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If the median of set A is 10 and 0 < x < y, what is the range of set A
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28 Jul 2017, 09:10
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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
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If the median of set A is 10 and 0 < x < y, what is the range of set A
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Updated on: 28 Jul 2017, 10:24
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 xy 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 = {xy, 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)  (xy) = 2x+2y or 2(x+y) = 2*20 = 40 Option D
Originally posted by niks18 on 28 Jul 2017, 09:45.
Last edited by niks18 on 28 Jul 2017, 10:24, edited 1 time in total.



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Re: If the median of set A is 10 and 0 < x < y, what is the range of set A
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28 Jul 2017, 10:09
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 xy 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 = {xy, 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)  (xy) = 2x+2y or 2(x+y) = 2*20 = 40 Option DKudos man !! Classic way to solve this problem. I also came down to answer but by a trivial method. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile app



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Re: If the median of set A is 10 and 0 < x < y, what is the range of set A
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29 Jul 2017, 21:50
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 {xy, 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
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31 Jul 2017, 09:54
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 => xy<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+yx+y=2x+2y= 2(x+y) We've found x+y=20 Therefore, Range = 20*2 = 40.



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If the median of set A is 10 and 0 < x < y, what is the range of set A
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06 Nov 2017, 07:25
For this problem, I chose numbers. Since we know x < y, then we know xy 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 xy and 3x+y, we get 2 and 38, respectively. So the range for this set is 40. 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
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06 Nov 2017, 07:39
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 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 Answer (D)
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