It is currently 22 Feb 2018, 06:50

TODAY:

MIT Sloan Releasing 1st Wave of Interview Invites - Join GMATClub CHAT for Live Updates

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

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

Track

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A set of 25 different integers has a median of 50 and a

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43864
Re: A set of 25 different integers has a median of 50 and a [#permalink]

Show Tags

17 Dec 2015, 08:45
Expert's post
1
This post was
BOOKMARKED
ArunpriyanJ 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}

Hope it's clear.

Thanks for the explanation Bunel...

Could you pls share some similar question for practice?

Thanks,
Arun

Similar topics:
three-straight-metal-rods-have-an-average-arithmetic-mean-148507.html
seven-pieces-of-rope-have-an-average-arithmetic-mean-lengt-144452.html
the-median-of-the-list-of-positive-integers-above-is-129639.html
in-a-certain-set-of-five-numbers-the-median-is-128514.html
given-distinct-positive-integers-1-11-3-x-2-and-9-whic-109801.html
set-s-contains-seven-distinct-integers-the-median-of-set-s-101331.html
three-boxes-have-an-average-weight-of-7kg-and-a-median-weigh-99642.html

Hope it helps.
_________________
Intern
Joined: 08 Jun 2016
Posts: 1
Re: A set of 25 different integers has a median of 50 and a [#permalink]

Show Tags

03 Aug 2016, 09:07
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}

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?
Board of Directors
Status: Aiming MBA
Joined: 18 Jul 2015
Posts: 3076
Location: India
Concentration: Healthcare, Technology
GPA: 3.65
WE: Information Technology (Health Care)
Re: A set of 25 different integers has a median of 50 and a [#permalink]

Show Tags

03 Aug 2016, 09:25
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}

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"
_________________

How I improved from V21 to V40! ?

How to use this forum in THE BEST way?

Retired Moderator
Joined: 12 Aug 2015
Posts: 2424
GRE 1: 323 Q169 V154
Re: A set of 25 different integers has a median of 50 and a [#permalink]

Show Tags

15 Dec 2016, 15:14
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

_________________

Getting into HOLLYWOOD with an MBA

Stone Cold's Mock Tests for GMAT-Quant(700+)

Intern
Joined: 06 Apr 2017
Posts: 8
Location: India
Re: A set of 25 different integers has a median of 50 and a [#permalink]

Show Tags

28 Sep 2017, 10:56
enigma123 wrote:
I think Bunuel , you meant "We want to maximize $$x_{25}$$, hence we need to minimize $$x_{1}$$. I still don't understand how did you get the below:

Since all integers must be distinct then the maximum value of $$x_{1}$$ will be $$median-12=50-12=38$$

How did we assue the least number to be 50-12? why it can't be 0 or 1 or 2 or 34 for that matter?
Re: A set of 25 different integers has a median of 50 and a   [#permalink] 28 Sep 2017, 10:56

Go to page   Previous    1   2   [ 25 posts ]

Display posts from previous: Sort by