If x 1/3?

Question

If x < 3, is \frac{x + 1}{x - 3} > 1/3? (1) x < 2 (2) x > -1 *Kudos for all correct solutions

Kurtosis · Answer

(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.

Answer: B

BrentGMATPrepNow · Answer

I think you may have incorrectly transcribed the question.

Kurtosis · Answer

Thanks. I have corrected it now. But the steps seem to be fine.

BrentGMATPrepNow · Answer

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) 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 Answer: B RELATED VIDEOS FROM OUR COURSE https://www.youtube.com/watch?v=MyG1K3ee69w https://www.youtube.com/watch?v=Od98BOicIq4

RogerRF · Answer

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

exc4libur · Answer

\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?

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