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:

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

Is |x-1| < 1?

Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Thanks, Just 1 question. can I solve the equation in 1. (x-1)^2<=1 by taking sq root on both sides? so the equaton becomes |x-1|<=1 or -1<=x-1<=1..

Since both parts of the inequality are nonnegative we can take square root and write as you did.

GENERAL RULES FOR THAT: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

Is |x-1| < 1?

Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunuel,

Is it not B?

x^2 - 1 > 0 --> x<-1 or x>1

x can be 2,3,4.... or -2,-3,-4...

So every time, |x-1| > 1.

Isn't this sufficient to answer that |x-1| will never be less than 1?

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

Is |x-1| < 1?

Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunuel,

Is it not B?

x^2 - 1 > 0 --> x<-1 or x>1

x can be 2,3,4.... or -2,-3,-4...

So every time, |x-1| > 1.

Isn't this sufficient to answer that |x-1| will never be less than 1?

We are not told that x is an integer. So, check for x=1.5.
_________________

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

Is |x-1| < 1?

Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunuel,

Is it not B?

x^2 - 1 > 0 --> x<-1 or x>1

x can be 2,3,4.... or -2,-3,-4...

So every time, |x-1| > 1.

Isn't this sufficient to answer that |x-1| will never be less than 1?[/quote]

We are not told that x is an integer. So, check for x=1.5.[/quote]

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

Is |x-1| < 1?

Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

Hope it's clear.

Hi, Could someone please explain how you arrived at (2) in the above statement. How is X^2 - 1>0 ---> X<-1 or X>1?? I am not understanding how you arrived at X<-1

Hi, Could someone please explain how you arrived at (2) in the above statement. How is X^2 - 1>0 ---> X<-1 or X>1?? I am not understanding how you arrived at X<-1

\(x^2-1>0, x^2>1\)

I have an old trick to solve those, basically you solve \(x^2=1, x=+-1\), now you look at the sign of x^2 and at the operator. If they are (+,>) or (-,<) you pick the external values : \(x>1 and x<-1\) You can insert numbers and you`ll see it s right. In the other two cases you pick the internal values, it would have been \(-1<x<1\)

Hope it helps
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Hi, Could someone please explain how you arrived at (2) in the above statement. How is X^2 - 1>0 ---> X<-1 or X>1?? I am not understanding how you arrived at X<-1

\(x^2-1>0, x^2>1\)

I have an old trick to solve those, basically you solve \(x^2=1, x=+-1\), now you look at the sign of x^2 and at the operator. If they are (+,>) or (-,<) you pick the external values : \(x>1 and x<-1\) You can insert numbers and you`ll see it s right. In the other two cases you pick the internal values, it would have been \(-1<x<1\)

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

A very good question. The question is asking you : Does x lie between 0 and 1.

From 1 ) Open the given condition and we have x(x-2) <=0. This says x lies between 0 and 2 including both end points. Not sufficient to answer the question. 2) this says x > 1 or x < -1 . Again Insufficient.

Combining 1 and 2 , we can't find the answer as x=2 , will prove the condition false and x = 1.5 will prove it to be true.

The stem tells us that 0 < x < 2 however #1 says that 0 ≤ x ≤ 2. In other words, the stem says x must be BETWEEN 0 and 2 while #1 says that x could be 0 or 2. INSUFFICIENT.

(2) x^2 - 1 > 0 x^2 - 1 > 0 x^2 > 1 |x| > 1

x>1 OR x<-1

The stem tells us that 0 < x< 2 but #2 says that x > 1 or x < -1 which means x could fall in the correct range but it might not also. INSUFFICIENT

1+2) is 0 < x < 2?

0 ≤ x ≤ 2 AND x>1 OR x<-1

Tells us that x is between zero and two inclusive and also that x > 1 which means x could = 2 which isn't allowed in the stem (0 < x < 2) INSUFFICIENT

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...