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02 Apr 2019, 03:59
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
59% (02:18) correct
41%
(02:35)
wrong
based on 79
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Not Attempted Yet
If \(x > y\) and \(x < 0\), then which of the following must be true?
I. \(\frac{1}{x} < \frac{1}{y}\)
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
III. \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
I. \(\frac{1}{x} < \frac{1}{y}\)
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
III. \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
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02 Apr 2019, 06:38
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Bunuel
If \(x > y\) and \(x < 0\), then which of the following must be true?
I. \(\frac{1}{x} < \frac{1}{y}\)
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
III. \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
I. \(\frac{1}{x} < \frac{1}{y}\)
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
III. \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
We're told that x is NEGATIVE, and we're told that y < x, so y is also NEGATIVE
Take each statement and manipulate it (by applying the rules of inequalities) to see which one(s) obey the restriction that y < x
I. \(\frac{1}{x} < \frac{1}{y}\)
Multiply both sides by x to get: \(1 > \frac{x}{y}\) [since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]
Multiply both sides by y to get: \(y < x\) [since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]
PERFECT! This obeys the restriction that y < x
Statement I is true.
Check the answer choices . . . ELIMINATE B and C
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
NOTE: Since x is NEGATIVE, we know that x-1 is NEGATIVE
Likewise, since y is NEGATIVE, we know that y-1 is NEGATIVE
Multiply both sides by \(x-1\) to get: \(1 > \frac{x-1}{y−1}\) [since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]
Multiply both sides by \(y-1\) to get: \(y-1 < x-1\) [since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]
Add 1 to both sides to get: \(y < x\)
PERFECT! This obeys the restriction that y < x
Statement II is true.
Check the remaining answer choices . . . ELIMINATE A and E
By the process of elimination, the correct answer is D
RELATED VIDEO FROM MY COURSE
03 Apr 2019, 03:46
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Bunuel
If \(x > y\) and \(x < 0\), then which of the following must be true?
I. \(\frac{1}{x} < \frac{1}{y}\)
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
III. \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
I. \(\frac{1}{x} < \frac{1}{y}\)
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
III. \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
given x<0 ; x is -ve so y has to be -ve as well
x=-1 and y=-2
test with options
only IMO E is correct
