Last visit was: 27 May 2024, 00:28 It is currently 27 May 2024, 00:28
Toolkit
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Senior Manager
Joined: 25 Jun 2011
Status:Finally Done. Admitted in Kellogg for 2015 intake
Posts: 395
Own Kudos [?]: 16802 [173]
Given Kudos: 217
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE:Information Technology (Consulting)
Math Expert
Joined: 02 Sep 2009
Posts: 93470
Own Kudos [?]: 627010 [70]
Given Kudos: 81958
Intern
Joined: 03 Feb 2012
Posts: 44
Own Kudos [?]: 56 [9]
Given Kudos: 12
Location: United States (WI)
Concentration: Other
Schools: University of Wisconsin (Madison) - Class of 2014
GMAT 1: 680 Q46 V38
GMAT 2: 760 Q48 V46
GPA: 3.66
WE:Marketing (Manufacturing)
General Discussion
Senior Manager
Joined: 25 Jun 2011
Status:Finally Done. Admitted in Kellogg for 2015 intake
Posts: 395
Own Kudos [?]: 16802 [1]
Given Kudos: 217
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE:Information Technology (Consulting)
Re: A set of 25 different integers has a median of 50 and a [#permalink]
1
Kudos
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 $$median-12=50-12=38$$
Math Expert
Joined: 02 Sep 2009
Posts: 93470
Own Kudos [?]: 627010 [9]
Given Kudos: 81958
Re: A set of 25 different integers has a median of 50 and a [#permalink]
4
Kudos
4
Bookmarks
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 $$median-12=50-12=38$$

Since $$x_{25}=50+x_{1}$$ then in order to maximize $$x_{25}$$ we need to maximize $$x_{1}$$, so everything is correct there.

As for the next part: $$median=x_{13}=50$$ and since all terms must be distinct integers then the maximum value of $$x_{1}$$ is 50-12=38: $$x_{1}=38$$, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, $$median=x_{13}=50$$. As you can see $$x_{1}$$ cannot possibly be more than 38.

Hope it's clear.
Intern
Joined: 13 Sep 2011
Posts: 3
Own Kudos [?]: 3 [3]
Given Kudos: 11
Re: A set of 25 different integers has a median of 50 and a [#permalink]
1
Kudos
2
Bookmarks
I am not a quant genius, but a basic learner. However, I found an answer to this question after passing through different hurdles virtually....I worked on the following question to get the answer for this one, because both are similar:

A set of 13 different integers has a median of 30 and a range of 30. What is the greatest possible integer that could be in this set?

A)36

B)43

C)54

D)57

E)60

I considered the below set of values to get the maximum value of the number in a set:

30, 30, 30, 30, 30, 30, 30, 35, 40, 45, 50, 55, 60.

Hence, median = range = 30
Therefore, the maximum value is 60.

I believe, we can apply the same thought process in the actual question and answer it. I am open to feedback. Please do let me know if I have gone wrong anywhere.
Intern
Joined: 19 Aug 2011
Posts: 15
Own Kudos [?]: 53 [0]
Given Kudos: 2
Concentration: Finance, Entrepreneurship
Re: A set of 25 different integers has a median of 50 and a [#permalink]
smileforever41 wrote:
I considered the below set of values to get the maximum value of the number in a set:

30, 30, 30, 30, 30, 30, 30, 35, 40, 45, 50, 55, 60.

Hence, median = range = 30
Therefore, the maximum value is 60.

I believe, we can apply the same thought process in the actual question and answer it. I am open to feedback. Please do let me know if I have gone wrong anywhere.

You missed the word 13 different integers
Intern
Joined: 28 Feb 2012
Posts: 29
Own Kudos [?]: 801 [4]
Given Kudos: 20
Re: A set of 25 different integers has a median of 50 and a [#permalink]
3
Kudos
1
Bookmarks
smileforever41:
In the style of the legend Bunuel:

Consider 13 numbers in ascending order to be $$x_1, x_2, x_3, ..., x_{13}$$.

