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# Mean and Median

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Senior Manager
Joined: 19 Mar 2008
Posts: 351

Kudos [?]: 68 [0], given: 0

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09 Aug 2008, 20:53
00:00

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Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 3 sessions

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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

Kudos [?]: 68 [0], given: 0

Director
Joined: 23 Sep 2007
Posts: 782

Kudos [?]: 236 [0], given: 0

Re: PS: Mean and Median [#permalink]

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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

Kudos [?]: 236 [0], given: 0

VP
Joined: 17 Jun 2008
Posts: 1374

Kudos [?]: 406 [0], given: 0

Re: PS: Mean and Median [#permalink]

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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
_________________

cheers
Its Now Or Never

Kudos [?]: 406 [0], given: 0

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