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M13-22

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M13-22  [#permalink]

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New post 15 Sep 2014, 23:49
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A
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C
D
E

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  35% (medium)

Question Stats:

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Re M13-22  [#permalink]

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New post 15 Sep 2014, 23:49
Official Solution:

If \(X\) is a positive integer, which of the following cannot be a median of set \(S = (X, 9, 2, 5)\)?

A. 3.0
B. 3.5
C. 4.0
D. 5.0
E. 6.5


\(X = 1\) or \(2\); the median is \(\frac{(2 + 5)}{2} = 3.5\)

\(X = 3\); the median is \(\frac{(3 + 5)}{2} = 4.0\)

\(X = 5\); the median is \(\frac{(5 + 5)}{2} = 5.0\)

\(X = 8\); the median is \(\frac{(5 + 8)}{2} = 6.5\)

\(X \gt 8\); the median is \(\frac{(5 + 9)}{2} = 7\)

The median cannot equal 3.0.


Answer: A
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Re: M13-22  [#permalink]

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New post 04 May 2018, 19:38
1
Another way to approach this by figuring out the range in which the median can lie i.e. in what conditions would you have the smallest and largest median.

[X 2 5 9 ] - when you know that X is the smallest - the median is 2+5=7/2=3.5
[2 5 9 X] - when you know that X is largest - the median will be 9+5=14/2= 7

Therefore, the median say y will be 3.5<y<7

Only A i.e. 3 is outside the range.
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Re: M13-22  [#permalink]

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New post 09 Jul 2018, 05:41
Hi Bunuel niks18 pikolo2510 Abhishek009 chetan2u VeritasPrepKarishma

I could not understand the logic of cherry picking numbers.
Why are 4,5, and 6 for e.g. not considered?

As per definition of median, it is the middle no in a set of numbers
arranged in an increasing / decreasing fashion.

Only restriction for X is \(X\geq{0}\)
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Re: M13-22  [#permalink]

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New post 09 Jul 2018, 05:54
adkikani wrote:
Hi Bunuel niks18 pikolo2510 Abhishek009 chetan2u VeritasPrepKarishma

I could not understand the logic of cherry picking numbers.
Why are 4,5, and 6 for e.g. not considered?

As per definition of median, it is the middle no in a set of numbers
arranged in an increasing / decreasing fashion.

Only restriction for X is \(X\geq{0}\)



Hi..

Since you are looking for the median that is the central value, the minimum value will be when x<2, 2 because that is the least number comedian will be(2+5)/2=3.5
So the min value is 3.5...3 therefore cannot be the answer..
A
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Re: M13-22  [#permalink]

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New post 09 Jul 2018, 06:47
adkikani wrote:
Hi Bunuel niks18 pikolo2510 Abhishek009 chetan2u VeritasPrepKarishma

I could not understand the logic of cherry picking numbers.
Why are 4,5, and 6 for e.g. not considered?

As per definition of median, it is the middle no in a set of numbers
arranged in an increasing / decreasing fashion.

Only restriction for X is \(X\geq{0}\)


As numbers keep increasing the median kept on increasing. There won't be a case when median decreases as you increase the value of X. Think aabout it . if X=3, then median=4. But when X=7, then median=6.

Now the lowest possible median is 2+5/2 = 3.5
All other possible values will be more than 3.5 as explained above. A quick glance at the options and we can see answer should be A

So there is no need to check for 4,5,6 or every possible value of X
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Re: M13-22  [#permalink]

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New post 09 Jul 2018, 16:17
Thanks pikolo2510 and chetan2u for your two cents.

Any reason for not selecting 1 in spite of it satisfying condition in question stem?
May I know why does X has to be either minimum or maximum value for the
elements in the set though I have no restriction in choosing X less than 2 ie 1
and greater than 9, ie 10.
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Re: M13-22 &nbs [#permalink] 09 Jul 2018, 16:17
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