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nga.kam.lai
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Is (x+1)/(x-3) < 0?

by nga.kam.lai Sun Jul 11, 2010 3:14 pm

Hi, can someone help me with the data sufficiency problem below?

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

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

thanks!
rohit801
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Re: Is (x+1)/(x-3) < 0?

by rohit801 Sun Jul 11, 2010 6:20 pm

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

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

So, let's see what the question is asking. Is the ratio of 2 quantities negative=> either the numerator OR the denominator must be negative for this to be true but not BOTH.

1) The numerator will be positive because x is greater than -1. The denominator will always be negative since the maximum x can get is very close to 1, but subtracting 3 from it, will result in a negative number. So, we have a negative result. SUFFICIENT.

2) X^2 <4 => -2<X<2. Now, the denomiator, again, will be always negative since the MAX x can get is very close to 2. BUT, now the numertor can be positive [for x>-1] or negative [for -2< x<-1]. Hence, we can get a postive or a negative end result. INSUFFICIENT.

Hope this helps....
adiagr
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Re: Is (x+1)/(x-3) < 0?

by adiagr Sun Jul 11, 2010 10:17 pm

nga.kam.lai Wrote:Hi, can someone help me with the data sufficiency problem below?

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

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

thanks!



Rohit has given very nice explanation.

I would do it by plugging numbers:

St. 1

Put x = 0, -(1/2) , (1/2), given expression Holds

St. 2

Given expression holds for x = 0

but for x = -1, given expression becomes equal to zero , thus expression does not hold. Insufficient.

So A is the answer.
funsho.olukade
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Re: Is (x+1)/(x-3) < 0?

by funsho.olukade Mon Jul 12, 2010 12:20 pm

Let Y=(x+1)/(x-3). There are two ways y can be less than zero (negative):
scenario1: Numerator is negative AND Denominator is positive
or
Senario2: Numerator is positive AND Denominator is negative

For Scenario1 x+1<0 implies x<-1 AND x-3>0 implies x>3 This condition can not be satisfied as x can not be less than -1 AND still greater than 3.

For scenario2 x+1>0 implies x>-1 AND x-3<0 implies x <3 which satisfies the condition -1<x<3 Call this the universal set.
This makes statement 1 sufficient since all values lies within the universal set

From Statement2, x^2<4 implies x<2 e.g -3 (substitute x=-3) which will return y positive, hence statement 2 does not provide sufficient information.
debmalya.dutta
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Re: Is (x+1)/(x-3) < 0?

by debmalya.dutta Mon Jul 12, 2010 8:41 pm

(x+1)/(x-3)< 0
From statement 1
-1 < x < 1 => 0 < x+1 < 2 and -4<x-3<-2
Surely (x+1)/(x-3)< 0
From Statement 2
x^2 - 4 < 0 => -2<x<2
-1<x+1<3 and -5<x-3<-1
(x-3) is always negative but (X+1) can be positive or negative.
Hence the statement is insufficient
Answer is A
RonPurewal
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Re: Is (x+1)/(x-3) < 0?

by RonPurewal Sun Aug 01, 2010 3:18 am

debmalya.dutta Wrote:(x+1)/(x-3)< 0
From statement 1
-1 < x < 1 => 0 < x+1 < 2 and -4<x-3<-2
Surely (x+1)/(x-3)< 0
From Statement 2
x^2 - 4 < 0 => -2<x<2
-1<x+1<3 and -5<x-3<-1
(x-3) is always negative but (X+1) can be positive or negative.
Hence the statement is insufficient
Answer is A


this is a nice analysis.