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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?

(A) 62 (B) 68 (C) 75 (D) 88 (E) 100

Any idea how to solve this question please?

Consider 25 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{25}\).

The median of a set with odd number of elements is the middle number (when arranged in ascending or descending order), so the median of given set is \(x_{13}=50\);

The range of a set is the difference between the largest and the smallest numbers of a set, so the range of given set is \(50=x_{25}-x_{1}\) --> \(x_{25}=50+x_{1}\);

We want to maximize \(x_{25}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-12=50-12=38\) and thus the maximum value of \(x_{25}\) is \(x_{25}=38+50=88\).

The set could be {38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 88}

Answer: D.

Hope it's clear.

Thanks for the explanation Bunel...

Could you pls share some similar question for practice?

Re: A set of 25 different integers has a median of 50 and a [#permalink]

Show Tags

03 Aug 2016, 10:07

Bunuel wrote:

enigma123 wrote:

A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?

(A) 62 (B) 68 (C) 75 (D) 88 (E) 100

Any idea how to solve this question please?

Consider 25 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{25}\).

The median of a set with odd number of elements is the middle number (when arranged in ascending or descending order), so the median of given set is \(x_{13}=50\);

The range of a set is the difference between the largest and the smallest numbers of a set, so the range of given set is \(50=x_{25}-x_{1}\) --> \(x_{25}=50+x_{1}\);

We want to maximize \(x_{25}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-12=50-12=38\) and thus the maximum value of \(x_{25}\) is \(x_{25}=38+50=88\).

The set could be {38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 88}

Answer: D.

Hope it's clear.

Having all lower 13 as 50, can still lead a distribution with a median of 50 and a range of 50, whereby the answer will be E. Any thoughts?

Re: A set of 25 different integers has a median of 50 and a [#permalink]

Show Tags

03 Aug 2016, 10:25

Zaraki22 wrote:

Bunuel wrote:

enigma123 wrote:

A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?

(A) 62 (B) 68 (C) 75 (D) 88 (E) 100

Any idea how to solve this question please?

Consider 25 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{25}\).

The median of a set with odd number of elements is the middle number (when arranged in ascending or descending order), so the median of given set is \(x_{13}=50\);

The range of a set is the difference between the largest and the smallest numbers of a set, so the range of given set is \(50=x_{25}-x_{1}\) --> \(x_{25}=50+x_{1}\);

We want to maximize \(x_{25}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-12=50-12=38\) and thus the maximum value of \(x_{25}\) is \(x_{25}=38+50=88\).

The set could be {38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 88}

Answer: D.

Hope it's clear.

Having all lower 13 as 50, can still lead a distribution with a median of 50 and a range of 50, whereby the answer will be E. Any thoughts?

No, Question says "set of 25 different integers"
_________________

Re: A set of 25 different integers has a median of 50 and a [#permalink]

Show Tags

15 Dec 2016, 16:14

Excellent Official Question. Here is what i did in this one=>

Median => 13th element = 50

Now range =50. Let a and b be the minimum value and maximum value in set.

As range is given=> To maximise b => We must maximise a. As the integers involved are different and all the element to the left of the median must be less than or equal to it=> 50 49 48 47 46 45 44 43 42 41 40 39 38

Hence the maximum value of a will be 38

Now b-38=50(Range) Hence b=50+38=88

Hence D

Note=>If the question didn't mention that all integers are different that a would have been 50 and b would have been 100 _________________