Problem Solving

BTGmoderatorLU wrote:Source: Magoosh

Set A:\(\{x, x, x, y, y, y, 3x+y, x-y\}\)

If the median of set A is 10 and \(0 < x < y\), what is the range of set A?

A. 10
B. 20
C. 30
D. 40
E. 60

The OA is D

Given \(0 < x < y\), we know that both x and y are positive and y is greater than x. Thus, the smallest number in Set A would be x - y and the largest would be 3x + y. Thus, range = (3x + y) - (x - y) = 2(x + y).

Arranging the elements of Set A, we get \(\{x-y, x, x, x, y, y, y, 3x+y\}\). There are 8 elements in the set. The median would be average of the 4th and 5th elements = (x + y)/2

Thus, (x + y)/2 = 10 => x + y = 20

Thus, range = 2(x + y) = 2*20 = 40.

The correct answer: D

Hope this helps!

-Jay
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BTGmoderatorLU wrote:Source: Magoosh

Set A:\(\{x, x, x, y, y, y, 3x+y, x-y\}\)

If the median of set A is 10 and \(0 < x < y\), what is the range of set A?

A. 10
B. 20
C. 30
D. 40
E. 60

The OA is D

Since both x and y are positive, x - y will be the smallest element, and 3x + y will be the largest element of set A.

Since x < y, in ascending order, we have:

x - y, x, x, x, y, y, y, 3x + y.

Since the median is 10, we have (x + y)/2 = 10 or x + y = 20.

The range of set A is:

3x + y - (x - y) = 2x + 2y = 2(x + y) = 2(20) = 40.

Answer: D