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

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Math Expert
Joined: 02 Sep 2009
Posts: 58396
a, b, and c are three different numbers, none of which equals the aver  [#permalink]

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07 Apr 2019, 21:35
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Difficulty:

25% (medium)

Question Stats:

83% (01:48) correct 17% (02:32) wrong based on 23 sessions

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$$a$$, $$b$$, 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. $$\frac{3}{4}$$

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Joined: 09 Apr 2018
Posts: 32
Re: a, b, and c are three different numbers, none of which equals the aver  [#permalink]

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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.

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

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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
Intern
Joined: 24 Jun 2018
Posts: 35
Re: a, b, and c are three different numbers, none of which equals the aver  [#permalink]

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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
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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