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From stmt2: (x-1)(x+1) > 0. This means, either (x-1)>0 or, (x-1) < 0 or, |x-1| > 0 and is no additional information (as modulus will always be greater than 0).

Re: m08 q11 IMO OA need to be be B) and not E) [#permalink]

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09 Dec 2008, 22:57

5

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Hi Cramya, this is one of the most confusing qs in GMAT, the absolute DS ones. Here's what i would have done to solve it.

First lets look at the qs statement and simplify it Is |X-1| < 1? |x-1| = -(x-1) or x-1 solving both -x+1 < 1 , 1-1 < x , 0 < x or x >0 x-1 < 1 , x < 2

so Is 0 < x < 2 ? is the actual qs .

Now lets look at the statements 1) (x-1) ^ 2 <= 1 x-1 <= +- 1 x <= 0 , x<= 2 So not sufficient

2) x^2 - 1 > 0 x^2 > 1 , x > +-1 , -1 < x < 1 Here x is < 2 so that satisfies one side of the qs, but x can be between -1 and 0 , so that fails the x > 0 part. Hence this is not sufficient

Together , they dont do much to satisfy the qs above, hence the answer should be E.

I guess the way you are trying to solve the statements is different, so i hope you got what i am trying to do here.

Correct me someone if they feel i am wrong here. I am still getting used to solving absolutes as well ... It will be good to know if i have reached the correct answer using incorrect methods afterall.

Re: m08 q11 IMO OA need to be be B) and not E) [#permalink]

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28 Dec 2008, 00:22

gamecode,

Looking at your statement:-

2) x^2 - 1 > 0 x^2 > 1 , x > +-1 , -1 < x < 1 Here x is < 2 so that satisfies one side of the qs, but x can be between -1 and 0 , so that fails the x > 0 part. Hence this is not sufficient

I always solve the stem first.In this we solve the two equations: x-1<1 and -(x-1)<1 Solving we get, x<2,x>0

QUERIE:Does the above two values represent: 1.Is the value of X<2 OR x>0 ?? 2.The actual question is whther 0<x<2 ??

Solving stmt 1: (x-1) ^ 2 <= 1 x-1 <= +- 1 x <= 0 , x<= 2 ..Hmm..is'nt this SUFF??It's telling us what the ques. stem is asking. Confused!!

You are forgetting the equality sign on statement 1.

Question asks if |x-1|<1, which is the same as asking if 0<x<2.

Stat1 gives 0<=x<=2. If x<2 the answer is yes; if x=2 the answer is no => not sufficient. Stat2 gives x<-1 or x>1 Combined statements give 1<x<=2. Still not sufficient.
_________________

I always solve the stem first.In this we solve the two equations: x-1<1 and -(x-1)<1 Solving we get, x<2,x>0

QUERIE:Does the above two values represent: 1.Is the value of X<2 OR x>0 ?? 2.The actual question is whther 0<x<2 ??

Solving stmt 1: (x-1) ^ 2 <= 1 x-1 <= +- 1 x <= 0 , x<= 2 ..Hmm..is'nt this SUFF??It's telling us what the ques. stem is asking. Confused!!

You are forgetting the equality sign on statement 1.

Question asks if |x-1|<1, which is the same as asking if 0<x<2.

Stat1 gives 0<=x<=2. If x<2 the answer is yes; if x=2 the answer is no => not sufficient. Stat2 gives x<-1 or x>1 Combined statements give 1<x<=2. Still not sufficient.

Totally agree, with you. The pitfall is here the difference in meaning of signs "less than" and "Less than or equal." I missed this part and thought the answer was A because the first statement repeats the stem, well kind of. So the main trick lies here. Well, what can I say, you have to be really careful.

1. not sufficient. If \((x - 1)^2 \le 1\), then \(-1< x <= 2\); 2. sufficient. -1 > x > 1

Ans: B

\(|z|<k\). It means, \(-k<z<k\) \(|z| \le k\). It means, \(-k \le z \le k\) \(|z|>k\). It means, \(z>k \hspace{3} OR \hspace{3} z<-k\) \(|z| \ge k\). It means, \(z \ge k \hspace{3} OR \hspace{3} z \le -k\) \(\sqrt{z^2}=|z|\)
_________________

"You can do it if you believe you can!" - Napoleon Hill "Insanity: doing the same thing over and over again and expecting different results." - Albert Einstein

There is no need to find the ranges for each inequality to solve this question.

(1 \((x - 1)^2\leq1\) --> since both sides of the inequality are non-negative then we can take square root from both parts: \(|x-1|\leq{1}\), so \(|x-1|\) can be less than 1 (answer YES), as well as equal to 1, for \(x=2\) or \(x=0\) (answer NO). Not sufficient. Notice that \(|x-1|\leq{1}\), means \(0\leq{x}\leq{2}\).

(2) \(x^2 - 1>0\) --> \(x^2>1\) --> again, since both sides of the inequality are non-negative then we can take square root from both parts: \(|x|>1\). If \(x=1.5\) then the answer is YES but if \(x=2\) then the answer is NO. Not sufficient.

(1)+(2) \(x=1.5\) and \(x=2\) satisfy both statements and give different answer to the question. Not sufficient.

(2) \(x^2 - 1>0\) --> \(x^2>1\) --> again, since both sides of the inequality are non-negative then we can take square root from both parts: \(|x|>1\). If \(x=1.5\) then the answer is YES but if \(x=2\) then the answer is NO. Not sufficient.

Did you mean 0.5 instead of 1.5 or did I miss something?
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(2) \(x^2 - 1>0\) --> \(x^2>1\) --> again, since both sides of the inequality are non-negative then we can take square root from both parts: \(|x|>1\). If \(x=1.5\) then the answer is YES but if \(x=2\) then the answer is NO. Not sufficient.

Did you mean 0.5 instead of 1.5 or did I miss something?

No, everything is correct there: if \(x=1.5\) then \(1.5^2>1\) (so the second statement is satisfied) and \(|1.5-1|=0.5<1\) (so the answer to the original question is YES).