Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Aug 2710:00 AM EDT
-11:00 AM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
Sep 0201:00 PM EDT
-11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 0410:00 AM PDT
-11:00 AM PDT
What does test anxiety really look like on the GMAT? For many test takers, it does not look like a panic attack. Instead, it shows up as brain fog, rereading, negative comparisons, mental noise....
Sep 0609:30 AM EDT
-12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.- Sep 07
12:01 AM PDT
-11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT - Sep 09
07:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway 
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2108:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
07 Aug 2026, 05:03
Kudos
Bookmarks
\(x, 6, 18, 21, 27, 27\)
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
07 Aug 2026, 05:03
Kudos
Bookmarks
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 \(6, 18, 21, 27, 27, x\), 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 \(x \geq 27\), the two middle terms are 21 and 27, so the median is:
\(\frac{21 + 27}{2} = 24\)
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:
\(\frac{x + 99}{6}\)
Case 1: \(\{21, 24, 27\}\)
\(\frac{x + 99}{6} = 21\)
\(x + 99 = 126\)
\(x = 27\)
Case 2: \(\{24, 27, 30\}\)
\(\frac{x + 99}{6} = 30\)
\(x + 99 = 180\)
\(x = 81\)
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 \(6, 18, 21, 27, 27, x\), 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 \(x \geq 27\), the two middle terms are 21 and 27, so the median is:
\(\frac{21 + 27}{2} = 24\)
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:
\(\frac{x + 99}{6}\)
Case 1: \(\{21, 24, 27\}\)
\(\frac{x + 99}{6} = 21\)
\(x + 99 = 126\)
\(x = 27\)
Case 2: \(\{24, 27, 30\}\)
\(\frac{x + 99}{6} = 30\)
\(x + 99 = 180\)
\(x = 81\)
So the possible values of \(x\) are 27 and 81.
The range of all possible values of \(x\) is:
\(81 - 27 = 54\)
Answer: B