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 $$x_{7}=30$$;

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 $$30=x_{13}-x_{1}$$ --> $$x_{13}=30+x_{1}$$;

We want to maximize $$x_{13}$$, hence we need to maximize $$x_{1}$$ (remember the restriction of range). Since all integers must be distinct then the maximum value of $$x_{1}$$ will be $$median-6=30-6=24$$ and thus the maximum value of $$x_{13}$$ is $$x_{13}=24+30=54$$.

The set could be {24,25,26,27,28,29,30,38,40,45,47,49,54}

Originally posted by Anonym on 03 Apr 2012, 06:08.
Last edited by Anonym on 04 Apr 2012, 02:32, edited 1 time in total.
Intern
Joined: 18 Jan 2012
Posts: 21
Own Kudos [?]: 747 [0]
Given Kudos: 27
Location: United States
Schools: IIM A '15 (A)
Re: A set of 25 different integers has a median of 50 and a [#permalink]

Hi..I believe that everyone has inadvertently overlooked the fact that the definition an integer includes negative numbers as well.

All The following sets easily satisfy the conditions stated in the problem.. (Median = 50 , Range is 50).

-50,-51,-52............50......100
-150,-51,-52............50.....200
-250,-51,-52............50.....200
Math Expert
Joined: 02 Sep 2009
Posts: 93470
Own Kudos [?]: 627010 [1]
Given Kudos: 81958
Re: A set of 25 different integers has a median of 50 and a [#permalink]
1
Kudos
hafizkarim wrote:

Hi..I believe that everyone has inadvertently overlooked the fact that the definition an integer includes negative numbers as well.

All The following sets easily satisfy the conditions stated in the problem.. (Median = 50 , Range is 50).

-50,-51,-52............50......100
-150,-51,-52............50.....200
-250,-51,-52............50.....200

Not so.

First of all set -50,-51,-52............50......100 is not ordered. In ascending order it would be {-52, -51, ..., 50, ..., 100} (assume there are 25 integer in the set). Now, the range of this set is {range}={largest}-{smallest}=100-(-52)=152, not 50.

Hope it's clear.
Intern
Joined: 07 Sep 2011
Posts: 40
Own Kudos [?]: 52 [1]
Given Kudos: 3
Location: United States
GMAT 1: 640 Q39 V38
WE:General Management (Real Estate)
Re: A set of 25 different integers has a median of 50 and a [#permalink]
1
Bookmarks
Hi Bunuel,

Had the question not mentioned "Distinct Integers", we would have taken all first 13 values as 50 and maximum would have been 100". Is my thinking correct?

Bunuel wrote:
hafizkarim wrote:

Hi..I believe that everyone has inadvertently overlooked the fact that the definition an integer includes negative numbers as well.

All The following sets easily satisfy the conditions stated in the problem.. (Median = 50 , Range is 50).

-50,-51,-52............50......100
-150,-51,-52............50.....200
-250,-51,-52............50.....200

Not so.

First of all set -50,-51,-52............50......100 is not ordered. In ascending order it would be {-52, -51, ..., 50, ..., 100} (assume there are 25 integer in the set). Now, the range of this set is {range}={largest}-{smallest}=100-(-52)=152, not 50.

Hope it's clear.
Math Expert
Joined: 02 Sep 2009
Posts: 93470
Own Kudos [?]: 627010 [2]
Given Kudos: 81958
Re: A set of 25 different integers has a median of 50 and a [#permalink]
2
Bookmarks
manjeet1972 wrote:
Hi Bunuel,

Had the question not mentioned "Distinct Integers", we would have taken all first 13 values as 50 and maximum would have been 100". Is my thinking correct?

Bunuel wrote:
hafizkarim wrote:

Hi..I believe that everyone has inadvertently overlooked the fact that the definition an integer includes negative numbers as well.

All The following sets easily satisfy the conditions stated in the problem.. (Median = 50 , Range is 50).

-50,-51,-52............50......100
-150,-51,-52............50.....200
-250,-51,-52............50.....200

Not so.

First of all set -50,-51,-52............50......100 is not ordered. In ascending order it would be {-52, -51, ..., 50, ..., 100} (assume there are 25 integer in the set). Now, the range of this set is {range}={largest}-{smallest}=100-(-52)=152, not 50.

Hope it's clear.

