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a, b, and c are three different numbers. None of the numbers
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07 Apr 2019, 07:49
07 Apr 2019, 07:49
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66% (01:39) correct
33% (02:08) wrong based on 21 sessions
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\(a\), \(b\), and \(c\) are three different numbers. None of the numbers equals the average of the other two. If \(\frac{x}{a+b-2c} = \frac{y}{b+c-2a} = \frac{z}{c+a-2b}\) , then \(x+y+z=\)
(A) 0
(B) 3
(C) 4
(D) 5
(E) 6
(A) 0
(B) 3
(C) 4
(D) 5
(E) 6
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Re: a, b, and c are three different numbers. None of the numbers
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20 Apr 2020, 15:50
20 Apr 2020, 15:50
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Carcass wrote:
\(a\), \(b\), and \(c\) are three different numbers. None of the numbers equals the average of the other two. If \(\frac{x}{a+b-2c} = \frac{y}{b+c-2a} = \frac{z}{c+a-2b}\) , then \(x+y+z=\)
(A) 0
(B) 3
(C) 4
(D) 5
(E) 6
(A) 0
(B) 3
(C) 4
(D) 5
(E) 6
x/(a+b-2c) = y/(b+c-2a) = z/(c+a-2b) = k (let us assume)
Thus, we have:
x = k(a+b-2c)
y = k(b+c-2a)
z = k(c+a-2b)
Adding: x+y+x = k(0) = 0
Answer A
Re: a, b, and c are three different numbers. None of the numbers
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28 May 2024, 05:02
28 May 2024, 05:02
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