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average= (3x+xy)/6=y+3 => (y+3) (x-6)=0 => x=6 because y>6 Can arrange the set in the following order: (x-4y,x,y,x+y,2y,xy) median= (y+x+y)/2=y+x/2 = y+3 ok
We can calculate the mean as Sum / count which means 3x + xy /6 = y +3 or x(3+y)=6(y+3). After recognizing this manipulation, we deduce x =6 and we have y = GT(6).
Using this information, we can arrange in increasing order to find the median. Note the count is even so the median is the average of the two middle terms.
The arranged increasing set on the number line is: 6 -4y ; 6 ; y ; 6 + y ; 2y ; 6y
The two middle terms are are y and 6 +y, so the average is 6+2y /2 or 3 + y.
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.
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