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Math Expert V
Joined: 02 Sep 2009
Posts: 61385
Re: A set of 25 different integers has a median of 50 and a  [#permalink]

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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

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

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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?
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Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3551
Re: A set of 25 different integers has a median of 50 and a  [#permalink]

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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"
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Joined: 12 Aug 2015
Posts: 2535
Schools: Boston U '20 (M)
GRE 1: Q169 V154 Re: A set of 25 different integers has a median of 50 and a  [#permalink]

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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

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Joined: 06 Apr 2017
Posts: 19
Location: India
GMAT 1: 640 Q47 V31
Re: A set of 25 different integers has a median of 50 and a  [#permalink]

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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?
Intern  B
Joined: 30 Nov 2017
Posts: 22
Re: A set of 25 different integers has a median of 50 and a  [#permalink]

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I didn't read the word "different" and assumed that the first 13 numbers were 50. Hence, applying a range of 50, I reached the highest number as 100.

100 was present as an answer choice and hence, I immediately chose it. Amazing how the question paper setters lay these traps.
Manager  B
Joined: 30 Jul 2014
Posts: 104
GPA: 3.72
Re: A set of 25 different integers has a median of 50 and a  [#permalink]

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I have done silly mistake while solving this question and marked option A... I have calculated the first element of the series right, but then messed up in adding the value 50 to the first number of the set...
X+12=50..
Number N should be X+50 not X+24

Regards
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Joined: 03 Jun 2019
Posts: 2013
Location: India
GMAT 1: 690 Q50 V34 WE: Engineering (Transportation)
Re: A set of 25 different integers has a median of 50 and a  [#permalink]

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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?

Given: A set of 25 different integers has a median of 50 and a range of 50.

Asked: What is the greatest possible integer that could be in this set?

To maximise greatest possible integer, all other numbers need to be minimised.

Let the number be {38,39,40,41,42,43,44,45,46,47,48,49,50,,,,,,,,,,,,,,,38+50}

Greatest possible integer = 38+50 = 88

IMO D
Senior Manager  G
Joined: 28 Jun 2019
Posts: 470
Re: A set of 25 different integers has a median of 50 and a  [#permalink]

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The phrase "different integers" changes the conditions. if it there was not we would say 100 is the answer but now option D is the answer.

Posted from my mobile device
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Intern  B
Joined: 07 Jul 2017
Posts: 17
GMAT 1: 660 Q49 V30
Re: A set of 25 different integers has a median of 50 and a  [#permalink]

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MY LOGIC

MEDIAN 50 RANGE 50 No of terms =25 MEDIAN >>13TH TERM

Z,X,C,V,B,B,,N,,F,D,S,A,(50),R,T,Y,U,I,O,P,Q,G, 12 TERMS ON EITHER SIDE SO

50-12=38 MUST BE THE FIRST TERM TO GET THE MAX..... SO RANGE=HT-LT>>>50=HT-38

*HT=88*

D

Posted from my mobile device Re: A set of 25 different integers has a median of 50 and a   [#permalink] 31 Dec 2019, 20:47

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