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Set A is {x,2x+y,2y,y+z,x+z}. If x,y, and z are positive [#permalink]
 11 Mar 2010, 06:40
Question Stats:
75% (03:22) correct
25% (00:37) wrong based on 4 sessions
Set A is {x,2x+y,2y,y+z,x+z}. If x,y, and z are positive integers, then what is the value of the mode of Set A? 1. x is equal to one half of y and twice of z 2. The median of set A is equal to 125. Please explain your answer
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Re: Set A is {x,2x+y,2y,y+z,x+z}. If x,y, and z are positive [#permalink]
11 Mar 2010, 13:17
IMHO C
From 1 statement:
x=y/2 and x=2z, So the given set A can be written in the following manner.
A=(x, 2x+2x, 4x, 2x+x/2, x+x/2)
A=(x,4x,4x,2.5x,1.5x)
Now all the no. positive integers, so A can be arranged in increasing order.
A=(x,1.5x,2.5x,4x,4x)
Mode will be 4x
We don't know the value of X, Statement 1 is Insufficient.
From Statement 2:
We don't know the relation between x,y,and Z. So arranging them in order and finding the most repeatitive term in not possible. Statement 2 is Insufficient.
From 1 and 2
Median is 2.5x=125 >>> x=50
so Mode is 4x= 200
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Re: Set A is {x,2x+y,2y,y+z,x+z}. If x,y, and z are positive [#permalink]
11 Mar 2010, 16:15
Great question. I had E for this originally but agree that it's C.
nverma, good job or re-arranging the order by using A). I didn't think of that.
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Re: Set A is {x,2x+y,2y,y+z,x+z}. If x,y, and z are positive
[#permalink]
11 Mar 2010, 16:15
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