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For the set of terms shown above, if y > 6 and the mean of

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For the set of terms shown above, if y > 6 and the mean of [#permalink] New post 02 Nov 2009, 03:08
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B
C
D
E

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Question Stats:

63% (03:06) correct 37% (02:36) wrong based on 84 sessions
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
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Feb 2012, 15:59, edited 1 time in total.
Edited the question and added the OA
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Re: Sequence [#permalink] New post 02 Nov 2009, 10:22
adarsh12345 wrote:
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

(x + y) 2

y + 3

y

3y/2

x /3 + y


adding up all the elements of set we get
(3x + xy)/6 = y+3 => solving this we get x=6
we have y>6 so we get
x=6,y>6,x+y >12, x-4y < -18,xy>36 and 2y >12
arranging these we get x-4y, x, y, x+y, 2y, xy

median = (y + x+y) /2 = (2y+6)/2 = y+3
will go with option 2
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Re: Sequence [#permalink] New post 02 Nov 2009, 11:22
adarsh12345 wrote:
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

(x + y) 2

y + 3

y

3y/2

x /3 + y


I get:

(x + y + x + y + x - 4y + xy + 2y)/6 = y + 3

Combine like terms and you get:

3x + xy = 6y + 18
x(3+y)=6(y+3)
3+y cancel
x = 6

Substitute 6 for x in the above equation
Also, since y>6 pick any number for y and solve. Let's say I pick a 7

the following six numbers are:
6, 7, 13, -22, 42, 14

median = (7+13)/2 = 10

my answer is B = y + 3
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Re: Sequence [#permalink] New post 11 Feb 2012, 15:33
Find the average and set it equal to the y+3 then find out that x=6
then you find that y and x+y are the middle 2 numbers and sub x for 6 then you find the median number is y+3
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Re: Sequence [#permalink] New post 11 Feb 2012, 15:57
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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

First of all: the median of a set with even # of terms is the average of two middle terms (when ordered in ascending/descending order).

Next, given that the mean of the set equals y+3: x+y+(x+y)+(x-4y)+xy+2y=6*(y+3) --> 3x+xy=6*(y+3) --> x(3+y)=6(y+3), notice that generally you cannot reduce by 3+y here as suggested above --> (3+y)(x-6)=0 --> x=6 or y=-3 (not a valid solution as y>6), so x=6;

Arrange the set in ascending order: {x-4y, x, y, x+y, 2y, xy} --> median=(y+x+y)/2=(2y+6)/2=y+3.

Answer: B.

Or: pick some number greater than 6 for y, let's say y=7. Then our set will be: {6, 7, 13, -22, 42, 14} --> arrange in ascending order: {-22, 6, 7, 13, 14, 42}.

Thus the median of the set is (7+13)/2=10 --> 10=7+3=y+3.

Answer: B.

Hope it helps.
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For the set of terms shown above, if y > 6 and the mean of the s [#permalink] New post 12 Dec 2012, 23:05
x y, x + y, x – 4y, xy, 2y

\frac{{x + y + (x+y) + (x-4y) + xy + 2y}}{{6}} = y+3
{x + y + (x+y) + (x-4y) + xy + 2y}=6(y+3)
3x + xy = 6(x+y)
x(y+3)=6(3+y)
x=6

Let y=7 and x=6: Set={-18, 6, 7, 13, 14, 42}
Thus, median is \frac{(7+13)}{2} = 10


A. (x + y)/2 = 13/2
B. y + 3 = 10
C. y = 7
D. (3y)/2 = 21/2
E. x/3 + y = 9

Answer: B
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For the set of terms shown above, if y > 6 and the mean of the s   [#permalink] 12 Dec 2012, 23:05
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