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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]

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06 May 2008, 12:07
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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
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06 May 2008, 12:12
i get 19..

K+m..+t=16*5

k+m+r..=80-40
k+m+r+s=40..lets make K+m as small as possible..say 1 and 1..

R+s=38..now..inorder to max R, we set r=s=19+19=38...

19 it is..
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06 May 2008, 12:15
hmm let me try this

the sum of the 5 integers = 16*5 = 80

we know that t is 40 .. so k+m+r+s = 40

we are asked to maximize the median of the 5 numbers which means that we are really looking for 'r', in that case lets try minimizing k&m. Since they have to be integers let use k=1, m=2 .. that means that r+s = 37 ... since we want to maximize r ... and s>r ... the possible set of numbers could be 18,19.

Therefore r = 18.

OA?
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06 May 2008, 12:44

yes r=18..
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06 May 2008, 12:51
puma 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

We want to make k,m = 1,2. We can't go any small b/c they must be integers and must be positive.

now we want to me s=s+1 b/c that way s is as great as it can be.

so now its just (1+2+40+r+r+1)/5=16 --> 80. 2r+44=80 --> 2r=36

B.

I originally made the same mistake and forgot that the integers cannot be the same.
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07 May 2008, 06:40
B for me too

t is 40 so the rest of 4 integers need to sum up to 40

thus D,E are gone now out A,B,C

if it is C then 40-19=21 will be the rest of the 3 integers sum and this is not possible

since r>s and needs to be atleast 20 that will force K to be 0 and negates the question

thus C is out

B perfectly fits in 40-18=22 even if and the other 3 # could be 1,2,19
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07 May 2008, 07:15
let r = 19 (plug it in ) => s = 20 (allw for max) => min man could be (40+20+19+2+1)/5 => 82/5 which is > 16.
so rule out c,d,e
now, consider r=18 => s => 19 => min mean (40+19+18+2+1)=> 80/5 => 16 !
=> ans is b.
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07 May 2008, 08:26
nice one. Agree with B
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08 May 2008, 12:17
puma 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

Good question to make you think.... well, at least me...

Anyway, you have five values ___, ___, ___, ___, 40

Since the mean of the 5 numbers is 16, 16 x 5 = 80. We know the last value is 40, so the average of the remaining 4 numbers to find is 40 as well.

Let's assume the following:

1, 2, ___, ___, 40

so now we have 43 total, out of a possible 80. 80 - 43 = 37. 37 / 2 = 18.5, or since we need two increasing numbers, 18 and 19. Filling in the remaining numbers in our patterns yields:

1, 2, 18, 19, 40. Thus the median number must be 18.

B.
Re: arithmetic mean   [#permalink] 08 May 2008, 12:17
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