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When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, 04 Jan 2023, 02:30
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When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, is the median of new 8 numbers equal to the average (arithmetic mean) of new 8 numbers?

(1) The median of new 8 numbers is 10
(2) The average (arithmetic mean) of new 8 numbers is 10


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When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, 04 Jan 2023, 03:03
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Bunuel
When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, is the median of new 8 numbers equal to the average (arithmetic mean) of new 8 numbers?

(1) The median of new 8 numbers is 10
(2) The average (arithmetic mean) of new 8 numbers is 10
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Couple of things to observer before we move into the statements.

  1. All the numbers in the set are equally spaced, i.e. the numbers are in arithmetic progression
    In a equally spaced set, the mean is same as the median.
  2. As the resultant set consists of even number of terms, the median will not be a member of the set as it will be an even number (for this set only).

Let's arrange the set in increasing order

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Statement 1

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

As the median is 10, 9 and 11 (indicated by red) are the middle two terms.

Therefore we must have removed one number whose value is less than 9 and one number whose value is greater than 11.

In other words, we must have removed one number from the set of numbers in green and the other from set of numbers indicated in blue.

However, we do not know which numbers were removed. If the first term i.e. 1 and last term i.e. 10 were removed, the members of the resultant set will still be equally spaced and the mean of the resultant set will be equal to the median of the resultant set.

If any other terms were removed, the members of the set will no longer remain equally spaced, and the mean will not be equal to the median.

The statement is not sufficient.

Statement 2

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Given that the average of the set is 10, hence the total of the resultant set is 8 * 10 = 80

Total of current set = 100

Hence the sum of the two terms that were removed = 100 - 80 = 20

Therefore, the terms that were removed were 1 and 19.

3, 5, 7, 9, 11, 13, 15, 17

The resultant members of the resultant set are equally spaced, hence the mean of the set = median of the set.

Sufficient.

Option B
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When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, 09 Jan 2023, 10:33
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Bunuel
When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, is the median of new 8 numbers equal to the average (arithmetic mean) of new 8 numbers?

(1) The median of new 8 numbers is 10
(2) The average (arithmetic mean) of new 8 numbers is 10

Couple of things to observer before we move into the statements.

  1. All the numbers in the set are equally spaced, i.e. the numbers are in arithmetic progression
    In a equally spaced set, the mean is same as the median.
  2. As the resultant set consists of even number of terms, the median will not be a member of the set as it will be an even number (for this set only).

Let's arrange the set in increasing order

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Statement 1

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

As the median is 10, 9 and 11 (indicated by red) are the middle two terms.

Therefore we must have removed one number whose value is less than 9 and one number whose value is greater than 11.

In other words, we must have removed one number from the set of numbers in green and the other from set of numbers indicated in blue.

However, we do not know which numbers were removed. If the first term i.e. 1 and last term i.e. 10 were removed, the members of the resultant set will still be equally spaced and the mean of the resultant set will be equal to the median of the resultant set.

If any other terms were removed, the members of the set will no longer remain equally spaced, and the mean will not be equal to the median.

The statement is not sufficient.

Statement 2

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Given that the average of the set is 10, hence the total of the resultant set is 8 * 10 = 80

Total of current set = 100

Hence the sum of the two terms that were removed = 100 - 80 = 20

Therefore, the terms that were removed were 1 and 19.

3, 5, 7, 9, 11, 13, 15, 17

The resultant members of the resultant set are equally spaced, hence the mean of the set = median of the set.

Sufficient.

Option B
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Quick question - For statement 2, how do we know 1 & 19 are removed? We can remove 3 & 17 (sum of 20) as well??
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When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, 09 Jan 2023, 13:44
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Quick question - For statement 2, how do we know 1 & 19 are removed? We can remove 3 & 17 (sum of 20) as well??
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Good catch ! Yes 17 and 3 can also be pair that could have been removed.

However the point to note here is whatever the pair might be, it will always be equidistant from the mean. This is required to keep the mean at 10.

The difference between Statement 1 and Statement 2 is that, statement 1 provides information with respect to median. Hence we are free to remove any term as long as we are removing one term is less and the other term is greater than the median. However, in the second statement imposes an additional constraint of distance. Hence, now we can no longer remove any term. We can only remove term which are equidistant on either side from mean.
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When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, 27 Aug 2025, 10:40
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I tried solving the question myself and followed the given logic. With Statement 1, removing 9 and 11 still gives a total of 80 and an average of 10, with 7 and 13 as the 4th and 5th numbers, so the median is also 10. I don’t see why only Statement 2 is sufficient?
Bunuel
When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, is the median of new 8 numbers equal to the average (arithmetic mean) of new 8 numbers?

(1) The median of new 8 numbers is 10
(2) The average (arithmetic mean) of new 8 numbers is 10


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When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, Updated on: 27 Aug 2025, 11:11
Originally posted by Bunuel on 27 Aug 2025, 11:09.
Last edited by Bunuel on 27 Aug 2025, 11:11, edited 1 time in total.
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KRNSW
I tried solving the question myself and followed the given logic. With Statement 1, removing 9 and 11 still gives a total of 80 and an average of 10, with 7 and 13 as the 4th and 5th numbers, so the median is also 10. I don’t see why only Statement 2 is sufficient?

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From (1), if we remove 17 and 19, we get {1, 3, 5, 7, 9, 11, 13, 15}, which has the median and the average of 8. So, this case gives a YES answer to the question. However, if we remove 11 and 13, we get {1, 3, 5, 7, 9, 15, 17, 19}, which also has the median of 8 but the average is now 9.5. So, this case gives a NO answer to the question.

From (2), the average of 8 numbers being 10 implies the sum is 80. The sum of the original 10 numbers is 100, so we removed 2 numbers totaling 20. This means we removed 1 and 19, 3 and 17, 5 and 15, 7 and 13, or 9 and 11. In any of these cases, the resulting set has the median of 8. So, we get a YES answer to the question.

P.S. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

  • Word Problems
  • Work Problems
  • Distance Problems
  • Mixture Problems
  • Percent and Interest Problems
  • Overlapping Sets Problems
  • Statistics Problems
  • Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.­
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When 2 numbers are removed from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19, 27 Aug 2025, 11:14
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Okay. Now I get it. We're getting both YES and NO situation from statement 1.
Thank you Bunuel!
Bunuel


From (1), if we remove 17 and 19, we get {1, 3, 5, 7, 9, 11, 13, 15}, which has the median and the average of 8. So, this case gives a YES answer to the question. However, if we remove 11 and 13, we get {1, 3, 5, 7, 9, 15, 17, 19}, which also has the median of 8 but the average is now 9.5. So, this case gives a NO answer to the question.

From (2), the average of 8 numbers being 10 implies the sum is 80. The sum of the original 10 numbers is 100, so we removed 2 numbers totaling 20. This means we removed 1 and 19, 3 and 17, 5 and 15, 7 and 13, or 9 and 11. In any of these cases, the resulting set has the median of 8. So, we get a YES answer to the question.

P.S. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

  • Word Problems
  • Work Problems
  • Distance Problems
  • Mixture Problems
  • Percent and Interest Problems
  • Overlapping Sets Problems
  • Statistics Problems
  • Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.­
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