Absolutely. If we were not told that the integers in the set must be distinct, then all terms from $$x_1$$ to $$x_{13}$$, could be 50 and in this case the greatest integer, which could be in the set, would be $$x_1+\{range\}=50+50=100$$.

Hope it's clear.
Intern
Joined: 08 Feb 2011
Posts: 11
Own Kudos [?]: 17 [1]
Given Kudos: 9
Re: A set of 15 different integers has a median of 25 [#permalink]
1
Kudos
The median (middle number in the set) is 25. It is stated that there are 15 different integers in the set, so there will be 7 different integers smaller than 25 and 7 different integers larger than 25 in the set.

To maximize the largest value in this set, we want to maximize the smallest value in the set.
Therefore, the 7 different integers in the set smaller than 25 will be 24, 23, 22, 21, 20, 19, 18.

18 (maximized smallest integer in set) + 25 (range of set) = largest possible integer in the set

Senior Manager
Joined: 27 Jun 2012
Posts: 324
Own Kudos [?]: 2478 [2]
Given Kudos: 185
Concentration: Strategy, Finance
Re: A set of 15 different integers has a median of 25 [#permalink]
1
Kudos
1
Bookmarks
Consider first number = x
Last number = x + 25
And we have median (8th number)=25
Series: x,...6 numbers..., 25 , ...6 numbers..., (x+25)

In order to have the "last number" (x+25) as the greatest possible, we have to maximize the "first number" (x) (under 25).

To maximize x, identify Integers under 25 such that they are consecutive in descending order.
i.e. 18, 19, 220, 21, 22, 23, 24, 25

Thus first number $$x=18$$ and largest number$$x + 25 = 18 +25 =43.$$

Hence choice(D) is correct.

The series would look like this: 18, 19, 220, 21, 22, 23, 24, 25, ... 6 numbers ..., 43
Note that choice (E) with 50 is a TRAP answer for someone who didn't notice the information "different" numbers.
Manager
Joined: 18 Oct 2011
Posts: 58
Own Kudos [?]: 355 [0]
Given Kudos: 0
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Re: A set of 25 different integers has a median of 50 and a [#permalink]
For these type of questions you know that since we have an odd number of terms in the set - The median is the 13th number. Since we are looking at the greatest possible number that could have a range of 50, we would want to maximize the 12 numbers that precede the median (50). Since all the numbers in the set are different the first term would have to be 38.

For a range of 50, only 88 would satisfy this condition. Answer: D
Director
Joined: 29 Nov 2012
Posts: 579
Own Kudos [?]: 6091 [0]
Given Kudos: 543
Re: A set of 25 different integers has a median of 50 and a [#permalink]
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.

Since the question mentions different we have to pick numbers as close to the median? What would happen if they didn't mention different numbers? so then we assign the value as median?
Manager
Joined: 14 Jan 2013
Posts: 114
Own Kudos [?]: 1538 [0]
Given Kudos: 30
Concentration: Strategy, Technology
GMAT Date: 08-01-2013
GPA: 3.7
WE:Consulting (Consulting)
Re: A set of 25 different integers has a median of 50 and a [#permalink]
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.

Why have we taken consecutive integers in this cases?
Math Expert
Joined: 02 Sep 2009
Posts: 93470
Own Kudos [?]: 627010 [0]
Given Kudos: 81958
Re: A set of 25 different integers has a median of 50 and a [#permalink]
Mountain14 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.

Why have we taken consecutive integers in this cases?

The set could be...
Director
Joined: 09 Jun 2010
Posts: 529
Own Kudos [?]: 525 [1]
Given Kudos: 916
Re: A set of 25 different integers has a median of 50 and a [#permalink]
1
Kudos
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.

I can not say a word for this wonderful explanation
Intern
Joined: 03 Aug 2015
Posts: 35
Own Kudos [?]: 17 [0]
Given Kudos: 219
Concentration: Strategy, Technology
GMAT 1: 680 Q48 V35
Re: A set of 25 different integers has a median of 50 and a [#permalink]
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
Re: A set of 25 different integers has a median of 50 and a [#permalink]
1   2
Moderator:
Math Expert
93470 posts