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D01-21

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New post 16 Sep 2014, 00:12
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A set of 11 different integers has a median of 25 and a range of 50. What is the greatest possible integer that could be in this set?

A. 65
B. 70
C. 75
D. 80
E. 85

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Re D01-21  [#permalink]

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New post 16 Sep 2014, 00:12
1
5
Official Solution:

A set of 11 different integers has a median of 25 and a range of 50. What is the greatest possible integer that could be in this set?

A. 65
B. 70
C. 75
D. 80
E. 85


Consider 11 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{11}\).

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 \(x_{6}=25\);

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 \(50=x_{11}-x_{1}\) \(\rightarrow\) \(x_{11}=50+x_{1}\);

We want to maximize \(x_{11}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-5=25-5=20\) and thus the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\).

The set could be {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 70}


Answer: B
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Re: D01-21  [#permalink]

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New post 18 Dec 2014, 14:23
median-5=25-5=20 - why did we subtract 5 from median?





Bunuel wrote:
Official Solution:

A set of 11 different integers has a median of 25 and a range of 50. What is the greatest possible integer that could be in this set?

A. 65
B. 70
C. 75
D. 80
E. 85


Consider 11 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{11}\).

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 \(x_{6}=25\);

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 \(50=x_{11}-x_{1}\) \(\rightarrow\) \(x_{11}=50+x_{1}\);

We want to maximize \(x_{11}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-5=25-5=20\) and thus the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\).

The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}


Answer: B

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Re: D01-21  [#permalink]

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New post 18 Dec 2014, 20:51
1
sunita123 wrote:
median-5=25-5=20 - why did we subtract 5 from median?





Bunuel wrote:
Official Solution:

A set of 11 different integers has a median of 25 and a range of 50. What is the greatest possible integer that could be in this set?

A. 65
B. 70
C. 75
D. 80
E. 85


Consider 11 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{11}\).

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 \(x_{6}=25\);

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 \(50=x_{11}-x_{1}\) \(\rightarrow\) \(x_{11}=50+x_{1}\);

We want to maximize \(x_{11}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-5=25-5=20\) and thus the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\).

The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}


Answer: B



2 things to be noted here

We need to find the largest number possible in the list and that number is x1

Option C,D and E can be ruled out because if median is 25 and range 50 that means number x1 to x5 will be 25 but we are told there are 11 different integers so this case is not possible.Likewise D and E not possible.

Now to maxmimize X11,you need to minimize other numbers..

Thus when you subtract from 5 median to get 20 as the smallest number then largest possible number will be Range (50)+20= 70..

Mind you even 65 can be an option but since we need to find largest integer possible 70 will be the answer
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Re: D01-21  [#permalink]

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New post 11 May 2015, 09:47
Bunuel wrote:
Official Solution:

A set of 11 different integers has a median of 25 and a range of 50. What is the greatest possible integer that could be in this set?

A. 65
B. 70
C. 75
D. 80
E. 85


Consider 11 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{11}\).

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 \(x_{6}=25\);

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 \(50=x_{11}-x_{1}\) \(\rightarrow\) \(x_{11}=50+x_{1}\);

We want to maximize \(x_{11}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-5=25-5=20\) and thus the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\).

The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}


Answer: B


Hi, how do you know the 11 numbers are consecutive numbers? What if X1 - X6 are all 25, which makes X11 75.
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Re: D01-21  [#permalink]

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New post 11 May 2015, 09:54
propcandy wrote:
Bunuel wrote:
Official Solution:

A set of 11 different integers has a median of 25 and a range of 50. What is the greatest possible integer that could be in this set?

A. 65
B. 70
C. 75
D. 80
E. 85


Consider 11 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{11}\).

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 \(x_{6}=25\);

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 \(50=x_{11}-x_{1}\) \(\rightarrow\) \(x_{11}=50+x_{1}\);

We want to maximize \(x_{11}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-5=25-5=20\) and thus the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\).

The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}


Answer: B


Hi, how do you know the 11 numbers are consecutive numbers? What if X1 - X6 are all 25, which makes X11 75.


