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Source: — Data Sufficiency |
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by MBACenter » Thu Nov 03, 2011 5:49 pm
The way you deal with statement 2 is thus:

Division by 0 is never possible, so you know that x cannot equal -1. Therefore, you can multiply both sides by x + 1:

(x^2 - 1)/(x + 1) = 1
x^2 - 1 = x + 1
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0

So, x = 2 or x = -1. And since we know that x cannot be -1 or we get an illegal operation (division by 0), then we know x = 2, and so statement 2 is sufficient.

To answer your other question, no, 0/0 is not equal to 1. You CAN NEVER divide by 0, period.
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by studentps2011 » Thu Nov 03, 2011 8:03 pm
If we re-write the first statement (x2-5x = -6) as

(x2-4)/(x-2) = 5, it will have the same issue as statement 2.

So isn't it true that we have to consider both roots of statement 2 as valid?
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by saketk » Thu Nov 03, 2011 8:30 pm
studentps2011 wrote:If we re-write the first statement (x2-5x = -6) as

(x2-4)/(x-2) = 5, it will have the same issue as statement 2.

So isn't it true that we have to consider both roots of statement 2 as valid?
To answer your question -- NO, we don't need to consider both roots of stmt 2. first of all you should not be getting 2 roots from statement 2.

This is stmt 2 -- (x^2-1)/(x+1) =1 or [(x-1)(x+1)]/(x+1) = 1 --------- (A^2-B^2) = (A+B)(A-B)
cancel out (x+1) from neumerator and denominator.. you will be left with -- (x-1) = 1
or, x =2. ONLY 1 ROOT.

SUFFICIENT

According to statement 1 --

X^2-5x+6 = 0
or (x-3) (x+2) = 0

Two different roots. X = 3, or -2

INSUFFICIENT.

ANSWER OPTION B
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by fcabanski » Thu Nov 03, 2011 8:40 pm
When dealing with data sufficiency problems, the correct answer has to lead to one and only one solution and you cannot assume additional information.

In the first statement the information given is that x^2-5x=-6. You can't assume that (x^2-4)/(x-2) = 5 because the statement does not contain that information.

x^2 - 5x = -6

x^2 - 5x + 6 = 0

(x+2)(x-3) = 0

x = -2
x = 3

Absent other information there are two solutions. Therefore Statement I is not sufficient.
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by GmatMathPro » Thu Nov 03, 2011 10:18 pm
studentps2011 wrote:What is the value of x2-1?

(1) x2-5x = -6
(2) (x2-1)/(x+1) = 1

(x2 is x squared)

OA is B, but I think it should be C. One question to the experts, is 0/0 = 1? Thanks.
0/0 does not equal 1 or anything else. Consider this: 12=6*2 and 12/2=6. 24=8*3 and 24/8=3. In general, if a*b=c, then a=c/b. Now, 0*1=0, so you might think 1=0/0. But 2*0=0, so we could just as easily say 2=0/0 or 5=0/0 or 98=0/0. There is no sensible way to attach a unique value to the expression 0/0. Hence, we say it is undefined.

When you have something like (x^2-1)/(x+1)=1, you have to assume right off the bat that x does not equal -1. Otherwise nothing makes any sense. Something over zero could never equal 1, and when you multiply both sides of the equation by (x+1), you're making the implicit assumption that x is not -1. Otherwise it's not a valid algebraic maneuver. So it doesn't make sense to do all these things that depend entirely on x not being negative one, and then at the end, saying, "hey! x equals -1!"

Above all remember that for a solution to be legitimate, you must be able to plug it in to the ORIGINAL equation and get a true statement.
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by vaibhavgupta » Fri Nov 04, 2011 2:04 am
studentps2011 wrote:What is the value of x2-1?

(1) x2-5x = -6
(2) (x2-1)/(x+1) = 1

(x2 is x squared)

OA is B, but I think it should be C. One question to the experts, is 0/0 = 1? Thanks.
take 2) (x-1)=1 so for x+1,

X-1+2= 1+2
X+1=3

Hence B it is!
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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