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a, b, and c are three different numbers, none of which equal
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07 Apr 2019, 08:14
07 Apr 2019, 08:14
Expert Reply
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Question Stats:
88% (01:20) correct
11% (03:17) wrong based on 43 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. \(\frac{3}{4}\)
A. \(0\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{4}\)
Re: a, b, and c are three different numbers, none of which equal
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15 Apr 2020, 13:06
15 Apr 2020, 13:06
Expert Reply
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
let a=1,b=2,c=3
so x=-1 ,y=2,z=-1
so \(x+y+z=\)
IMO A
Re: a, b, and c are three different numbers, none of which equal
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27 May 2020, 13:23
27 May 2020, 13:23
2
Let \(\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}\) = k
Then,
x = kb-kc
y = kc-ka
z = ka-kb
So, x+y+z= kb-kc+kc-ka+ka-kb.. you see it? All cancel out and hence 0
Then,
x = kb-kc
y = kc-ka
z = ka-kb
So, x+y+z= kb-kc+kc-ka+ka-kb.. you see it? All cancel out and hence 0
