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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

(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

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 _________________

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.

Re: For the set of terms shown above, if y > 6 and the mean of [#permalink]
14 Oct 2014, 04:37

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Re: For the set of terms shown above, if y > 6 and the mean of [#permalink]
15 Apr 2015, 20:20

Bunuel 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 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.

Just a small clarification needed. The highlighted portion: Is it implying that (y+3) on both sides cannot be discarded? if so, then why exactly because as per my understanding when the term in question can be ZERO, only then we must avoid cancelling it on both LHS and RHS.

Re: For the set of terms shown above, if y > 6 and the mean of [#permalink]
08 Jun 2015, 20:55

Expert's post

Hi Shreks1190,

As you continue to study, you're going to find that TESTing VALUES is a great approach for many types of GMAT questions (and not just Stats questions....DS questions, in particular, can often be beaten in this way). To spot 'TEST VALUES' questions, you should be on the lookout for variables, especially those that involve 'descriptions' (odd, even, >, <, integer, positive, negative, etc.). You'll also come to find that trying this approach during your studies (you can go back and re-attempt past questions) can help you to better define when to TEST VALUES and when to just 'do the math.'

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