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?

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Kudos for correct solution.

Using the fact that x<y, we can order our set increasingly as {x-y x x x y y y 3x-y}. Then, the fact that the median is 10 can be rewritten as (x+y)/2 = 10 or x+y = 20.

The range of the set is computed as 3x-y-(x-y) = 2x+2y = 2(x+y) = 2*20 = 40.

The range is 40!

Bump for further discussion

Solved in 1:30

We are given the median. Since A is composed of 9 numbers, the median will be the average of the 4th and 5th value.
So we need to find the 4th and 5th value.
We know that x-y must be some negative number, so it will be the first
then all the xs, then all the ys, and then 3x+y
the 4th value is x, and 5th value is y.
So we know that (x+y)=20
the range is (max-min)=(3x+y-(x-y))=(2x+2y)
we know that x+y, so twice that is just 40

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