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M15 Q 27 - DS

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Manager
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M15 Q 27 - DS [#permalink]

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New post 19 Jul 2009, 23:07
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Set S is composed of consecutive multiples of 3. Set T is composed of consecutive multiples of 6. If each set contains more than one element, is the median of Set S larger than the median of Set T ?

1. The least element in either set is 6.
2. Set T contains twice as many elements as Set S.

I dont quite agree with the solution. Anyone, please enlighten, thanks.

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Re: M15 Q 27 - DS [#permalink]

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New post 20 Jul 2009, 00:56
What is the official solution?

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New post 20 Jul 2009, 01:03
Is it C?
since
a) Since the numbers are consecutive multiple, we need to know with which number the series begins with - satisfied by a
b) How many numbers are present in the set, T contains twice of S - answered by B,
i.e.
T {6,12,18,24,30,36} S{3,6,9}
therefore Tmedian > Smedian

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Re: M15 Q 27 - DS [#permalink]

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New post 20 Jul 2009, 02:53
I also got C as the ans.
What is the OA?

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Re: M15 Q 27 - DS [#permalink]

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New post 22 Jul 2009, 03:34
klb15 plz post the OA.

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Re: M15 Q 27 - DS   [#permalink] 22 Jul 2009, 03:34
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