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# 3 k 2 8 3 m If the arithmetic mean of the list of numbers

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CEO
Joined: 21 Jan 2007
Posts: 2672
Location: New York City
3 k 2 8 3 m If the arithmetic mean of the list of numbers  [#permalink]

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27 Jan 2008, 04:23
1
1
1
3 k 2 8 3 m

If the arithmetic mean of the list of numbers above is 4, and k and m are integers, what is the median of the list?

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Director
Joined: 12 Jul 2007
Posts: 849
Re: OG Quant #161 - Median  [#permalink]

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27 Jan 2008, 07:43
5
3
{2, 3, 3, 8, m, k} has an average of 4.

2+3+3+8= 16

$formdata=\frac{16+k+m}{6}=4$

Meaning that k+m = 8.

Here's our list without K and M. {2, 3, 3, 8}

K and M could be 4+4: {2, 3, 3, 4, 4, 8} median = 3.5 impossible due to restraints in the question stem
K and M could be 5+3: {2, 3, 3, 3, 5, 8} median = 3
K and M could be 6+2: {2, 2, 3, 3, 6, 8} median = 3
K and M could be 7+1: {1, 2, 3, 3, 7, 8} median = 3

Since K cannot be the same as M the median will be 3 for any integers you put in there to add up to 8.

You left out the line about k not = m, that makes all the difference here. Especially since both 3 and 3.5 are listed as answers.

SVP
Joined: 29 Aug 2007
Posts: 2410
Re: OG Quant #161 - Median  [#permalink]

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27 Jan 2008, 07:57
bmwhype2 wrote:
3 k 2 8 3 m

If the arithmetic mean of the list of numbers above is 4, and k and m are integers, what is the median of the list?

sum = 3 + k + 2 + 8 + 3 + m
24 = 16 + k + m
so, K + m = 8

if k and m are different integers, the median is 3. otherwise, it is 3.5.

is it DS or PS question?

eschn3am wrote:
Since K cannot be the same as M

why? since k and m are different symbols/letters?
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Director
Joined: 12 Jul 2007
Posts: 849
Re: OG Quant #161 - Median  [#permalink]

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27 Jan 2008, 08:17
Quote:
why? since k and m are different symbols/letters?

I looked it up in the OG Quant book and it says in the question stem that they cannot be the same integer.
CEO
Joined: 21 Jan 2007
Posts: 2672
Location: New York City
Re: OG Quant #161 - Median  [#permalink]

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27 Jan 2008, 13:31
eschn3am wrote:
{2, 3, 3, 8, m, k} has an average of 4.

2+3+3+8= 16

$formdata=\frac{16+k+m}{6}=4$

Meaning that k+m = 8.

Here's our list without K and M. {2, 3, 3, 8}

K and M could be 4+4: {2, 3, 3, 4, 4, 8} median = 3.5 impossible due to restraints in the question stem
K and M could be 5+3: {2, 3, 3, 3, 5, 8} median = 3
K and M could be 6+2: {2, 2, 3, 3, 6, 8} median = 3
K and M could be 7+1: {1, 2, 3, 3, 7, 8} median = 3

Since K cannot be the same as M the median will be 3 for any integers you put in there to add up to 8.

You left out the line about k not = m, that makes all the difference here. Especially since both 3 and 3.5 are listed as answers.

thanks. couldnt quite think at 6 in the morning.
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Re: 3 k 2 8 3 m If the arithmetic mean of the list of numbers  [#permalink]

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20 Oct 2017, 10:43
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Re: 3 k 2 8 3 m If the arithmetic mean of the list of numbers &nbs [#permalink] 20 Oct 2017, 10:43
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