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If x, y, and z are integers, and x < y < z, is z y = y

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Re: If x, y, and z are integers, and x < y < z, is z y = y  [#permalink]

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New post 22 Nov 2019, 06:06
kairoshan wrote:
If x, y, and z are integers, and x < y < z, is z – y = y – x?

(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}.
(2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.


\((x,y,z)=integers…z-y=y-x…z+x=2y…y=(z+x)/2…y=avg(x,y)?\)

(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. insufic

\(\frac{x+y+z+4}{4}>\frac{x+y+z}{3}…x+y+z<12\)

(2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}. sufic

\(median(x,y,z,4)<median(x,y,z)=y\)
\(med(4,x,y,z)=(x+y/2)<y…x<y=valid…4<x<y<z\)
\(med(x,4,y,z)=(4+y/2)<y…4<y=valid…x≤4<y<z\)
\(med(x,y,z,4)=(y+z/2)<y…z<y=invalid\)

(1&2) sufic

\(x+y+z<12…x<y…(4<x<y<z)…x+y+z=5+6+7=18>12…invalid\)
\(x+y+z<12…4<y…(x≤4<y<z)…x+y+z=x+5+6<12…x<1…y≠avg(x,y)\)

Ans (C)
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Re: If x, y, and z are integers, and x < y < z, is z y = y   [#permalink] 22 Nov 2019, 06:06

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