We are given that the set consists of 11 different integers .
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Re: D01-21  [#permalink]

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New post 26 Nov 2015, 14:11
1
we can solve it by drawing some lines:

_ _ _ _ _ 25 _ _ _ _ _
the range is 50
thus the last one would be x
and the highest one would be y
for the range to be 50, x can be 20 and y 70.
note that if x is lower than 20, then y would be lower.
and x can't be higher than 20 since all the numbers should be different.
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Re: D01-21  [#permalink]

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New post 25 Jan 2016, 11:29
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If all numbers before 25 are also 25 :

25,25,25,25,25,25,X1,X2 Then also median will be 25 and Max value will be 25+50 = 75
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Re: D01-21  [#permalink]

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New post 02 Oct 2016, 15:01
I fell for the very very annoying GMAT Trap which turns this problem in to a 700+ level problem.

The questions mentions

"A set of 11 different integers"
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Re: D01-21  [#permalink]

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New post 13 Oct 2016, 22:59
Shouldn't the questions state that the numbers are placed in ascending order? Median is just the mid value, the values can be spread anyway. For example median of a set of 3 numbers {12,4,24} is 4. Why arrange it in ascending order unless explicitly specified?
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Re: D01-21  [#permalink]

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New post 14 Oct 2016, 01:49
satyakrishna wrote:
Shouldn't the questions state that the numbers are placed in ascending order? Median is just the mid value, the values can be spread anyway. For example median of a set of 3 numbers {12,4,24} is 4. Why arrange it in ascending order unless explicitly specified?


No, the median of {12,4,24} is 12, not 4.
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Re: D01-21  [#permalink]

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New post 06 Aug 2017, 05:55
n=11 (all diff + integers)
Med = 25
Range = 50

so n6 = 25
1st 6 numbers are an AP with d=1
so a + 5d = 25
a=25-5(1)
a=20
add the range .... so 20 + 50 = 70
ans.
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New post 07 Aug 2017, 19:38
I may be missing a basic point but why shouldn't we consider negative numbers as part of the set ?
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New post 08 Aug 2017, 01:08
arvind9733 wrote:
I may be missing a basic point but why shouldn't we consider negative numbers as part of the set ?


We want to find the greatest possible integer in the set. It turns out that the greatest possible integer is 70, and in this case the set is {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}, so no negative integers could be in the set if the greatest integer is 70, median is 25 and the range is 50.
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Re: D01-21  [#permalink]

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New post 08 Aug 2017, 07:18
Thank you. I made a subtraction mistake and got this wrong. Silly me !! I considered negative numbers to be part of the set and added two numbers, instead of subtracting, to calculate the range.

Good question !!
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New post 15 Aug 2017, 21:48
Why can the lowest number not be 1? The question does not say anything about different consecutive integers. Just that they are different.

So why can it not be 1,2,3,4,5,25,26,27,28,29,51. I'm sorry, am I missing something obvious here?
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New post 15 Aug 2017, 21:50
SKas07 wrote:
Why can the lowest number not be 1? The question does not say anything about different consecutive integers. Just that they are different.

So why can it not be 1,2,3,4,5,25,26,27,28,29,51. I'm sorry, am I missing something obvious here?


Nvm, I just realised that in order to maximize you would need to consider 20 and not 1. 70>51.

Thanks anyways,
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New post 15 Aug 2017, 21:55
SKas07 wrote:
Why can the lowest number not be 1? The question does not say anything about different consecutive integers. Just that they are different.

So why can it not be 1,2,3,4,5,25,26,27,28,29,51. I'm sorry, am I missing something obvious here?


We want to build a set which will have median of 25 and the range of 50 so that to maximise the greatest integer. We go that the the greatest integer could be 70 and in this case the smallest integer turns out to be 20.
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Re: D01-21  [#permalink]

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New post 11 Jan 2018, 05:59
1
Let the 11 integers be =>

x1
x2
x3
x4
x5
25
x7
x8
x9
x10
x1+50


Now to maximise last we must maximise the first and minimise others
NOTE -> Different integers.

Hence
x5=24
x4=23
x3=22
x2=21
x1=20
So last term = 20+50 = 70


Smash that B.

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