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Set X has 5 numbers and its average is greater than its

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Manager
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Set X has 5 numbers and its average is greater than its [#permalink] New post 18 Nov 2006, 17:36
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A
B
C
D
E

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I'm looking for the best approach to solve this problem...I don't need the answer, I already have it.

Thanks...

Set X has 5 numbers and its average is greater than its median. Set Y has 7 numbers and its average is greater than its median also. If the 2 sets have no common number and are combined to a new set, is the average of the new set greater than its median?

(1) The average of Y is greater than the average of X
(2) The median of Y is greater than the median of X
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 [#permalink] New post 18 Nov 2006, 23:23
I can guess the answer to be B.

I roughly guess it using a skewed bell curve.
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 [#permalink] New post 19 Nov 2006, 05:43
Tennis,

Thanks for the input, but it looks like you paid no attention to my post. I have the answer (not B by the way), and am much more interested in an approach to solve this problem than the answer itself.
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 [#permalink] New post 19 Nov 2006, 09:00
Set X has 5 numbers and its average is greater than its median. Set Y has 7 numbers and its average is greater than its median also. If the 2 sets have no common number and are combined to a new set, is the average of the new set greater than its median?

(1) The average of Y is greater than the average of X
(2) The median of Y is greater than the median of X
=======================================
I tried solving the following way, but did not really reach at any answer:

Set X = {X1, X2, X3, X4, X5} ; X3 is median
Set Y = {Y1, Y2, Y3, Y4, Y5, Y6, Y7}; Y4 is median

Given: Sum X > 5X3
Sum Y > 7Y4
If the 2 sets are combined:
Set XY {X1, X2, X3, X4, X5, Y1, Y2, Y3, Y4, Y5, Y6, Y7}

Trying to arrange the set in increasing order:
We know that X1, X2 and X3 will be less than Y4 based on 2

(i) If X3 > Y3, but less than Y4, and X4 is greater than Y4, the set will be something like:
XY = {X1, Y1, X2, Y2, Y3, X3, Y4, X4, X5, Y5, Y6, Y7}
Here we do not care about the respective positions of X1, X2, X4 and X5.
The median will be X3+Y4/2
Average = SumX+SumY /12
Is X3+Y4/2 < Sum X+SumY/12
i.e 6X3+6Y4 < Sum X + Sum Y
i.e 5X3+X3+6Y4 < Sum X + Sum Y
This is true because (5X3 < Sum X. 7Y4 < Sum Y so based on (1), X3+6Y4 < Sum Y)

After this I am stuck. I am unable to prove for other combinations, eg if X3 is less than Y3, then what about median Y3+Y4/2. or if X5 is also less than Y4 etc

I would say the answer is either C or E, but since I can't prove it for the other combinations, I would just select E
  [#permalink] 19 Nov 2006, 09:00
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