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Manager
Joined: 21 May 2008
Posts: 85
Followers: 1
Kudos [?]:
0
[0], given: 4
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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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Manager
Joined: 27 Jun 2008
Posts: 163
Followers: 1
Kudos [?]:
14
[0], given: 11
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What is the official solution?
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Manager
Joined: 27 Jun 2008
Posts: 163
Followers: 1
Kudos [?]:
14
[0], given: 11
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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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Manager
Joined: 17 Apr 2009
Posts: 161
Followers: 0
Kudos [?]:
6
[0], given: 0
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I also got C as the ans. What is the OA?
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Manager
Joined: 17 Apr 2009
Posts: 161
Followers: 0
Kudos [?]:
6
[0], given: 0
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klb15 plz post the OA.
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