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# A set of 15 different integers has a median of 25 and a

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CEO
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A set of 15 different integers has a median of 25 and a [#permalink]

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24 Mar 2008, 07:26
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A set of 15 different integers has a median of 25 and a range of 25. What is the largest possible integer that could be in the set?
32
37
40
43
50
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Manager
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24 Mar 2008, 07:53
bmwhype2 wrote:
A set of 15 different integers has a median of 25 and a range of 25. What is the largest possible integer that could be in the set?
32
37
40
43
50

median is the 8th term and it must be 25
so to maximize the largest integer, we have to maximize the smallest integer
we get the smallest term to be 18 (deduct one each from 8th term of 25 to find 1st term)
so max value = 18+range = 43
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24 Mar 2008, 08:19
it doesn't say these are consecutive integers, why would we increment each term by 1?
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24 Mar 2008, 08:33
jimmyjamesdonkey wrote:
it doesn't say these are consecutive integers, why would we increment each term by 1?

from median of 25 (8th), if we do not decrease by one each to find the previous term (i.e. 24 for 7th term), the the minimum will be less than 18.

If minimum is less than 18, maximum will be less than 43. However, question asks for maximum possible value.
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24 Mar 2008, 08:37
Is there a way to solve this via equation? I started working it but did not hit the result...I had:

x - y = 25 (range)

(x + y)/2 = 25 (median)

Can anyone tell me where my logic failed?
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24 Mar 2008, 08:39
jimmyjamesdonkey wrote:
Is there a way to solve this via equation? I started working it but did not hit the result...I had:

x - y = 25 (range)

(x + y)/2 = 25 (median)

Can anyone tell me where my logic failed?

median is not the average between max and min
median is the middle value. In this case, when there are 15 numbers and we put them in order from min to max, median is the middle term (i.e. 8th term). If there are 16 numbers, then median will be the average between term 8th and 9th.
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24 Mar 2008, 08:41
So is there anyway to solve this via equation?
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24 Mar 2008, 08:44
jimmyjamesdonkey wrote:
So is there anyway to solve this via equation?

I dont think so, or if there is, it would be so hard to create because this type of question is not "single solution" question. I mean the question asks for maximum possible value so we have to somewhat have idea in our mind of how to maximize the answer.
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24 Mar 2008, 08:47
I'm kinda of lost about how we maximize the top value, by max the bottom....can you provide more explaination for this?
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24 Mar 2008, 08:54
jimmyjamesdonkey wrote:
I'm kinda of lost about how we maximize the top value, by max the bottom....can you provide more explaination for this?

This is because the "range" is given and it is fixed by the question. No matter what the minimum value is, the maximum value will be min+25. This is why, in order to come up with maximum possible value of greatest integer, we have to try to find the maximum possible value of least integer.

to illustrate, see example

8th term is fixed at 25
if we reduce one for each previous term
7th = 24
6th = 23
5th = 22
4th = 21
3rd = 20
2nd = 19
1st = 18
greatest = 18+25 = 43

if we reduce two for each previous term
23 21 19 17 15 13 11
then greatest =11+25 = 36

since question asks for maximum possible value, 43 would be the answer
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24 Mar 2008, 08:59
i guess i'm confusing median and mean....i figured if 18 is the lowest, and median - 18 = 7, then median + 7 = highest #. But the median isn't equi-distance i suppose from the bottom and top #s?
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24 Mar 2008, 09:04
jimmyjamesdonkey wrote:
i guess i'm confusing median and mean....i figured if 18 is the lowest, and median - 18 = 7, then median + 7 = highest #. But the median isn't equi-distance i suppose from the bottom and top #s?

mean - we just add all and divide by number of elements
median - just rank from highest to lowest and pick the middle one

eg. 1 2 3 4 15
mean = sum/5 = 25/5 = 5
median = 3
CEO
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24 Mar 2008, 09:20
pretty straight forward.

OA is D
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26 Mar 2008, 01:09
Hey guys it has to be E in my opinion.

I believe if you have a set of 3 numbers ( 4,4,4) then mean,median,mode are all equal to 4.

So that said you can have 8 of 15 numbers all 25 which will maximize answer to 50.
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26 Mar 2008, 01:40
kyatin wrote:
Hey guys it has to be E in my opinion.

I believe if you have a set of 3 numbers ( 4,4,4) then mean,median,mode are all equal to 4.

So that said you can have 8 of 15 numbers all 25 which will maximize answer to 50.

question asks for 15 "different" integers
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26 Mar 2008, 10:30
Thanks 'prov'

I hope I read questions well in exam. missed on that 'different'
Re: GMATprep median   [#permalink] 26 Mar 2008, 10:30
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