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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.