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a, b, and c are three different numbers, none of which equals the aver

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a, b, and c are three different numbers, none of which equals the aver  [#permalink]

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New post 07 Apr 2019, 21:35
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

88% (01:41) correct 12% (02:37) wrong based on 17 sessions

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New post 07 Apr 2019, 22:07
In x+y+z, substituting for x and z as y(b-c)/(c-a) and y(a-b)/(c-a)
Solving for this equation we get 0/c-a.

Answer A

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Re: a, b, and c are three different numbers, none of which equals the aver  [#permalink]

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New post 08 Apr 2019, 01:22
Bunuel wrote:
\(a\), \(b\), and \(c\) are three different numbers, none of which equals the average of the other two. If \(\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}\), then \(x+y+z=\)

A. \(0\)

B. \(\frac{1}{2}\)

C. \(\frac{1}{3}\)

D. \(\frac{2}{3}\)

E. \(\frac{3}{4}\)


using expression \(\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}\),
let a=1,b=2,c=3
so x=-1 ,y=2,z=-1
so \(x+y+z=\)

IMO A;0
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Re: a, b, and c are three different numbers, none of which equals the aver  [#permalink]

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New post 08 Apr 2019, 10:28
From the question, we can also write,

x=y(b-c)/(c-a)
z=y(a-b)/(c-a)

Hence x + y + z can be written as:

y(b-c)/(c-a) + y + y(a-b)/(c-a)
= [y(b-c) + y(c-a) + y(a-b)]/(c-a)

=[yb - yc + yc - ya + ya - yb]/(c-a)

= 0
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Re: a, b, and c are three different numbers, none of which equals the aver   [#permalink] 08 Apr 2019, 10:28
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