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If 25 different integers have a median of 50, then the 13th largest integer in the set is 50 - 12 less than 50, 12 greater than 50 or { 12 different integers less than 50, 50, 12 different integers greater than 50 }.

To get to "the greatest possible integer in the set" - you need to first find the highest possible set of integers in 12 integers less than 50. This is {38, 39, ..., 48, 49}.

So 38 is the low point for the range of 50, leaving you with a high point of 88. The set could be many variations of {38, 39, ... , 50, 11 integers greater than 50 and less than 88, 88}

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}

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.

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

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03 Apr 2012, 04:10

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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:

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.

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.

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

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03 Apr 2012, 06:08

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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}

Answer: 54.
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Last edited by Anonym on 04 Apr 2012, 02:32, edited 1 time in total.

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

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04 Apr 2012, 00:42

Anonym wrote:

smileforever41: In the style of the legend Bunuel:

Consider 25 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}

Answer: 54.

This explanation has been an exemplary help....thanks a ton

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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.

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

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17 Aug 2012, 04:44

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Hi Bunuel,

Please confirm;

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).

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.

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).

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 \(x_1+\{range\}=50+50=100\).

Re: A set of 15 different integers has a median of 25 [#permalink]

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09 Jan 2013, 18:03

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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

Re: A set of 15 different integers has a median of 25 [#permalink]

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09 Jan 2013, 18:23

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

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10 Jan 2013, 07:58

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

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

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26 Feb 2013, 08:36

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.

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

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18 Nov 2013, 16:20

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.

Why have we taken consecutive integers in this cases?
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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.

Why have we taken consecutive integers in this cases?

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

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