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The median value of a set of 9 positive integers is 6. If

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The median value of a set of 9 positive integers is 6. If  [#permalink]

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New post 24 Feb 2005, 17:27
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The median value of a set of 9 positive integers is 6. If the only mode in the distrubution is 25 what is the least possible mean of the distribution.

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New post Updated on: 24 Feb 2005, 18:19
9

numbers shud be....1,2,3,4,6,7,8,25,25 = 81/9 = 9

Originally posted by banerjeea_98 on 24 Feb 2005, 17:40.
Last edited by banerjeea_98 on 24 Feb 2005, 18:19, edited 1 time in total.
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New post 24 Feb 2005, 18:19
MA wrote:
banerjeea_98 wrote:
9

numbers shud be....1,2,3,4,5,6,7,8,25,25 = 81/9 = 9


baner, 5 cannot be in the series.


sorry that was a typo....ans remains the same
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New post 25 Feb 2005, 03:31
Hmm... In the beginning I couldn't und the q at all but after seeing banerjee's post, I think he is right.
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New post 25 Feb 2005, 05:01
2,3,4,5,6,25,25,25,25 ;

Even this could be a plausible series. Mode can be repeated any number of times.

There is a formula which relates mean, median and mode -
3 * mean - 2 * mode = median

The answer is 17 ( minimum possible value of the mean)
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New post 25 Feb 2005, 06:21
HSWKM wrote:
2,3,4,5,6,25,25,25,25 ;

Even this could be a plausible series. Mode can be repeated any number of times.

There is a formula which relates mean, median and mode -
3 * mean - 2 * mode = median

The answer is 17 ( minimum possible value of the mean)


Stem asks for the least possible mean, therefore, we need to keep 25s to the minimum.
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New post 25 Feb 2005, 07:52
i get 9 as well

1+2+3+4+6+7+8+25+25=81
81/9=9
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New post 25 Feb 2005, 10:50
The series is 1,2,3,4,6,7,8,25,25
So the mean is 81/9=9
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New post 28 Feb 2005, 22:00
MA wrote:
banerjeea_98 wrote:
9

numbers shud be....1,2,3,4,5,6,7,8,25,25 = 81/9 = 9


baner, 5 cannot be in the series.


Please could you explain why 5 cannot be a part of the series?

Thanks
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New post 28 Feb 2005, 22:39
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Problem says mean should be least possible.So we should try to keep the numbers as small as possible.

6 is the median,middle number.So there are 4 numbers greater than 6 and 4 numbers less than 6.

Mode is 25 only.So one of the numbers is 25.But it is the only mode,so there should be one more 25 to make it "the only mode".

Still we donot know the 4 numbers less than 6 and the 2 numbers greater than 6.We need to keep the numbers as small as possible.So, the 4 numbers that are less than 6 can be 1,2,3,4 . cant go smaller than that.Similarly missing two numbers that are greater than 6 are 7,8 (smallest possible).

So numbers are 1,2,3,4,6,7,8,25,25.
phew....lengthy explanation.
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New post 01 Mar 2005, 02:32
Yep. Silly mistake that. Question is asking minimum value of mean. Answer is indeed 9. Thanks banerjee
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New post 01 Mar 2005, 03:23
one more for 9

first mode :
two 25

second the median : 6

we are left with : 6 postive integer to pick with the objective of making the mean the smallest

so we have to pick from both side of the median the smallest positive integers :

1, 2 , 3, 4
7,8, 25 , 25

mean 9
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New post 01 Mar 2005, 09:47
700Plus wrote:
Problem says mean should be least possible.So we should try to keep the numbers as small as possible.

6 is the median,middle number.So there are 4 numbers greater than 6 and 4 numbers less than 6.

Mode is 25 only.So one of the numbers is 25.But it is the only mode,so there should be one more 25 to make it "the only mode".

Still we donot know the 4 numbers less than 6 and the 2 numbers greater than 6.We need to keep the numbers as small as possible.So, the 4 numbers that are less than 6 can be 1,2,3,4 . cant go smaller than that.Similarly missing two numbers that are greater than 6 are 7,8 (smallest possible).

So numbers are 1,2,3,4,6,7,8,25,25.
phew....lengthy explanation.


lengthy but clear. Thanks for taking time to break it down because now i do understand it.
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New post 04 Mar 2005, 12:54
Yes that's a good and detailed explanation
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New post 12 Mar 2005, 07:58
banerjeea_98 wrote:
9

numbers shud be....1,2,3,4,6,7,8,25,25 = 81/9 = 9



why it can't be numbers shud be...0,.1,2,3,,6,7,8,25,25. isn't zero an integer?
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New post 12 Mar 2005, 17:04
mirhaque wrote:
banerjeea_98 wrote:
9

numbers shud be....1,2,3,4,6,7,8,25,25 = 81/9 = 9



why it can't be numbers shud be...0,.1,2,3,,6,7,8,25,25. isn't zero an integer?


0 is not +ve or -ve, ques says +ve integers only
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Re: Find the Mean  [#permalink]

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New post 28 Dec 2009, 17:40
where does it say you can't repeat?

couldn't it be

1,1,1,1,6,6,6,25,25

for 72/9 = 8

What is the source?
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New post 28 Dec 2009, 17:42
oh mode is 25 nvrmnd
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New post 29 Dec 2009, 23:50
700Plus wrote:
Problem says mean should be least possible.So we should try to keep the numbers as small as possible.

6 is the median,middle number.So there are 4 numbers greater than 6 and 4 numbers less than 6.

Mode is 25 only.So one of the numbers is 25.But it is the only mode,so there should be one more 25 to make it "the only mode".

Still we donot know the 4 numbers less than 6 and the 2 numbers greater than 6.We need to keep the numbers as small as possible.So, the 4 numbers that are less than 6 can be 1,2,3,4 . cant go smaller than that.Similarly missing two numbers that are greater than 6 are 7,8 (smallest possible).

So numbers are 1,2,3,4,6,7,8,25,25.
phew....lengthy explanation.



Thanks for clear explanation.
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New post 30 Dec 2009, 07:27
mirhaque wrote:
banerjeea_98 wrote:
9

numbers shud be....1,2,3,4,6,7,8,25,25 = 81/9 = 9



why it can't be numbers shud be...0,.1,2,3,,6,7,8,25,25. isn't zero an integer?


Sure it's an integer but not positive: "9 positive integers"

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Re: &nbs [#permalink] 30 Dec 2009, 07:27

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