Events & Promotions
|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 3008:00 AM PDT
-11:59 PM PDT
Join us June 30th for 2 weeks of GMAT questions (just 8 per day) to help with your prep and to compete with your fellow GMAT Clubbers for a chance to win a prize fund of $30,000 for you and your team-mates!
Jul 0901:00 PM EDT
-02:00 PM EDT
More Target Test Prep students are earning 100th percentile GMAT scores. Don’t settle for less when the Target Test Prep course gives you everything you need to score high on the GMAT. Start your 5-day free trial today.
Jul 1206:30 AM PDT
-08:30 AM PDT
Join our webinar to master GMAT RC Main Point Questions! Learn effective strategies to decipher passage themes and purposes. Senior Verbal Expert- Sunita Singhvi will equip you with tools to navigate complexities and avoid common traps set by the GMAT.
Jul 1211:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
Jul 1311:00 AM EDT
-12:30 PM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Jul 1508:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Jul 1611:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
16 Aug 2016, 21:25
Kudos
Bookmarks
SJ007
Hi, I tried solving the problem by multiplying both sides by |x| since |x| will always be positive. This makes the equation as x<x|x|.
Case 1: x>0 => x<sq(x)=>x>1
Case 2:x<0=>x<-sq(x)=>x<-1
I believe I am going wrong in multiplying both sides by |x|, however, I fail to understand how since |x| will always be positive and hence I could multiply it.
Please assist.
Thanks
Notice that we are asked to find which of the options MUST be true, not COULD be true.
Let's see what ranges does \(\frac{x}{|x|}< x\) give for \(x\). Two cases:
If \(x<0\) then \(|x|=-x\), hence in this case we would have: \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that we consider the range \(x<0\), so \(-1<x<0\);
If \(x>0\) then \(|x|=x\), hence in this case we would have: \(\frac{x}{x}<x\) --> \(1<x\).
So, \(\frac{x}{|x|}< x\) means that \(-1<x<0\) or \(x>1\).
Only option which is ALWAYS true is B. ANY \(x\) from the range \(-1<x<0\) or \(x>1\) will definitely be more the \(-1\).
Answer: B.
As for other options:
A. \(x>1\). Not necessarily true since \(x\) could be -0.5;
C. \(|x|<1\) --> \(-1<x<1\). Not necessarily true since \(x\) could be 2;
D. \(|x|>1\) --> \(x<-1\) or \(x>1\). Not necessarily true since \(x\) could be -0.5;
E. \(-1<x<0\). Not necessarily true since \(x\) could be 2.
P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Please pay attention to the rules #3 and 6. Thank you.
Case 1: x>0 => x<sq(x)=>x>1
Case 2:x<0=>x<-sq(x)=>x<-1
I believe I am going wrong in multiplying both sides by |x|, however, I fail to understand how since |x| will always be positive and hence I could multiply it.
Please assist.
Thanks
Bunuel
Marcab
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?
(A) \(x>1\)
(B) \(x>-1\)
(C) \(|x|<1\)
(D) \(|x|>1\)
(E) \(-1<x<0\)
Explanations required for this one.
Not convinced at all with the OA.
My range is -1<x<0 and x>1.
(A) \(x>1\)
(B) \(x>-1\)
(C) \(|x|<1\)
(D) \(|x|>1\)
(E) \(-1<x<0\)
Explanations required for this one.
Not convinced at all with the OA.
My range is -1<x<0 and x>1.
Notice that we are asked to find which of the options MUST be true, not COULD be true.
Let's see what ranges does \(\frac{x}{|x|}< x\) give for \(x\). Two cases:
If \(x<0\) then \(|x|=-x\), hence in this case we would have: \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that we consider the range \(x<0\), so \(-1<x<0\);
If \(x>0\) then \(|x|=x\), hence in this case we would have: \(\frac{x}{x}<x\) --> \(1<x\).
So, \(\frac{x}{|x|}< x\) means that \(-1<x<0\) or \(x>1\).
Only option which is ALWAYS true is B. ANY \(x\) from the range \(-1<x<0\) or \(x>1\) will definitely be more the \(-1\).
Answer: B.
