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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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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
