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.
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:
Join IIMU Director to gain an understanding of DEM program, its curriculum & about the career prospects through a Q&A chat session. Dec 11th at 8 PM IST and 6:30 PST
Enter The Economist GMAT Tutor’s Brightest Minds competition – it’s completely free! All you have to do is take our online GMAT simulation test and put your mind to the test. Are you ready? This competition closes on December 13th.
Attend a Veritas Prep GMAT Class for Free. With free trial classes you can work with a 99th percentile expert free of charge. Learn valuable strategies and find your new favorite instructor; click for a list of upcoming dates and teachers.
Does GMAT RC seem like an uphill battle? e-GMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days.
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
18 Jun 2012, 16:15
57
40
Good question. +1.
Is x> 1?
(1) (x+1) (|x| - 1) > 0. Consider two cases:
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: \(|5|=5\).
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
19 Jun 2012, 12:22
Bunuel wrote:
Good question. +1.
Is x> 1?
(1) (x+1) (|x| - 1) > 0. Consider two cases:
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?
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
20 Jun 2012, 01:07
kashishh wrote:
Bunuel wrote:
Good question. +1.
Is x> 1?
(1) (x+1) (|x| - 1) > 0. Consider two cases:
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.
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
21 Jun 2012, 02:41
pavanpuneet wrote:
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.
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
27 Jan 2013, 23:17
2
fozzzy wrote:
Is x>1
1) (x+1)(lxl-1) > 0 2) lxl < 5
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.
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
06 Feb 2013, 03:21
andrew40 wrote:
Bunuel wrote:
Good question. +1.
Is x> 1?
(1) (x+1) (|x| - 1) > 0. Consider two cases:
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
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
18 May 2013, 05:56
Bunuel wrote:
Good question. +1.
Is x> 1?
(1) (x+1) (|x| - 1) > 0. Consider two cases:
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?
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
23 May 2013, 02:14
smartyman wrote:
Please help,
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).
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
02 Jun 2013, 10:59
Hi guys,
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
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
23 Jun 2013, 17:59
So here is my question:
For #1 we have two cases: positive and negative
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
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
24 Jun 2013, 02:11
1
WholeLottaLove wrote:
So here is my question:
For #1 we have two cases: positive and negative
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
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
28 Aug 2013, 13:08
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.
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
29 Aug 2013, 04:34
ygdrasil24 wrote:
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.
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
Show Tags
04 Apr 2014, 22:15
enigma123 wrote:
Is x > 1?
(1) (x+1) (|x| - 1) > 0
(2) |x| < 5
Sol: (x+1) (|x| - 1) > 0
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
final sol is A
gmatclubot
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5
[#permalink]
04 Apr 2014, 22:15
close
61% (01:53) correct 
