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01 Nov 2016, 07:56
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A
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Difficulty:
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Question Stats:
56% (02:37) correct
44%
(02:39)
wrong
based on 151
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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
*Kudos for all correct solutions
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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
*Kudos for all correct solutions
Video solution here: https://www.gmatprepnow.com/module/gmat ... /video/818
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01 Nov 2016, 11:10
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Since 0<x<y,
x-y<x=x=x<y=y=y<3x+y (since x-y is negative)
(1) The median of set A is 10
median of set a = (x+y)/2
(x+y)/2=10 ->x+y=20
Now range=3x+y-(x-y) -> 3x+y-x+y ->2x+2y->2(x+y) ->2*20=40
Sufficient.
(2) The average (arithmetic mean) of set A is 9
=(x+x+x+y+y+y+(x-y)+3x+y)/8=9
=7x+3y=72,this is not sufficient to find range of set A.
Insufficient.
Hence,A.
x-y<x=x=x<y=y=y<3x+y (since x-y is negative)
(1) The median of set A is 10
median of set a = (x+y)/2
(x+y)/2=10 ->x+y=20
Now range=3x+y-(x-y) -> 3x+y-x+y ->2x+2y->2(x+y) ->2*20=40
Sufficient.
(2) The average (arithmetic mean) of set A is 9
=(x+x+x+y+y+y+(x-y)+3x+y)/8=9
=7x+3y=72,this is not sufficient to find range of set A.
Insufficient.
Hence,A.
01 Nov 2016, 12:49
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GMATPrepNow
If 0 < x < y the set is, in order, A: { x-y, y, y, y, x, x, x, 3x+y,}
therefore, the Median is \(\frac{(x+y)}{2}\) and the average is \(\frac{7x+3y}{8}\)
the range is equal to \(3x+y-(x-y)=2(x+y)\) Thus the value of x+y would be sufficient.
(1) The median of set A is 10 ====> x+y=20 SUFFICIENT
(2) The average (arithmetic mean) of set A is 9 \(\frac{7x+3y}{8}\)=9====> you can't find the value of x+y INSUFFICIENT
answer A
