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13 Jul 2026, 08:00
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B
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Difficulty:
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Question Stats:
57% (02:17) correct
43%
(03:03)
wrong
based on 221
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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
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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M46-42
13 Jul 2026, 09:00
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GMAT Club Official Solution:
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
In the list 6, 18, 21, 27, 27, the number 27 appears twice, and each other number appears once. So 27 is already a mode. However, for the list to have a unique mode, x cannot be 6, 18, or 21, because any of those values would then also appear twice, and 27 would no longer be the unique mode.
Since the average, median, and mode are consecutive multiples of 3, and the unique mode is 27, the three values could be:
{21, 24, 27}
{24, 27, 30}
{27, 30, 33}
Now consider the median. Since the list has an even number of terms, the median is the average of the two middle terms.
If x < 27, then the median is less than 24. Among the possible sets above, the only value less than 24 that could be the median is 21. But a median of 21 would require x = 21, which is not allowed because then 21 and 27 would both appear twice.
So x cannot be less than 27.
When x >= 27, the two middle terms are 21 and 27, so the median is:
(21 + 27)/2 = 24
Thus, only the following two sets remain possible:
{21, 24, 27}
{24, 27, 30}
So the average must be either 21 or 30.
The sum of the six integers is:
x + 6 + 18 + 21 + 27 + 27 = x + 99
So the average is:
(x + 99)/6
Case 1: {21, 24, 27}
(x + 99)/6 = 21
x + 99 = 126
x = 27
Case 2: {24, 27, 30}
(x + 99)/6 = 30
x + 99 = 180
x = 81
So the possible values of x are 27 and 81.
The range of all possible values of x is:
81 - 27 = 54
Answer: B.
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
In the list 6, 18, 21, 27, 27, the number 27 appears twice, and each other number appears once. So 27 is already a mode. However, for the list to have a unique mode, x cannot be 6, 18, or 21, because any of those values would then also appear twice, and 27 would no longer be the unique mode.
Since the average, median, and mode are consecutive multiples of 3, and the unique mode is 27, the three values could be:
{21, 24, 27}
{24, 27, 30}
{27, 30, 33}
Now consider the median. Since the list has an even number of terms, the median is the average of the two middle terms.
If x < 27, then the median is less than 24. Among the possible sets above, the only value less than 24 that could be the median is 21. But a median of 21 would require x = 21, which is not allowed because then 21 and 27 would both appear twice.
So x cannot be less than 27.
When x >= 27, the two middle terms are 21 and 27, so the median is:
(21 + 27)/2 = 24
Thus, only the following two sets remain possible:
{21, 24, 27}
{24, 27, 30}
So the average must be either 21 or 30.
The sum of the six integers is:
x + 6 + 18 + 21 + 27 + 27 = x + 99
So the average is:
(x + 99)/6
Case 1: {21, 24, 27}
(x + 99)/6 = 21
x + 99 = 126
x = 27
Case 2: {24, 27, 30}
(x + 99)/6 = 30
x + 99 = 180
x = 81
So the possible values of x are 27 and 81.
The range of all possible values of x is:
81 - 27 = 54
Answer: B.
General Discussion
13 Jul 2026, 08:14
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Here's my solution for this question
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13 Jul 2026, 08:18
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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
average of given sequence is
x+6+18+21+27+27 /6 ; 99+x/6
avg, median , mode all have to multiple of consecutive multiples of 3
test with values of x in
(99+x) / 6 ; the value has to be such that it is multiple of 3 and is twice divisible
smallest value of x will be 27
99+27 = 126 ; avg is 21
6,18,21,27,27,27
avg 21
median 24 mode is 27
(21, 24,27)
and largest value of x will be at 81
99+81 = 180 ; avg 30
6,18,21,27,27,81
avg 30 , median is 24 mode is 27 ( 24,27,30)
range of value of x 81-27 =54
OPTION B is correct
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
average of given sequence is
x+6+18+21+27+27 /6 ; 99+x/6
avg, median , mode all have to multiple of consecutive multiples of 3
test with values of x in
(99+x) / 6 ; the value has to be such that it is multiple of 3 and is twice divisible
smallest value of x will be 27
99+27 = 126 ; avg is 21
6,18,21,27,27,27
avg 21
median 24 mode is 27
(21, 24,27)
and largest value of x will be at 81
99+81 = 180 ; avg 30
6,18,21,27,27,81
avg 30 , median is 24 mode is 27 ( 24,27,30)
range of value of x 81-27 =54
OPTION B is correct
