Set A: {x, x, x, y, y, y, 3x+y, x-y}

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Set A: {x, x, x, y, y, y, 3x+y, x-y}

by BTGmoderatorDC » Fri Dec 08, 2017 2:50 am
Set A: {x, x, x, y, y, y, 3x+y, x-y}

If 0 < x < y, then what is the range of set A?

(1) The median of set A is 10
(2) The average (arithmetic mean) of set A is 9

Can some experts find the best Option?

OA A

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by GMATGuruNY » Fri Dec 08, 2017 5:12 am
lheiannie07 wrote:Set A: {x, x, x, y, y, y, 3x+y, x-y}

If 0 < x < y, then what is the range of set A?

(1) The median of set A is 10
(2) The average (arithmetic mean) of set A is 9
Since 0 < x < y:
The smallest value in the set is x-y (which will be NEGATIVE).
The greatest value in the set is 3x+y (the sum of two POSITIVE values).
Listing the values in ascending order, we get:
x-y, x, x, x, y, y, y, 3x+y.
Range = biggest - smallest = (3x+y) - (x-y) = 2x + 2y = 2(x+y).
Thus, to calculate the range, we need to know the value of x+y.
Question stem, rephrased:
What is the value of x+y?

Statement 1:
The median of the set in blue is the average of the two middle values:
(x+y)/2.
Since the median is 10, we get:
(x+y)/2 = 10
x+y = 20.
SUFFICIENT.

Statement 2:
Since the average of the 8 values in set A is 9, we get:
Sum of the values in A = (count)(average) = 8*9 = 72.
Thus:
x + x + x + y + y + y + (3x+y) + (x-y) = 72
7x + 3y = 72.
No way to determine the value of x+y.
INSUFFICIENT.

The correct answer is A.
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