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For the set of terms shown above, if y > 6 and the mean of [#permalink]
 02 Nov 2009, 04:08
Question Stats:
61% (02:57) correct
38% (02:40) wrong based on 6 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
Last edited by Bunuel on 11 Feb 2012, 16:59, edited 1 time in total.
Edited the question and added the OA
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Manager
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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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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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Manager
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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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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 beA. (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]
13 Dec 2012, 00:05
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]
13 Dec 2012, 00:05
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