A set of 11 positive integers have an average of 25. Which

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BTGmoderatorLU: Moderator; Posts: 2544; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

A set of 11 positive integers have an average of 25. Which

by BTGmoderatorLU » Tue Feb 05, 2019 2:26 pm

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Source: EMPOWERgmat

A set of 11 positive integers have an average of 25. Which of the following is the greatest possible value for the median of this set?

A. 25
B. 30
C. 36
D. 45
E. 46

The OA is D

Source: Beat The GMAT — Problem Solving | Tue Feb 05, 2019 2:26 pm

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Wed Feb 06, 2019 2:11 am

BTGmoderatorLU wrote:Source: EMPOWERgmat

A set of 11 positive integers have an average of 25. Which of the following is the greatest possible value for the median of this set?

A. 25
B. 30
C. 36
D. 45
E. 46

The OA is D

Given the average of 11 positive integers is 25, we have the sum of the 11 positive integers = 11*25 = 275.

Again, given that there are 11 numbers in the set, the median would be the 6th number.

Since we need to maximize the median, we must have the values of 1st to 5th numbers smallest. The smallest possible values of 1st to 5th numbers would be 1 each.

Again, since median would be the 6th numbers, let's assume that 6th to 11th numbers are same.

So, the 11 numbers would be 1, 1, 1, 1, 1, M, M, M, M, M, M

Thus, the sum of 11 numbers = 5*1 + 6*M = 275

=> M = 45.

The correct answer: D

Hope this helps!

-Jay
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Scott@TargetTestPrep: GMAT Instructor; Posts: 8132; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Feb 06, 2019 6:32 pm

BTGmoderatorLU wrote:Source: EMPOWERgmat

A set of 11 positive integers have an average of 25. Which of the following is the greatest possible value for the median of this set?

A. 25
B. 30
C. 36
D. 45
E. 46

The OA is D

The sum of the 11 positive integers is 11 x 25 = 275. If we let m = smallest integer and n = largest integer, we can have 5 values of m and 6 values of n such that the median is n. Since we want the median to be as large as possible, we want the 5 values of m to be as small as possible. Since the smallest positive integer is 1, we can let m be 1. Thus, we have:

5 x 1 + 6n = 275

6n = 270

n = 45

Answer: D

Scott Woodbury-Stewart
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