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Source: — Data Sufficiency |
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Re: Kaplan Question

by jayhawk2001 » Tue Apr 17, 2007 9:41 pm
rajesh_ctm wrote:What is the value of x^2-1?

(1) The value of x^2 - 5x is -6


(2) The value of (x^2 - 1)/(x + 1) is 1
1 - insufficient. x^2 - 5x + 6 = 0 implies x = 2 or x = 3

2 - insufficient. (x+1)(x-1)/(x+1) = 1. We cannot cancel out
x+1 from numerator and denominator unless we know x != -1.

Alternatively, x^2 - 1 = x + 1. So, x^2 -x -2 = 0 implies
x = 2 or x = -1

Combining 1 and 2, we have x = 2.

Is C the answer ?
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by ns88 » Sun May 06, 2007 12:14 pm
I'm getting B.

x^2-1= (x+1) (x-1)


(x+1) (x-1)/x+1= 1


x-1=1

x=2


what's the OA?
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by jayhawk2001 » Sun May 06, 2007 1:08 pm
ns88 wrote:I'm getting B.
x^2-1= (x+1) (x-1)
(x+1) (x-1)/x+1= 1

x-1=1

x=2

what's the OA?
ns88, you cannot cancel out x+1 from the numerator and denominator
unless you know x != -1.

If x=-1, then essentially you are doing 0/0 = 1 which is incorrect.
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