6

Average - 21
x+99=126
x=27

Average - 24
x+99=144
x=45

Average - 30
x+99=180
x=81

Range- 81-27=54
Answer B IMO

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14 Jul 2026, 00:20

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Here's how I approached it:

The unique mode is always 27.
Since the mean, median and mode are consecutive multiples of 3, the 3 values must be either
-> 21, 24, 27 (if 27 is the largest)
-> 24, 27, 30 (if 27 is the middle)

Now lets see where x can lie
-> If x < 21, the median is < 21, so it cannot be 24
-> If 21 < x < 27, the median is (21 + x)/2, which is never 24
-> If x >= 27, the median is (21+27)/2 = 24
So only x >= 27 works

Mean = (99+x)/6
Since the median is 24 and the mode is 27, mean must be either 21 or 30.
-> Mean = 21 -> x = 27
-> Mean = 30 -> x = 81

Possible values of x : 27, 81
Range = 81-27 = 54

Ans : B

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14 Jul 2026, 00:23

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There are multiple ways to think, though the most straightforward is through observation

Looking at the set, the only possible mode for this could be 27 - unique mode (bimodal or trimodal when x = 27)

giving 2 possible cases to check for {mean, median, and mode}

If x = 27, the median is (21+27)/2 = 24
If x =< 18, the median is (18+21)/2 = not a miltiple of 3
If x =< 6, the median is again not a multiple
If x > 27, the median is again 24

So we know mode is 27 and median is 24
The 2 cases we have with both 27 and 24 in the sets are

21,24,27
and
24,27,30

In both cases, the mean is 21 and 30, respectively

Sum of the given set is 99+x
mean is (99+x)/6
If the mean is 21, then x is 27 (min)
If the mean is 30, then x is 81 (max)

Therefore, 81-27 is the range (max-min) = 54

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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14 Jul 2026, 00:32

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The values are x, 6, 18, 21, 27, 27.
27 will be the mode unless x = 6, 18, or 21, but then we would have two modes which wouldn't be conducive to this problem. so since the mode is 27, the median and mode are one of 4 numbers since mean, median and mode are consecutive multiples of 3.

21, 24, 30, 33.

We can test if x is in the median calculation.

For (x+21)/2 with the only value of x that makes a median that is a multiple of 3 are 21 (median 21)

If x is 27, the mean is (6+18+21+27+27+27)/6 = 30, the mode is 27, and the mean is 21 which are three consecutive multiples of 3. So 27 is a possible x value.

When x>= 27, the median is 24 (27+21)/2 = 48/2 =48.

We can derive a mean from this information of having mode of 27 and a median of 24. Mean must either be 21 or 30. If the mean is 21,

(6+18+21+27+27+x)/6 = 21 x= 27

If the mean is 30, (6+18+21+27+27+x)/6 = 30, x =81.

The possible values for x are 81 and 27. 81-27 = 54 the answer is B.

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14 Jul 2026, 00:34

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In this question we need to take a note of a fixed value. The mode would not be changed as it shall be unique.

The value of x could be adjusted so the 3 consecutive multiples of 3 shall be around 27, our definitive mode.
Possible cases:
[21,24,27] - Case 1
[24,27,30] - Case 2
[27,30,33] - Case 3

Let us check for Case 1: we know (3+6+18+21+27+27+x)/6 = Avg. = (99+x)/6 ; and since there are 6 terms then the middle terms 3 and 4 would give us median.
Check for average, suppose x would be 21. Then,
x+99 = 6 * 21
x = 27 and for x = 27
the median, would be middle terms sum/2 = 21+27/2 = 24.
Hence we got one verified value of x which gives us 3 consecutive multiples of 3 of mean, median and mode in the order [21,24,27]

Case 2, Suppose mean is 30 now,
x+99 = 6 * 30
x = 81.
for x = 81,
the median would be again - (21 + 27)/2 = 24.
We got another value of x which would give 3 consecutive multiples of 3 of median, mode and mean in the order [24, 27, 30]

Case 3, Suppose mean is now 33
x+99 = 6 * 33
x = 99.
The mean would again be = 24.
if we notice this value of x would not give us 3 consecutive multiples of 3.

