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# The average (arithmetic mean) of the 5 positive integers

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The average (arithmetic mean) of the 5 positive integers [#permalink]  28 Dec 2010, 17:54
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Question Stats:

51% (03:15) correct 48% (01:43) wrong based on 2 sessions
The average (arithmetic mean) of the 5 positive integers k, m, r, s, and t is 16, and k < m < r < s < t. If t is 40, what is the greatest possible value of the median of the 5 integers?

A. 16
B. 18
C. 19
D. 20
E. 22

Please could someone explain why we cant take the smallest value as 0.
[Reveal] Spoiler: OA

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Re: The average (arithmetic mean) of the 5 positive integers k, [#permalink]  28 Dec 2010, 18:15
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ajit257 wrote:
The average (arithmetic mean) of the 5 positive integers k, m, r, s, and t is 16, and k < m < r < s < t. If t is 40, what is the greatest possible value of the median of the 5 integers?
A. 16
B. 18
C. 19
D. 20
E. 22

Please could someone explain why we cant take the smallest value as 0.

Note that k, m, r, s, and t are positive integers, thus neither of them can be zero.

Given: 0<k<m<r<s<t=40 and average=\frac{k+m+r+s+40}{5}=16. Question: median_{max}=? As median of 5 (odd) numbers is the middle number (when arranged in ascending or descending order) then the question basically asks to find the value of r_{max}.

average=\frac{k+m+r+s+40}{5}=16 --> k+m+r+s+40=16*5=80 --> k+m+r+s=40. Now, we want to maximize r:

General rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.

So to maximize r we should minimize k and m, as k and m must be distinct positive integers then the least values for them are 1 and 2 respectively --> 1+2+r+s=40 --> r+s=37 --> r_{max}=18 and s=19 (as r and s also must be distinct positive integers and r<s). So, r_{max}=18

Hope it's clear.
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Re: The average (arithmetic mean) of the 5 positive integers k, [#permalink]  28 Dec 2010, 21:19
Same approach as Bunuel's.

Bunnel - Do we have more similar questions like this one for practice?
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Re: The average (arithmetic mean) of the 5 positive integers k, [#permalink]  29 Dec 2010, 00:58
Got 19 as the answer because forgot to consider that the integers need to be distinct and are not equal per the stem.
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Re: The average (arithmetic mean) of the 5 positive integers k, [#permalink]  29 Dec 2010, 01:30
MasterGMAT12 wrote:
Same approach as Bunuel's.

Bunnel - Do we have more similar questions like this one for practice?

Min/max questions: search.php?search_id=tag&tag_id=63
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Statistics [#permalink]  12 Dec 2011, 07:59

The arithmetic mean of 5 positive integers a,b,c,d and e is 22 , and a<b<c<d<e.if e is 40,what is the greatest possible value of the median of the 5 integers?

Ans is : 33
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Kudos ???!@#$$!!! I just love them Last edited by Bunuel on 12 Jan 2012, 07:24, edited 1 time in total. NOTE: this question has the different average value from that of in the initial post, though the questions are basically the same. Thus merged. Senior Manager Joined: 30 Aug 2009 Posts: 296 Location: India Concentration: General Management Followers: 2 Kudos [?]: 65 [3] , given: 5 Re: Statistics [#permalink] 12 Dec 2011, 08:24 3 This post received KUDOS sairajesh063 wrote: Please help me out in this question. The arithmetic mean of 5 positive integers a,b,c,d and e is 22 , and a<b<c<d<e.if e is 40,what is the greatest possible value of the median of the 5 integers? Ans is : 33 Given , mean is 22 total sum = 5 * mean = 5* 22 = 110 e is 40 so a+b+c+d = 70 and a<b<c<d<e Median will be the middle term so value of "c" will be median let a = 1 and b = 2 as we need to find the maximum value of median c+d = 70 -(a+b) = 70 -3 = 67 so c < 33.5 and d >33.5 (on dividing 67 by 2) . As c and d are integers c = 33 and d =34 Hence median is 33 Manager Joined: 31 Oct 2011 Posts: 52 Location: India Followers: 0 Kudos [?]: 13 [0], given: 11 Re: Statistics [#permalink] 12 Dec 2011, 09:33 Thanks for the explanation kp _________________ Regards, Rajesh Helping hands are anytime better than praying hearts Kudos ???!@#$$ !!! I just love them

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Re: Statistics [#permalink]  12 Dec 2011, 23:06
Arithmeti mean is 22

so sum of numbers = 5* 22 = 110

e is 40 so a+b+c+d = 70 and a<b<c<d<e

Median will be the middle term so value of "c" will be median
since a+b+c+d = 70, sum is fix. to get C max a and b need to be min

let us chose a = 1 and b = 2 as we need to find the maximum value of median

c+d = 70 -(a+b) = 70 -3 = 67
As c and d are integers in increasing order so

c = 33 and d =34
Hence median is 33
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Re: Statistics [#permalink]  13 Dec 2011, 07:25
sandeeepsharma wrote:
Arithmeti mean is 22

so sum of numbers = 5* 22 = 110

e is 40 so a+b+c+d = 70 and a<b<c<d<e

Median will be the middle term so value of "c" will be median
since a+b+c+d = 70, sum is fix. to get C max a and b need to be min

let us chose a = 1 and b = 2 as we need to find the maximum value of median

c+d = 70 -(a+b) = 70 -3 = 67
As c and d are integers in increasing order so

c = 33 and d =34
Hence median is 33

Thank you Sandeep.

Cheers,
Rajesh
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Re: Statistics [#permalink]  13 Dec 2011, 20:49
Thanks rajesh it is my pleasure that i am able to help you
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Re: Statistics [#permalink]  22 Dec 2011, 01:15

thanks for posting these questions.
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Re: Statistics [#permalink]  03 Jan 2012, 00:51
Nice question. Thanks. 33 is the answer. Thanks for the explanations as well guys.
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Re: Statistics [#permalink]  09 Jan 2012, 15:09
To find the greatest median, you need have smallest values for the numbers below the median.

That is a=1, b=2. Now, subtract from the sum of numbers = (22 x 5) - (1 + 2 + 40) = 68.

So, we need to fit in c and d within a sum of 68 such that c is the largest it can be. We get c = 33 and d = 35.
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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]  28 Jun 2012, 11:01
Why cant we take the smallest number as zero
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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]  28 Jun 2012, 11:03
Eshaninan wrote:
Why cant we take the smallest number as zero

Because stem says that k, m, r, s, and t are positive integers and zero is not a positive number.
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Re: The average (arithmetic mean) of the 5 positive integers   [#permalink] 28 Jun 2012, 11:03
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