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:
x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1
lets suppose x be 0.5 which is more than -1. 0.5/l0.5l=1 which is not less than 0.5.
ya but in no case x can be < -1.....it has to be > -1....but not bet 0 and 1.....x > 1 is wrong.....as the eqn also satisfies when x = -1/2
baner, unclear to me.... pls be specific.
Let's imagine a number line with points -1, 0 and 1.....as u can see x can't be <-1...e.g. x = -2......-1 < -2....will not satisfy....
Now let's see -1<x<0......x = -1/2.....-1 < -1/2.....satisfies
Now for 0<x<1.....x = 1/2.....1 < 1/2.....will not satisfy....
For x > 1....x = 2.....1 < 2....satisfies
As u see the soln lies in the following categories:
-1<x<0
x > 1
In all case x > -1 even tho it can't be 0<x<1.....altho I think the ques is lil weird.
x/|x|<x if x>0, then 1/x<1 ie, x>1 if x<0, then 1/(-x)>1, ie, x>-1
So combine these two, we know that x must be greater than -1.
baner is right, (B).
i think we need AkamaiBrah.
if so, then why 0<x>1 does not fit in the inequality x/|x|<x? i mean why all the values for x>-1 donot satisfy the inequality x/|x|<x? pls do explain........................
Last edited by MA on 06 Feb 2005, 21:50, edited 1 time in total.
It doesn't say that all x greater than -1 would satisfy the equation. We get two range of x that satisfies the equation (-1,0) and (1,infinite). All that the answer says is that every x in our solution ranges are greater than -1, which is true.
I believe we're debating solution B, which says x>-1 must be always true.
I can pick positive 1 as an integer that is greater than -1, but does not satisfy x/|x|<x.
If x = +1,
x/|x| = 1 which is not < 1. (1=1)
Now, if there's just one value of x > -1 that does not satisfy the inequality, then we can't say x>-1 is always true.
If you look at my previous post, I've not said I'm solving for x, i'm just stating values of x that hold true for the inequality. I came up with x<-1, and x>1.
Let me give you another example. Lets say we have two sets of integers.
S1=(0), S2=(2,3,4)
Now we know that x belongs to S1 or S2. Ask what must be true for x:
(A) x>1
(B) x>-1
(A) is obviously wrong
Because x could be an element of S1 or S2. If x belongs to S1, then it would be 0, and it is NOT greater than 1.
(B) is, however, correct
No matter which set x belongs, x is greater than -1.
You cannot say that well what about x=1? It satisfies x>-1, but is not in either S1 or S2. The mistake you are having here, is instead of A->B, you are trying to say B->A.
Let me give you another example. Lets say we have two sets of integers. S1=(0), S2=(2,3,4) Now we know that x belongs to S1 or S2. Ask what must be true for x: (A) x>1 (B) x>-1
(A) is obviously wrong Because x could be an element of S1 or S2. If x belongs to S1, then it would be 0, and it is NOT greater than 1. (B) is, however, correct No matter which set x belongs, x is greater than -1. You cannot say that well what about x=1? It satisfies x>-1, but is not in either S1 or S2. The mistake you are having here, is instead of A->B, you are trying to say B->A.
Glad to see that I am not the only one in minority here, I am sticking with my ans "B" till we get the OA.
My example is completely parallel to the question, only in the integer and limited form.
Look:
HongHu wrote:
(A) cannot be correct, people.
Let me give you another example. Lets say we have two sets of integers. S1=(0), S2=(2,3,4) Now we know that x belongs to S1 or S2. Ask what must be true for x: (A) x>1 (B) x>-1
Compare it with:
S1=(-1,0), S2=(1,infinity)
We know x belongs to S1 or S2. What must be true for x?
x/|x|<x, which of the following must be true about x ?
A. x>1 B. x>-1 C. |x|<1 D. |x|=1 E. |x|^2>1
We are given that \(\frac{x}{|x|}<x\) (this is a true inequality), so first of all we should find the ranges of \(x\) for which this inequality holds true.
\(\frac{x}{|x|}< x\) multiply both sides of inequality by \(|x|\) (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too) --> \(x<x|x|\) --> \(x(|x|-1)>0\): Either \(x>0\) and \(|x|-1>0\), so \(x>1\) or \(x<-1\) --> \(x>1\); Or \(x<0\) and \(|x|-1<0\), so \(-1<x<1\) --> \(-1<x<0\).
So we have that: \(-1<x<0\) or \(x>1\). Note \(x\) is ONLY from these ranges.
Option B says: \(x>-1\) --> ANY \(x\) from above two ranges would be more than -1, so B is always true.
Answer: B.
nonameee wrote:
Quote:
x/|x|<x. which of the following must be true about x ?
a) x>1 b) x>-1 c) |x|<1 d) |x|=1 e) |x|^2>1
Bunuel, I think there's a mistake in the question or in the answer choices:
Here's my solution: 1) x<0: -x/x < x -1<x<0
2) x>=0 x/x<x x>1
The solution of the inequality is then: -1<x<0 union x>1
The answer can't be B, since let x=1/2. We get: (1/2)/(1/2) < (1/2) 1< 1/2 Contradiction.
I think the answer should be A since it satisfies all the scenarios.
Can you please clarify?
The options are not supposed to be the solutions of inequality \(\frac{x}{|x|}<x\).
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...