As for other options:
A. \(x>1\). Not necessarily true since \(x\) could be -0.5;
C. \(|x|<1\) --> \(-1<x<1\). Not necessarily true since \(x\) could be 2;
D. \(|x|>1\) --> \(x<-1\) or \(x>1\). Not necessarily true since \(x\) could be -0.5;
E. \(-1<x<0\). Not necessarily true since \(x\) could be 2.
P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Please pay attention to the rules #3 and 6. Thank you.
Here is the problem:
Case 2:x<0=>x<-sq(x)=>x<-1
If x is negative, |x| = -x
x < x|x| becomes x < -x^2
Now if you divide both sides by x, remember that x is negative so inequality sign will flip.
1 > -x
Multiply both sides by -1 and that will flip the sign again.
-1 < x
So you get that x should be greater than -1 in case x is negative.
So in any case, x will always be greater than -1.
20 Jun 2017, 07:32
Kudos
Bookmarks
Marcab
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?
(A) \(x>1\)
(B) \(x>-1\)
(C) \(|x|<1\)
(D) \(|x|>1\)
(E) \(-1<x<0\)
(A) \(x>1\)
(B) \(x>-1\)
(C) \(|x|<1\)
(D) \(|x|>1\)
(E) \(-1<x<0\)
We can simplify the given inequality:
x/|x| < x
x < (x)|x|
(x)|x| > x
If x is positive, we can divide both sides by x and obtain |x| > 1.
If x is negative, we can also divide both sides by x, but we have to switch the inequality sign, so we have |x| < 1.
We see that if x is positive, |x| > 1, which is choice C, and if x is negative, |x| < 1, which is choice D. However, since we don’t know whether x is positive or negative, both choice C and choice D “can be true,” not “must be true.”
Let’s analyze further. If x is positive, |x| = x. So, |x| > 1 means x > 1, which is choice A. If x is negative, |x| = -x. So, |x| < 1 means -x < 1 or x > -1. However, because x is negative, we have -1 < x < 0, which is choice E. Again, since we don’t know whether x is positive or negative, both choice A and choice E “can be true,” not “must be true.”
This leaves choice B as the correct answer. In fact, it’s the correct choice because the inequality x > -1 includes both x > 1 and -1 < x < 0. So, regardless of whether x is positive or negative, we can say x > -1.
Answer: B
himanshukamra2711
10 Aug 2017, 22:25
Kudos
Bookmarks
Bunuel
Marcab
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?
(A) \(x>1\)
(B) \(x>-1\)
(C) \(|x|<1\)
(D) \(|x|>1\)
(E) \(-1<x<0\)
Explanations required for this one.
Not convinced at all with the OA.
My range is -1<x<0 and x>1.
(A) \(x>1\)
(B) \(x>-1\)
(C) \(|x|<1\)
(D) \(|x|>1\)
(E) \(-1<x<0\)
Explanations required for this one.
Not convinced at all with the OA.
My range is -1<x<0 and x>1.
Notice that we are asked to find which of the options MUST be true, not COULD be true.
Let's see what ranges does \(\frac{x}{|x|}< x\) give for \(x\). Two cases:
If \(x<0\) then \(|x|=-x\), hence in this case we would have: \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that we consider the range \(x<0\), so \(-1<x<0\);
If \(x>0\) then \(|x|=x\), hence in this case we would have: \(\frac{x}{x}<x\) --> \(1<x\).
So, \(\frac{x}{|x|}< x\) means that \(-1<x<0\) or \(x>1\).
Only option which is ALWAYS true is B. ANY \(x\) from the range \(-1<x<0\) or \(x>1\) will definitely be more the \(-1\).
Answer: B.
As for other options:
A. \(x>1\). Not necessarily true since \(x\) could be -0.5;
C. \(|x|<1\) --> \(-1<x<1\). Not necessarily true since \(x\) could be 2;
D. \(|x|>1\) --> \(x<-1\) or \(x>1\). Not necessarily true since \(x\) could be -0.5;
E. \(-1<x<0\). Not necessarily true since \(x\) could be 2.
P.S. Please read carefully and follow: https://gmatclub.com/forum/rules-for-pos ... 33935.html Please pay attention to the rules #3 and 6. Thank you.
If in case I multiply |x| both sides then inequality will not change and then if i approach like this then how the inequality should be solved :-
X<X|X|
x|x|-x>0
x(|x|-1)>0
now either x>0 or |x|>1,x<0 |x|<1
how to proceed further to solve it to get the range as per the question.
Since it's a 
