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If the median of set A is 10 and 0 < x < y, what is the range of set A [#permalink]

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28 Jul 2017, 08:45

2

This post received KUDOS

Bunuel wrote:

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

As x & y are positive and x<y, we can arrange the numbers of set A in ascending order to arrive at the median value. Smallest element of set A will be x-y as it is the only element in set A that is negative. Largest element will be 3x+y as it is summation of two positive elements.

so set A in ascending order = {x-y, x, x, x, y, y, y, 3x+y} so the median of this set is = (x+y)/2 = 10 or x+y = 20 Range of the set A will be = (3x+y) - (x-y) = 2x+2y or 2(x+y) = 2*20 = 40

Option D

Last edited by niks18 on 28 Jul 2017, 09:24, edited 1 time in total.

Re: If the median of set A is 10 and 0 < x < y, what is the range of set A [#permalink]

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28 Jul 2017, 09:09

niks18 wrote:

Bunuel wrote:

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

As x & y are positive and x>y, we can arrange the numbers of set A in ascending order to arrive at the median value. Smallest element of set A will be x-y as it is the only element in set A that is negative. Largest element will be 3x+y as it is summation of two positive elements.

so set A in ascending order = {x-y, x, x, x, y, y, y, 3x+y} so the median of this set is = (x+y)/2 = 10 or x+y = 20 Range of the set A will be = (3x+y) - (x-y) = 2x+2y or 2(x+y) = 2*20 = 40

Option D

Kudos man !! Classic way to solve this problem. I also came down to answer but by a trivial method.

Re: If the median of set A is 10 and 0 < x < y, what is the range of set A [#permalink]

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29 Jul 2017, 20:50

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This post was BOOKMARKED

Bunuel wrote:

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

Since 0 < x < y , Set A can be arranged in ascending order as {x-y, x, x, x, y, y, y, 3x+y} Now , median is (x+y)/2 = 10 => x+y = 20

Range of set A = 3x+y -x+y = 2x+2y = 2(x+y) = 40

Answer D
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Re: If the median of set A is 10 and 0 < x < y, what is the range of set A [#permalink]

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31 Jul 2017, 08:54

Bunuel wrote:

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

This question revolves around the median concept, so a sorted list must come into the mind. Therefore, what matters is: 0<x<y, which will help sort our list.

Since, x<y => x-y<0 and will be the smallest, followed by three 'x' and three 'y' as x<y. 3x+y is the greatest. Now, median for the list of 8 elements = 4th + 5th/2 = x+y/2=10 => x+y=20

Range = 3x+y-x+y=2x+2y= 2(x+y) We've found x+y=20 Therefore, Range = 20*2 = 40.

If the median of set A is 10 and 0 < x < y, what is the range of set A [#permalink]

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06 Nov 2017, 06:25

For this problem, I chose numbers. Since we know x < y, then we know x-y is a number smaller than x. We also know 3x+y is a number bigger than Y, making it the largest number in the set. When we put the numbers in order from least to greatest, our 4th and 5th terms are X and Y, so X=9 and Y=11 making the median 10. Now plugging these into the equations for both x-y and 3x+y, we get -2 and 38, respectively. So the range for this set is 40. D

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

First step is to arrange the set in increasing order of numbers.

Since x and y are both positive and x < y, x - y will be the smallest (it will be negative). Then you will have the 3 x's,t hen 3 y's and finally 3x + y.

{x - y, x, x, x, y, y, y, 3x+y}

Median of 8 numbers is the average of middle 2. So (x+y)/2 = 10 (x + y) = 20

Now, let's assume values for x and y which satisfy all given constrains. Say x = 1 and y = 19 x - y = - 18 and 3x + y = 22 Range = 22 - (-18) = 40