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# If x > y and x < 0, then which of the following must be true?

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Math Expert
Joined: 02 Sep 2009
Posts: 61189
If x > y and x < 0, then which of the following must be true?  [#permalink]

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02 Apr 2019, 02:59
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Difficulty:

95% (hard)

Question Stats:

62% (02:21) correct 38% (02:08) wrong based on 41 sessions

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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

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Posts: 4318
Re: If x > y and x < 0, then which of the following must be true?  [#permalink]

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02 Apr 2019, 05:38
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Top Contributor
Bunuel wrote:
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

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

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Re: If x > y and x < 0, then which of the following must be true?  [#permalink]

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03 Apr 2019, 02:46
Bunuel wrote:
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

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
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Joined: 18 Oct 2018
Posts: 91
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GMAT 1: 710 Q50 V36
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Re: If x > y and x < 0, then which of the following must be true?  [#permalink]

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03 Apr 2019, 10:06
The 3rd statement won't stand for x=-1, y=-2
In this case, it will be 1/x= 1/(-1+1)= infinity
While, 1/y= 1/(-2+1)=-1

Hence answer should be D, statement 1 and 2 are true
Re: If x > y and x < 0, then which of the following must be true?   [#permalink] 03 Apr 2019, 10:06
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