Hence only 2 cases possible, x = 27 and x = 81
Answer: 81 - 27 = 54

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14 Jul 2026, 01:28

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Mean, median and unique mode are three consecutive multiples of 3. The mode is 27. So I wrote down three consecutive multiples of 3, each containing 27. Look at my ilustrative solution.

Attachments

IMG_7709.jpeg [ 1.58 MiB | Viewed 121 times ]

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14 Jul 2026, 03:49

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Given: 6, 18, 21, 27, 27, X

Since 27 already appears twice, the mode is 27.

The average, median, and mode are consecutive multiples of 3, so the only possibile sets are:

- 21, 24, 27
- 24, 27, 30

If x is >= 27:

Median: (21+27)/2=24
So, the average must be 30.
(x+99)/6 = 30
-> x = 81

If x = 27

Median = 24
The three values are 21, 24, 27, so the average 21.
(x+99)/6 =21
-> x = 27

No value of x <27 gives a median that is a multiple of 3.

So, the possible values of x are 27 and 81.

Range = 81 - 27 = 54

Answer: (B) - 54

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14 Jul 2026, 04:49

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x, 6, 18, 21, 27, 27
x can’t be 6, 18, or 21 as they have a unique mode. If x is any other number, then Mode is definitely 27.
If x is less than 18, then Median= (18+21)/2=19.5
If value of x is between 18 and 27 exclusive, (x can be 24 only) Median= (21+24)/2=22.5
If x is more than 24, then Median= (21+27)/2=24.
Except 24, all other median values are invalid. Only possible median is 24.
Median=24 and Mode=27
They are consecutive integers so mean can be either 21 or 30.
Mean= x+6+18+21+27+27=21*6=126....... x=27
Mean= x+99=30*6=180......... x=81
The range for possible values of x= 81-27=54

B. 54

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14 Jul 2026, 05:02

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Let's try fixing the values. The easiest is mode. Since the question says that there is a unique mode and 27 already appears twice, 27 has to be the mode. If x takes a value other than whatever is in the list, 27 will still be the mode. And due to there being a unique mode, x cannot take any value other than 27, in which case again, 27 will be mode. Hence mode = 27.

Given mode is 27 and the three measures (mean, mode, median) have to be consecutive multiples of 3, there can only be 6 cases:
Case 1: Mean<Median<Mode : 21<24<27
Case 2: Median<Mean<Mode : 21<24<27
Case 3: Mean<Mode<Median : 24<27<30
Case 4: Median<Mode<Mean : 24<27<30
Case 5: Mode<Median<Mean : 27<30<33
Case 6: Mode<Mean<Median : 27<30<33

Next, check possible values for median, which will depend on where x is placed in terms of value
Case A: x<=18 : In this case, Median will be (18+21)/2 = 19.5, None of the cases (1-6) discussed above have 19.5 as a possible value and hence this is not possible.
Case B: 18<x<21 : In this case, x can be 19 or 20, hence median = 20 or 20.5. Again, not covered in cases 1-6
Case C: x=21: In this case, median = 21. Then, mean = 20. Again, these values of mean and median do not fit any case 1-6
Case D: 21<x<27: Here, median will range from 21.5 to 23.5 and hence not a valid value
Case E: x=27 : Median = 24. Mean = 21. This is covered in Case 1.
Case F: x>27: In all these cases, Median will be fixed at 24 and Mode at 27. Only possible case is Case 4. In this case, mean = 30. Hence the total of the list has to be (30*6=180). The total of all values other than x in the list is 99 and hence x will be = 180-99 = 81.

Range of values = 81-27 = 54

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14 Jul 2026, 05:14

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IMO : B = 54
since the given integers are ==> x,6, 18,21,27,27 =====> sum =99+x
we can see that the mode can only be 27
hence we just need to figure out mean and median now
also as per the question ---> mean, median and mode are consecutive integers of 3
therefore possible cases will be among these values
3*7, 3*8, 3*9, 3*10, 3*11
now since we already have 3*9 = 27 as mode we just need to check from 21/24/30/33
lets take each case individually,
case 1 : average =21 , median =24, mode = 27
now , 99+x = 6*21
x = 27 ----with this yes we will get median =24 hence valid

case 2: average =24 , mode =27
99+x=24*6
x =45 --> not valid since median will be 24

case 3 : avg=30, mode =27
99+x=30*6
x =81
median = 24 -- hence valid
case 4 : avg =33, mode =27
99+x=198
x = 99
not valid since median will be 24

