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12 Apr 2018, 21:44
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
13% (01:52) correct
87%
(01:34)
wrong
based on 111
sessions
History
Date
Time
Result
Not Attempted Yet
If xy ≠ 0 and x ≠ y, what is the value of K?
(1) \(Kx + x - Ky - y = 0\)
(2) \((\frac{x}{y})^k = \frac{y}{x}\)
(1) \(Kx + x - Ky - y = 0\)
(2) \((\frac{x}{y})^k = \frac{y}{x}\)
12 Apr 2018, 21:50
Kudos
Bookmarks
Bunuel
Question: K = ?
Statement 1: \(Kx + x - Ky - y = 0\)
i.e. K (x-y) +(x-y) = 0
i.e. (K+1)(x-y) = 0
i.e. Either K = -1 or x=y but sine given that x ≠ y therefore K = 1
SUFFICIENT
Statement 2: \((\frac{x}{y})^k = \frac{y}{x}\)
i.e. (x/y)^K = (x/y)^(-1)
i.e. K = -1
SUFFICIENT
Answer: Option D
24 Aug 2023, 06:57
Kudos
Bookmarks
Bunuel
(1) \(Kx + x - Ky - y = 0…….x(K+1)-y(K+1)=0……….(x-y)(K+1)=0\)
But we are given x ≠ y, so only way above expression is valid is if K=-1.
Sufficient
(2) \((\frac{x}{y})^k = \frac{y}{x}\)
\((\frac{x}{y})^k = \frac{y}{x}=(\frac{x}{y})^{-1}\)
We know that xy≠0, that is neither of x or y is 0.
Equating two sides, we get K=-1.
But say x=1 and y=-1, then \((\frac{x}{y})^k = \frac{y}{x}\) would become
\((\frac{1}{-1})^k = \frac{-1}{1}\)
\((-1)^k=-1\). Thus k could be 1 or -1
Insufficient
A
Bunuel please check the OA. It should be A and not D.
