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01 Dec 2023, 02:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (02:01) correct
30%
(02:05)
wrong
based on 184
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If x and y are distinct integers so that xy ≠ 0, which of the following must be true?
(A) \(1 - \frac{(1 - x)}{x} = 1\)
(B) \(\frac{1}{x} + \frac{1}{y} = \frac{xy}{(y + x)}\)
(C) \(\frac{y}{(y - x)} + \frac{x}{(x - y)} = 1\)
(D) \(\frac{(x - y)}{(y - x)} = 1\)
(E) \((1 + \frac{1}{x})*(\frac{1}{x + 1}) = 1\)
(A) \(1 - \frac{(1 - x)}{x} = 1\)
(B) \(\frac{1}{x} + \frac{1}{y} = \frac{xy}{(y + x)}\)
(C) \(\frac{y}{(y - x)} + \frac{x}{(x - y)} = 1\)
(D) \(\frac{(x - y)}{(y - x)} = 1\)
(E) \((1 + \frac{1}{x})*(\frac{1}{x + 1}) = 1\)
03 Dec 2023, 07:33
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The answer is C.
(A) 1 - (1 - x)/x = (x - 1 + x)/x = (2x - 1)/x, which is not necessarily = 1 --> NOT TRUE
(B) (1/x) + (1/y) = (y + x)/xy --> NOT TRUE
(C) (y/y-x) + (x/x - y) = (y/y - x) - (x/y - x) = (y - x)/(y - x) = 1 --> TRUE
(D) (x - y)/(y - x) = -1 --> NOT TRUE
(E) (x + 1)/x * 1/(x + 1) = 1/x --> NOT TRUE
(A) 1 - (1 - x)/x = (x - 1 + x)/x = (2x - 1)/x, which is not necessarily = 1 --> NOT TRUE
(B) (1/x) + (1/y) = (y + x)/xy --> NOT TRUE
(C) (y/y-x) + (x/x - y) = (y/y - x) - (x/y - x) = (y - x)/(y - x) = 1 --> TRUE
(D) (x - y)/(y - x) = -1 --> NOT TRUE
(E) (x + 1)/x * 1/(x + 1) = 1/x --> NOT TRUE
06 Jan 2024, 01:45
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Bunuel
If x and y are distinct integers so that xy ≠ 0, which of the following must be true?
(A) \(1 - \frac{(1 - x)}{x} = 1\)
(B) \(\frac{1}{x} + \frac{1}{y} = \frac{xy}{(y + x)}\)
(C) \(\frac{y}{(y - x)} + \frac{x}{(x - y)} = 1\)
(D) \(\frac{(x - y)}{(y - x)} = 1\)
(E) \((1 + \frac{1}{x})*(\frac{1}{x + 1}) = 1\)
(A) \(1 - \frac{(1 - x)}{x} = 1\)
(B) \(\frac{1}{x} + \frac{1}{y} = \frac{xy}{(y + x)}\)
(C) \(\frac{y}{(y - x)} + \frac{x}{(x - y)} = 1\)
(D) \(\frac{(x - y)}{(y - x)} = 1\)
(E) \((1 + \frac{1}{x})*(\frac{1}{x + 1}) = 1\)
One can solve this question very easily by plugging in some values for x and y or use algebraic approach:
A. \(1 - \frac{(1 - x)}{x} = 1\) --> \(LHS=1 - \frac{(1 - x)}{x}=\frac{x-(1-x)}{x}=\frac{2x-1}{x}\). Discard.
B. \(\frac{1}{x} + \frac{1}{y} = \frac{xy}{(y + x)}\) --> \(LHS=\frac{1}{x} + \frac{1}{y}=\frac{x+y}{xy}\). Discard.
C. \(\frac{y}{(y - x)} + \frac{x}{(x - y)} = 1\) --> \(LHS=\frac{y}{(y - x)} + \frac{x}{(x - y)}=\frac{y}{(y - x)} - \frac{x}{(y - x)}=\frac{y-x}{y-x}=1\). BINGO.
D. \(\frac{(x - y)}{(y - x)} = 1\) --> \(LHS=\frac{(x - y)}{(y - x)}=-\frac{x-y}{x-y}=-1\). Discard.
E. \((1 + \frac{1}{x})*(\frac{1}{x + 1}) = 1\) --> \(LHS=(1 + \frac{1}{x})*(\frac{1}{x + 1})=\frac{x+1}{x}*\frac{1}{x+1}=\frac{1}{x}\). Discard.
Answer: C.