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14 Jul 2026, 05:15

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First AV Mean is (x+99)/6, Mode has to be 27 unique value)
For the median we can try different cases
If x is equal or less than 18 then the median is 19.5
If x is greater than or equal to 27 then the median is 24
If x is between 18 and 27 then the median is (x+21)/2 Only x=21 or 27 yield a multiple of 3 but x cannot be 21 coz of the unique mode meaning x is 27 meaning the median is fixed at 24
So we have two cases Mean 21 (21,24,27) or Mean 30 (24,27,30)
In case 1 x becomes 27 and in case 2 x becomes 81
So range is 81-27= 54

Bunuel

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14 Jul 2026, 05:19

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The Answer is B
mode is either ways 27.
So we need mean and median to be 21/24 and 24/30.
Only 27 and 81 help satisfy the condition
Range=81-27=54

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14 Jul 2026, 05:24

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The question mentions that the mean, median and mode are 3 consecutive integers.
In the given series, number 27 appears twice. Hence unless x is an integer which repeats from the given number, I am assuming mode to be 27.
If mode is 27, the 3 consecutive pairs can be:
i) 24, 27 and 30
ii) 27, 30 and 33
iii) 21, 24 and 27

For finding Median - there can be the below options -
a) Order 1 - x, 6, 18, 21, 27, 27
Median = 18+21/2 = 19.5 - not an integer
b) Order 2 - 6, x, 18, 21, 27, 27 - here also median is not an integer
c) Order 3 - 6, 18, x, 21, 27, 27 - here median = x+21/2 - Now since x+21/2 is a multiple of 3, x+21 is a multiple of 6. Hence amongst the option i, ii, iii above - the possible values are 24 and 30. But id x - 24 or 30, then x+ 21 will not fit in the range of order 3. Hence this is also not true
d) Order 4 - 6,18, 21, x, 27, 27 - This also has the same thoughtprocess as above - but we get a value here - x = 27 (Since x + 21 = multiple of 6, 27+21 = 48 /2 = 24 is in the range Hence X = 27
e) Order 5 - 6, 18, 21, 27, 27, x
Median = 24 Here mode is 27 and hence average = 30 For average to be 30, we back calculate and get x = 81

Hence in all the above orders, we get 2 possible value of X => 81 and 27 and the range is 81-21 = 54 . Hence Option B

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14 Jul 2026, 05:40

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The answer is B.54
Mode is 27. The three values must be consecutive multiples of three. Mode is fixed at 27, so mean and median are 21 and 24 in some order.
Mean 24 gives sum of 144, so X equals 45, and the median must be 21. But sorting 6, 18, 21,27, 45 gifts median 24, not 21. So that case fails, but check the other assignment.
Mean 21 gives sum 126, so X equals 27. That breaks the unique model. Fails too.
So instead of allow the three consecutive multiples to be other triples: mean, median, mode must be consecutive multiples of three, but mode need not be the largest. Since 27 appears twice and is only repeat. Mode equals 27, and the triple containing 27 can be 21, 24, 27, 30,33.
Working through these, the valid values are -9 and 45 giving a range of 54

Bunuel

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14 Jul 2026, 05:43

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Since there must be an unique mode and 27 appears twice, the mode must be 27, and x cannot be 6, 18 or 21.

average = (6+18+21+27+27+x)/6 = (99+x)/6
x = 6*average - 99

if 27 is the mode, then the consecutive multiples of 3 can be:

+ 21, 24, mode=27
average=21 -> x=6*21-99=27 -> the numbers are 6, 18, 21, 27, 27, 27 -> median=24 -> possible
average=24 -> x=6*24-99=45 -> the numbers are 6, 18, 21, 27, 27, 45 -> median=24 -> not possible (should be 21)

+ 24, mode=27, 30
average=24 -> x=6*24-99=45 -> the numbers are 6, 18, 21, 27, 27, 45 -> median=24 -> not possible (should be 30)
average=30 -> x=6*30-99=81 -> the numbers are 6, 18, 21, 27, 27, 81 -> median=24 -> possible

+ mode=27, 30, 33
average=30 -> x=6*30-99=81 -> the numbers are 6, 18, 21, 27, 27, 81 -> median=24 -> not possible (should be 33)
average=33 -> x=6*33-99=99 -> the numbers are 6, 18, 21, 27, 27, 99 -> median=24 -> not possible (should be 30)

x can be 27 or 81 -> range 81-27=54

IMO B

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14 Jul 2026, 06:08

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The unique mode is 27 because it is the only repeated number. The other numbers can be at most twice (if x is 6, 18 or 21) and, as the mode in unique, it is impossible.

