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Mean and Median

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Mean and Median  [#permalink] New post 09 Aug 2008, 20:53
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Mean and Median

--------------------------------------------------------------------------------

x, y, x + y, x – 4y, xy, 2y

For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be

(A) (x+y)/2
(B) y+3
(C) y
(D) 3y/2
(E) x/3+y
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Re: PS: Mean and Median [#permalink] New post 09 Aug 2008, 21:00
judokan wrote:
Mean and Median

--------------------------------------------------------------------------------

x, y, x + y, x – 4y, xy, 2y

For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be

(A) (x+y)/2
(B) y+3
(C) y
(D) 3y/2
(E) x/3+y


B

solve the average equation to get to x = 6
substitute x into each number and re-arrange from small to big, the 2 middle terms are y and 6+y

the average = (y + (6 + y))/2 = (2y + 6)/2 = y + 3, which is the median
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Re: PS: Mean and Median [#permalink] New post 10 Aug 2008, 06:16
judokan wrote:
Mean and Median

--------------------------------------------------------------------------------

x, y, x + y, x – 4y, xy, 2y

For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be

(A) (x+y)/2
(B) y+3
(C) y
(D) 3y/2
(E) x/3+y




here we know mean=y+3 calculating and solving ,
x=6 =>
x=6,y>6
hence arrange the give terms in the increasing order

x-4y,x,y,x+y,2y,xy => median = x+2y /2 = 3+y

hence IMO B
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Its Now Or Never

Re: PS: Mean and Median   [#permalink] 10 Aug 2008, 06:16
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