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# My appologies if this has been asked before But can someone

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My appologies if this has been asked before But can someone [#permalink]  25 Sep 2006, 16:08
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My appologies if this has been asked before
But can someone pls explain this,pls c the attachment for details

thanks
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DS-AvgQ.doc [68.5 KiB]

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Joined: 25 May 2006
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Yes/No problem.

Given: the sum of S and T are =

St1: average of S is less than average of T. This means that S has more numbers than list T.
Example: if the sum is 10, then:
S â€“-> 10/2=5 (S has two numbers)
T â€“-> 10/1=10 (T has one number)
Hence, YES to the question.

St2: Median of S > Median of T: INSUFF: S can have the same number of integers than T or more or less integers than T.
Hence, YES/NO to the question.

A it is.
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