Median must be 24:
+ if x<18 the median is (18+21)/2=19.5 (not an integer), so impossible
+ if 18<x<21, x=19 or x=20 the median would be (19+21)/2=20 (not multiple of 3) or (20+21)/2=20.5 (not an integer), so impossible
+ if 21<x<27, x=23 or x=25 (choose only odd numbers to get median is an integer) the median would be (23+27)/2=25 (not multiple of 3) or (25+27)/2=26 (not multiple of 3), so impossible

So x>=27 and median=(21+27)/2=24

mode=27, median=24, so average can be 21 or 30

average=(x+6+18+21+27+27)/6=(x+99)/6
x=6*average-99

if average=21, x=27
if average=30, x=81

range=81-27=54

Answer B

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14 Jul 2026, 06:17

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Mode is 27 as it appears twice.

Medial is 24

Therefore mean,median & mode must be consecutive & multiples of 3

Possiblity 1: Mean 21

(x + 99)/6 = 21

x = 27

Possibility 2: Mean 30

x = 81

81 - 27 = 54

B

Bunuel

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14 Jul 2026, 06:32

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Mode = 27 because it's the only repeated number in the list and mode must be unique.

Average = (6+18+21+27+27+x)/6 = (99+x)/6

Mode = 27 means 3 possibilities:
33, 30 and 27
30, 27 and 24
27, 24 and 21

x = 6*Average-99
If Average = 33, x = 6*Average-99 = 99, Median = 24 -> not consecutive multiples of 3
If Average = 30, x = 6*Average-99 = 81, Median = 24 -> consecutive multiples of 3
If Average = 24, x = 6*Average-99 = 45, Median = 24 -> not consecutive multiples of 3
If Average = 21, x = 6*Average-99 = 27, Median = 24 -> consecutive multiples of 3

x = 27 or x = 81
Range = 81 - 27 = 54

The answer is B

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Kudos: 248

[1]

14 Jul 2026, 06:46

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x,6,18,21,27,27

AM (Arithmetic mean or average), Media and unique mode when in increasing order are consecutive multiples of 3

=>Unique mode is 27 (If x = 6,18,21 then it wont be a unique mode, so x cannot be 6,18 and 21)

Now lets see at median
Case 1: x < 18
=> Median = 18+21/2 = 19.5
=> Not a multiple of 3
=> Not possible

Case 2: 18 < x < 21
=> Median = x + 21/2 = Lies between 19.5 and 21
=> Not a mulitple of 3
=> Not possible

Case 3: 21<x<27
Median = 21+x/2 = Lies between 21 and 24
=> Not a multiple of 3
=> Not possible

Case 4: x >= 27
Median = 21+27/2 = 24
=> Multiple of 3
=> Possible

We know average,median and mode are consecutive multiples of 3
=> Average must be
21 since (21,24,27) OR
30 since (24,27,30)

We know,
Sum of known numbers = 6+18+21+27+27 = 99
=>Average = x+99/6

When Average = 21
x+99/6 = 21
=> x = 27

When average = 30
x+99/6 = 30
=> x = 81

Possible values are 27 and 81

Range = 81-27 = 54

B. 54

patnaik1988

Joined: 05 Oct 2024

Last visit: 03 Sep 2026

Posts: 78

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Given Kudos: 66

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14 Jul 2026, 07:19

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If Avg, Median and Unique Mode are consecutive multiples of 3, then they can be written in the form of 3K, 3K+3, 3K+6 respectively.
Now since the sequence has only unique mode and 27 repeats twice, Mode has to be 27.
This gives 3K+6=27 ; K=7
Avg=21=(X+99)/6 ; X=27
Median=24
Since the value of X is limited by Mode=27.
Range of X=27.
Ans Choice: A.

Bunuel

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