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

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to pre-think assumptions and solve the most challenging questions in less than 2 minutes.

Want to solve 700+ level Algebra questions within 2 minutes? Attend this free webinar to learn how to master the most challenging Inequalities and Absolute Values questions in GMAT

A set of 13 integers has a median of 10 and an average of 15.
[#permalink]

Show Tags

12 May 2018, 09:38

9

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

36% (02:32) correct 64% (03:09) wrong based on 72 sessions

HideShow timer Statistics

A set of 13 integers has a median of 10 and an average of 15. If the highest term in the set is a unique, non-repeating integer, what is the lowest possible value of the highest integer in the set? A. 81 B. 25 C. 22 D. 21 E. 20

Re: A set of 13 integers has a median of 10 and an average of 15.
[#permalink]

Show Tags

16 May 2018, 06:05

1

Here's my method of finding it via concepts: 13 integers are present, and the median is 10. In order to minimize the value of X, the largest integer, we have to maximize the value of all integers till 10. The max value for each integer till the median is 10. So, our list has 7 integers, with value, '10'. The sum of our list is 195. 10*number of integers till median = 10*7=70. Apart from these, we have 6 remaining integers. Substract 70 from 195, we get 125. So, the sum of the remaining 6 integers must be 125. Now, in order to minimize the highest unique value, we must maximize the integers. If we assume these to be 20, we have 6*20=120 which is less than 125. Now assume them to be 21, we have 6*21=126, which is greater than 125. Now write, 20 20 20 20 20 25(X) ...Matches our answer choice, but let's see if we can lessen the value of X, somehow. 20 20 20 20 21 24 20 20 20 20 22 23 20 20 20 21 21 22 After this, we will not be able to minimize the value of X, because then X would not be a UNIQUE integer in the list. Hence, Option C is our answer.

A set of 13 integers has a median of 10 and an average of 15. If the highest term in the set is a unique, non-repeating integer, what is the lowest possible value of the highest integer in the set? A. 81 B. 25 C. 22 D. 21 E. 20

A set of 13 integers has a median of 10 and an average of 15.
[#permalink]

Show Tags

12 May 2018, 11:47

benejo wrote:

A set of 13 integers has a median of 10 and an average of 15. If the highest term in the set is a unique, non-repeating integer, what is the lowest possible value of the highest integer in the set? A. 81 B. 25 C. 22 D. 21 E. 20

To make X- the last 13th member of the set, we have to maximize other values in the set.

1) Median (7th member) = 10, then we can max 1-6th values also to 10 2) Numbers in the set are integers, so X will be unique min, when 8-12th numbers = X-1 3) Sum of the set = avarage*number of set members = 15*13 We got an uquality:

10*7 + 5*(x-1) + x = 13*15 3x=65 x= 21.7 Since x- is integer we can say it is 21 or 22, so 3x=66 or 3x=63. In second case (63) we have to give extra 2 points to any number/numbers of the set which will a)move median b) will make other numbers = or > than the X. So the only scenario is possible to "take" extra 1 point from any other number of the set. Hence Answer is 22 (C)

Re: A set of 13 integers has a median of 10 and an average of 15.
[#permalink]

Show Tags

12 May 2018, 12:22

benejo wrote:

A set of 13 integers has a median of 10 and an average of 15. If the highest term in the set is a unique, non-repeating integer, what is the lowest possible value of the highest integer in the set? A. 81 B. 25 C. 22 D. 21 E. 20

To make X- the last member of the set, we have to maximize other values in the set.

We know that median is the middle value. 7th term is 10. We need to minimize the 13th term.

1) 6 terms below Median can be not more than 10. So, 10 is the maximum value. 2) 7th term is 10. Given in question. 3) From 8th to 12th term can take maximum value of - One less than 13th term.

Numbers in the set are integers, Let X will be unique min 13th term, and 8th -12th numbers be X-1 each Sum of the set = average*number of set members = 15*13

10*7 + 5*(x-1) + x = 13*15 5*(x-1) + x = 195 - 70 5*(x-1) + x = 125---------------------------> 1 5x-5+x = 195 - 70 6x = 125+5

x = 130/6 = 21.7

Since x- is integer we can say it is 21 or 22,

Let's assume x =21,

8th term - 12th term is 20. And 13th term is 21. Sum is 20*5 + 21= 121. But it should add up to 125. See eq 1

4 less than the value it should add up to. This 4 can be adjusted by increasing the values of last four terms ( 10th, 11th, 12th and 13th term).

So now, 10th term is 21, 11th term is 21, 12th term is 21 and 13th term is 22.