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Is (x + 1)/(x - 3) < 0 ?

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Is (x + 1)/(x - 3) < 0 ?

by NandishSS » Fri Dec 23, 2016 6:11 am
Is (x + 1)/(x - 3) < 0 ?

(1) -1 < x < 1
(2) x^2 - 4 < 0

OA:A

Stat 1[spoiler]Tested for x=0.5 & x=-0.5 gives -ve[/spoiler]

Stat 2[spoiler]Tested for x=1.5 Gives -ve and for x=-1.5 gives +ve[/spoiler]

Hi GMATGuruNY/[email protected]

Are these values enough to test? I was just practising the problems by Testing methods.

Thanks
Nandish
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Source: — Data Sufficiency |

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by [email protected] » Fri Dec 23, 2016 9:36 am
Hi NandishSS,

Certainly the work that you did for Fact 2 was enough to prove its sufficiency/insufficiency. Since this question clearly involves Number Properties, it sometimes helps to think about the 'limits' of the numbers involved (in this case, how big could X really get? And how small? What about 0? Etc.). Remember that DS questions 'test' a number of different skills, including the 'thoroughness' of your thinking - so doing a bit more work on Fact 1 would be okay.

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by GMATGuruNY » Fri Dec 23, 2016 12:46 pm
NandishSS wrote:Is (x + 1)/(x - 3) < 0 ?

(1) -1 < x < 1
(2) x^2 - 4 < 0
The CRITICAL POINTS are x=-1 and x=3.
These are the only values where the left side is either 0 or undefined.
To determine where (x + 1)/(x - 3) < 0, test one value to the left of -1, one value between -1 and 3, and one value to the right of 3.

If x=-2, then (x+1)/(x-3) = (-2+1)/(-2-3) = 1/5.
If x=0, then (x+1)/(x-3) = (0+1)/(0-3) = -1/3.
If x=4, then (x+1)/(x-3) = (4+1)/(4-3) = 5.

Only the blue case -- which tests a value between -1 and 3 -- satisfies (x+1)/(x-3) < 0.
Implication:
(x+1)/(x-3) < 0 only if x is between -1 and 3.
Question stem, rephrased:
Is x between -1 and 3?

Statement 1: -1<x<1
Since -1<x<1, we know that x is between -1 and 3.
SUFFICIENT.

Statement 2: x² < 4
Case 1: x=0
In this case, x is between -1 and 3.
Case 2: x = -3/2
In this case, x is NOT between -1 and 3.
INSUFFICIENT.

The correct answer is A.
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by ceilidh.erickson » Mon Dec 26, 2016 6:11 am
NandishSS wrote:
Are these values enough to test? I was just practising the problems by Testing methods.

Thanks
Nandish
Instead of beginning by testing values, begin by REPHRASING the question. Otherwise you may not ever be sure if you've tested enough values.

Begin by asking yourself "when would (x + 1)/(x - 3) be negative?"
To get a negative fraction, we must have a negative numerator and positive denominator, or vice versa.

(x + 1) is negative when x < -1
(x - 3) is positive when x > 3
These cannot occur together, so we'll never have a negative numerator, positive denominator

(x + 1) is positive when x > -1
(x - 3) is negative when x < 3
Thus, the only time we'll get a "yes" answer to our question is when -1 < x < 3.

Target question: is -1 < x < 3 ? In other words, is x between -1 and 3?

Now we may not need to test any values at all if the statements directly answer this question.

(1) -1 < x < 1
This directly answers the target question. If x is between -1 and 1, then it is definitely within the larger range of -1 to 3. Sufficient.

(2) x^2 - 4 < 0
If we REPHRASE this statement, we get:
x^2 < 4
-2 < x < 2
This does not answer our question, since this range lies partially within and partially outside of our question range. -1.5 would give us a "no" answer to our question, but 0.5 would give us a "yes" answer. Insufficient.

The answer is A.
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by ceilidh.erickson » Mon Dec 26, 2016 6:16 am
Here are some more examples of when REPHRASING before testing values is a good idea (and how to use them in conjunction):
https://www.beatthegmat.com/factorizatio ... tml#581386
https://www.beatthegmat.com/inequality-t ... tml#695923
https://www.beatthegmat.com/is-a-0-data- ... tml#758157
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