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:

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\).

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Dear Bunuel, i got (x+1)^2<0 . and further solved to x+1<0 giving x <-1 and ended up in wrong answer E. square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Dear Bunuel, i got (x+1)^2<0 . and further solved to x+1<0 giving x <-1 and ended up in wrong answer E. square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?

It doesn't matter that \(x\) is a variable, it's still some number and so is \(x+1\). So, \((x+1)^2\) is a square of that number and it cannot be negative.

Also your way of solving is not correct: \((x+1)^2<0\) does not mean \(x+1<0\) it means that \(|x+1|<0\). From that you could deduce the same: since absolute value cannot be negative then this equation has no solution.

Hi Bunuel, I understood your solution but can you please tell me where am I going wrong?

When I see the expression, (x+1)(|x|-1)>0, I immediately think that these two brackets must be either positive or negative.

Hence, if that take both of them are positive, then x>-1 and and x>1 & x<-1

For these two ranges we realize when we start plugging in number that only for x>1 the equation hold true.--------(1)

Similarly, if we take both of them to be negative, then x<-1 and x<1 & x>-1

for x<-1 for example, -2, one of the brackets turn out to be positive, which does not satisfy our initial assumption that both must be negative.

for x<1 ex 0, one bracket is +ve and one is -ve <Not Desired>; for 0, the inequality is not > 0 <not desired>

if we go still less it follows the first case

for x>-1 pretty much follows like the above. Hence nothing desired.--------(2)

Hence, x>1

A is sufficient.

First of all I wouldn't recommend to solve this question the way you are doing.

Next, when you consider both multiples to be negative and get \(x<-1\) from the first one, then the second multiple automatically transformes to \((-x-1)\), since if \(x<-1<0\) then \(|x|=-x\). So, we have that \(-x-1<0\) must also be true or \(x>-1\), which contradicts the case for the first multiple (\(x<-1\)). So, both \(x+1\) and \(|x|-1\) can not be negative.

Also I think you got x<1 & x>-1 from |x|<1, and if yes, then it's not correct: \(|x|<1\) means that \(-1<x<1\). So, again \(x<-1\) (for the first multiple to be negative) and \(-1<x<1\) (for the second multiple to be negative) cannot simultaneously be true.

Statement 1: For (x+1)(lxl-1) > 0, we should have either (x+1)>0 and (lxl-1) > 0 or (x+1)<0 and (lxl-1) < 0 when (x+1)>0 and (lxl-1) > 0 (x+1)>0 => x>-1 (lxl-1) > 0 => x>1 or x <-1 From above two, possible solution is x>1 when (x+1)>0 and (lxl-1) < 0 (x+1)<0 => x<-1 (lxl-1) < 0 => -1<x<1 Both of these can not be satisfied by any value of x. Hence we get only 1 solution, x>1. which is what we wanted to ascertain. Sufficient.

Statement 2: |x| <5 => -5<x<5 Clearly not sufficient to tell whether x>1 or not.

so you just have to find the common region if its done on a number line and we ignore one of the cases, since there isn't a common region? _________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Last edited by fozzzy on 28 Jan 2013, 00:25, edited 1 time in total.

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Sorry, I don't understand why we should consider the range where x>0. because of the absolute value?

We need to get rid of the modulus in the expression to solve it and this is the way to do that. Check here: is-x-134652.html#p1097668

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Hi,

This is my first post so was little conscious to ask my doubt. In the above question, we took the roots as 1<x>1. However, in this question as mentioned below (unable to post the link as per new member rule)

is ((X-3)^2)^1/2 = 3-X ?

1) X does not = 3

2) -X|X| > 0

the roots are 3<x>3. Can you please explain the difference?

When I try this question for statement 1: (x+1) (|x| - 1) > 0 (x+1)|x|-(x+1)>0 (x+1)|x| >(x+1) |x| > 1

x>1 ; x<-1

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

So you cannot reduce both parts of inequality (x+1)|x|>(x+1) by x+1 as you don't know the sign of x+1: if x+1>0 you should write |x|>1 BUT if x+1<0 you should write |x|<1 (flip the sign).

I just want to present an easier way to prove that I is sufficient.

For I to be positive what are the conditions?

1. x+1 > 0 this means x >-1 2.|x| -1 > 0 this means x > 1 or x < -1

then you must draw the number line ----------------[b]-1[b]-----0------1--------------> then draw this inequalities on this line and look for any unity. then you will fine x > 1

for the other side, I mean, x+1 <0 and |x| - 1 <0 you wont find any unity. so I is sufficient. Hope it clears. cheers

For x≥0 (x+1)(|x|-1)>0 (x+1)(x-1)>0 x^2-1>0 x^2>1 Of course, x could be 2 or negative two for all we know, so it seems like this is insufficient as we are testing for the range of x>1

For x≤0 (x+1)(|x|-1)>0 (x+1)(-x-1)>0 -x^2-1>0 -x^2>-1

Of course, x could be 2 or negative two for all we know, so it seems like this is insufficient as we are testing for the range of x>1

x can never be -2, as because you have assumed that \(x\geq0\). Also, from the inequality you have correctly arrived at, i.e. \(x^2>1\) \(\to\) x>1 OR x<-1. As assumption was \(x\geq0\). thus only x>1 condition is valid. Also,as x>1 automatically makes \(x\geq0\), thus the correct range is x>1. Sufficient.

Quote:

For x≤0 No need to include equality with zero twice. (x+1)(|x|-1)>0 (x+1)(-x-1)>0 This leads to \(-(x+1)^2>0\) and this is not possible for any real value of x. So,there is no solution for this. -x^2-1>0 -x^2>-1

I mean to say, we define |X| as +x when x is greater than equal to zero, and -x when x is less than zero. But I have noticed you at times have taken x as -x based on x less than equal to zero ? I mean equality cant be on both sides right ? Infact I saw you explained definition of mod function keeping equality on both +x and - x, I hope you are able to make out what doubt I have.

I mean to say, we define |X| as +x when x is greater than equal to zero, and -x when x is less than zero. But I have noticed you at times have taken x as -x based on x less than equal to zero ? I mean equality cant be on both sides right ? Infact I saw you explained definition of mod function keeping equality on both +x and - x, I hope you are able to make out what doubt I have.

The point is that |0|=-0=0. So, in that definition we can include = sign in both cases.

The RHS can only be positive if both are positive or both are negative lets take -5 as value of x===> -ve *+ve ===>not possible let's take 0 as value of x ==> 1* (-1) ===>not possible let's take 1 as value (+ve) ===> 2* (0)===> not possible let's take -1 as value (-ve) ====>0 *0 ====>not possible so x != -ve, x != 0, x != 1 conditions apply. only remaining option is x>1 ....sufficient

2> ----(-5)@@@@@@(5)--------- -5<x<5 ...not sufficient since x is less than 1 here

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...