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In the list above, if n is an integer between 1 and 10, inclusive, the

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In the list above, if n is an integer between 1 and 10, inclusive, the  [#permalink]

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New post 16 Sep 2018, 22:28
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A
B
C
D
E

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91% (01:27) correct 9% (01:13) wrong based on 64 sessions

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Director
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Re: In the list above, if n is an integer between 1 and 10, inclusive, the  [#permalink]

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New post 16 Sep 2018, 22:36
2+4+6+8+3+5+7+9+n = 44+n

If n is 1 then median is 45/9 = 5

If n is 10 then median is 54/9 = 6

B is the answer.
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In the list above, if n is an integer between 1 and 10, inclusive, the  [#permalink]

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New post 16 Sep 2018, 23:02
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Afc0892 wrote:
2+4+6+8+3+5+7+9+n = 44+n

If n is 1 then median is 45/9 = 5

If n is 10 then median is 54/9 = 6

B is the answer.

Afc0892
Your method is perfect if n must equal 1 OR 10, because then the integers are consecutive and median = mean.

But median \(\neq\) mean if n = 2,3,4,5,6,7,8, or 9
n is an integer between 1 and 10, inclusive.

If n = 2, mean = \(\frac{46}{9}=5.1111\)
If n = 7, mean = \(\frac{51}{9}= 5.67\)

Just an issue with the method (absent a mention of extremes and why they work) :)
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Re: In the list above, if n is an integer between 1 and 10, inclusive, the  [#permalink]

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New post 16 Sep 2018, 23:05
generis wrote:
Afc0892 wrote:
2+4+6+8+3+5+7+9+n = 44+n

If n is 1 then median is 45/9 = 5

If n is 10 then median is 54/9 = 6

B is the answer.

Afc0892
Your method is perfect if n must equal 1 OR 10, because then the integers are consecutive and median = mean.

But median \(\neq\) mean if n = 2,3,4,5,6,7,8, or 9
n is an integer between 1 and 10, inclusive.

If n = 2, mean = \(\frac{46}{9}=5.1111\)
If n = 7, mean = \(\frac{51}{9}= 5.67\)

Just an issue with the method (absent a mention of extremes and why they work) :)


Got it. Thanks :-)
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Re: In the list above, if n is an integer between 1 and 10, inclusive, the   [#permalink] 16 Sep 2018, 23:05
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