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Set X has 5 numbers and its average is greater than its
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18 Nov 2006, 18:36
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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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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Re: Set X has 5 numbers and its average is greater than its
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19 Nov 2006, 06:43
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.
Re: Set X has 5 numbers and its average is greater than its
[#permalink]
19 Nov 2006, 10:00
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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
gmatclubot
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