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GMAT Club Legend  V
Joined: 11 Sep 2015
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If x < 3, is (x + 1)/(x - 3) > 1/3?  [#permalink]

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12 00:00

Difficulty:   65% (hard)

Question Stats: 56% (01:55) correct 44% (02:04) wrong based on 162 sessions

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If x < 3, is $$\frac{x + 1}{x - 3}$$ > 1/3?

(1) x < 2

(2) x > -1

*Kudos for all correct solutions

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Marshall & McDonough Moderator D
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Re: If x < 3, is (x + 1)/(x - 3) > 1/3?  [#permalink]

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(x + 1)/(x - 3) - 1/3 > 0?

2(x + 3)/3(x -3) > 0? --> We have to find out whether the expression (x + 3)/(x - 3) is positive.

St1: x < 2 --> The expression can be positive, negative or zero.
Insufficient.

St2: x > -1 --> -1 < x < 3 --> Numerator is always positive and denominator is always negative. So the expression is negative.
Sufficient.

Originally posted by Vyshak on 10 Feb 2017, 09:58.
Last edited by Vyshak on 10 Feb 2017, 10:23, edited 1 time in total.
Typo
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4320
Re: If x < 3, is (x + 1)/(x - 3) > 1/3?  [#permalink]

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Top Contributor
Vyshak wrote:
(x + 1)/(x - 1) - 1/3 > 0?

2(x + 3)/3(x -3) > 0? --> We have to find out whether the expression (x + 3)/(x - 3) is positive.

St1: x < 2 --> The expression can be positive, negative or zero.
Insufficient.

St2: x > -1 --> -1 < x < 3 --> Numerator is always positive and denominator is always negative. So the expression is negative.
Sufficient.

I think you may have incorrectly transcribed the question.
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Marshall & McDonough Moderator D
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Re: If x < 3, is (x + 1)/(x - 3) > 1/3?  [#permalink]

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GMATPrepNow wrote:
Vyshak wrote:
(x + 1)/(x - 1) - 1/3 > 0?

2(x + 3)/3(x -3) > 0? --> We have to find out whether the expression (x + 3)/(x - 3) is positive.

St1: x < 2 --> The expression can be positive, negative or zero.
Insufficient.

St2: x > -1 --> -1 < x < 3 --> Numerator is always positive and denominator is always negative. So the expression is negative.
Sufficient.

I think you may have incorrectly transcribed the question.

Thanks. I have corrected it now. But the steps seem to be fine.
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4320
Re: If x < 3, is (x + 1)/(x - 3) > 1/3?  [#permalink]

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GMATPrepNow wrote:
If x < 3, is $$\frac{x + 1}{x - 3}$$ > 1/3?

(1) x < 2

(2) x > -1

Given: x < 3

Target question: Is (x + 1)/(x - 3) > 1/3?
This is a good candidate for rephrasing the target question.
Aside: See below for a video with tips on rephrasing the target question

Since x < 3, we know that (x - 3) will always be NEGATIVE.
So, let's take (x + 1)/(x - 3) > 1/3, and multiply both sides by (x - 3)
We get: x + 1 < (1/3)(x - 3) [ASIDE: since we multiplied both sides by a NEGATIVE value, we REVERSED the inequality sign]
Now take x + 1 < (1/3)(x - 3) and multiply both sides by 3 to get: 3(x + 1) < x - 3
Expand to get: 3x + 3 < x - 3
Subtract x from both sides: 2x + 3 < -3
Subtract 3 from both sides: 2x < -6
Divide both sides by 2 to get: x < -3

This equivalent inequality is much easier to work with. So, ....
REPHRASED target question: Is x < -3?

Statement 1: x < 2
There are several values of x that satisfy statement 1. Here are two:
Case a: x = -7, in which case x < -3
Case b: x = 0, in which case x > -3
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x > -1
If x is greater than -1, then we can be 100% certain that x IS NOT less than -3
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

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If x < 3, is (x + 1)/(x - 3) > 1/3?  [#permalink]

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Step 1-
(x-3)+4
X-3
gives
1+4/(x-3)>1/3
(2/(x-3))>-1/3
Now, option b , since x>-1 & <3,LHS can never be greater than -1/3
So B is sufficient
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If x < 3, is (x + 1)/(x - 3) > 1/3?  [#permalink]

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GMATPrepNow wrote:
If x < 3, is $$\frac{x + 1}{x - 3}$$ > 1/3?

(1) x < 2

(2) x > -1

$$\frac{x+1}{x-3}>1/3…\frac{(x+1)3-(x-3)1}{3(x-3)}>0…\frac{3x+3-x+3}{3x-9}>0…\frac{2(x+3)}{3(x-3)}>0$$
$$\frac{2(x+3)}{3(x-3)}>0…\frac{(x+3)}{x-3}>0…true:x<-3$$
$$(ie).x=-7:\frac{2(-7+3)}{3(-7-3)}>0…\frac{2(-4)}{3(-10)}>0…\frac{-8}{-30}>0$$

(1) x < 2 insufic

(2) x > -1 sufic

Ans (B) If x < 3, is (x + 1)/(x - 3) > 1/3?   [#permalink] 19 Nov 2019, 04:25
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