Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

(1) \((x - 1)^2 \le 1\). Since both sides of the inequality are non-negative then we can take square root from both parts: \(|x-1| \le 1\), 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 \(|x-1| \le 1\), means \(0 \le x \le 2\).

(2) \(x^2 - 1 \gt 0\). Rearrange: \(x^2 \gt 1\). Again, since both sides of the inequality are non-negative then we can take square root from both parts: \(|x| \gt 1\). 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.

Is |x−1|<1 ? (1) (x−1)^2≤1 (2) x^2−1>0 First,though it sounds absurd, try to test the condition whether sufficient or not even without using (1) or (2),. Place 5, 4<1, answer is no. Again, place -2, 3<1..still no. Now, place 2,1<1..still no.Now place 1, now we get the answer yes for 0<1. again, place 0, we get 1<1, no. So, once yes, once no suggests we need more data! Now take (1) whether we get constant yes or constant no. By rooting, we get |x−1|≤1, means 0≤x≤2. Now we have restricted freedom to choose value. We can't choose any value less than 0 or more than 2. Lets try within this boundary. Place 0 or 2, we get 1 but is 1 less than 1?,NO. Try 1, |x−1|=0 (<1), YES,now less than 1. So, insufficient because of both outcome. We need either yes or no, not both simultaneously. Think solely on (2) now.x^2−1>0;|x|>1. so, we have to take such values either greater than 1 or less than -1. Any less than -1 value makes |x−1| more than 1, so NO. In addition, placing any value 2 or more than 2 gives us same NO. Whatif, we try placing values within 0 to 1;say,1.5. we get |x−1|=.5(<1),so YES. Once yes,once no suggests insufficient.[if x were integer, (2) would be sufficient alone] Now combine (1) and (2). (1) restricts the freedom to choose within 0≤x≤2. (2) restricts it by |x|>1.Even after combining, nothing changes.Still we can put 1<x≤2. Try putting 2, answer NO. Try putting 1.5, YES. So, combining still fails to give a reliable or constant answer.

Hi - For this question, Stmt1 solutions assumed inequality is less than equal to. Hence solutions states 0<= x <= 2 but it should be 0 < x < 2 which will make stmt 1 Sufficient. Please confirm?

Hi - For this question, Stmt1 solutions assumed inequality is less than equal to. Hence solutions states 0<= x <= 2 but it should be 0 < x < 2 which will make stmt 1 Sufficient. Please confirm?

Why should it be 0 < x < 2 ? Both 0 and 2 satisfy \((x−1)^2≤1\).
_________________

Hi - For this question, Stmt1 solutions assumed inequality is less than equal to. Hence solutions states 0<= x <= 2 but it should be 0 < x < 2 which will make stmt 1 Sufficient. Please confirm?

Why should it be 0 < x < 2 ? Both 0 and 2 satisfy \((x−1)^2≤1\).

Sorry should have clarified. I was referring to question stem (following part from the solution)

Hi - For this question, Stmt1 solutions assumed inequality is less than equal to. Hence solutions states 0<= x <= 2 but it should be 0 < x < 2 which will make stmt 1 Sufficient. Please confirm?

Why should it be 0 < x < 2 ? Both 0 and 2 satisfy \((x−1)^2≤1\).

Sorry should have clarified. I was referring to question stem (following part from the solution)

Notice that |x−1|≤1, means 0≤x≤2.

But question has |x−1|<1 only

Please read the solution carefully:

Statement 1 - \((x−1)^2≤1\) is equivalent to \(|x−1|≤1\), which translates to \(0≤x≤2\).
_________________

Am I correct, that if statement 1) mentions \((x-1)^2<1\) instead of \((x-1)^2≤1\) that both statements together would be sufficient?

In this case the first statement itself would be sufficient because \((x-1)^2<1\) is the same as \(|x-1|<1\) (by taking the square root).
_________________

why the negative value of x in case of 2nd statement i.e x < -1 not considered while solving? for eg wen x > 1, it is not sufficient. but when x< -1, in every case the eq exceeds 1 and therefore it is sufficient to solve. By this logic optn 2 shud be sufficient right?

why the negative value of x in case of 2nd statement i.e x < -1 not considered while solving? for eg wen x > 1, it is not sufficient. but when x< -1, in every case the eq exceeds 1 and therefore it is sufficient to solve. By this logic optn 2 shud be sufficient right?

In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".

We can get both an YES and a NO answers for the second statement (even not considering negative numbers) hence the statement is not sufficient.
_________________