|
|
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.
11 Feb 2023, 03:06
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
58% (02:29) correct
42%
(02:30)
wrong
based on 192
sessions
History
Date
Time
Result
Not Attempted Yet
If \(|x| ≠ 1\), is \(\frac{x}{x + 1} > \frac{-x}{1 - x}\) ?
(1) \(x > 1\)
(2) \(\frac{x}{x + 1} < 1\)
(1) \(x > 1\)
(2) \(\frac{x}{x + 1} < 1\)
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
11 Feb 2023, 12:33
Kudos
Bookmarks
Is \(\frac{x}{(x+1)}>−\frac{x}{(1−x) }\)?
|x|≠1 implies x ≠ 1 or -1
(1) x > 1
We can check the above inequality for different values of x(say) = 1.5 or 2 or 10.
x = 1.5 . Inequality is \(\frac{1.5}{2.5} > \frac{-1.5}{(-0.5)}\) i.e., 0.6 > 3 which is not true. Similarly we can prove that for any value of x >1, the above inequality is false.
Hence statement 1 is sufficient
(2) \(\frac{x}{(x+1)}\)<1
this inequality is true for all values of x>-1
For ex - \(\frac{-0.5}{0.5 }< 1\) is true; 0/1 < 1 is true ; \(\frac{5}{6 }< 1\) is true.
we have already seen that for x >1 our main inequality is not satisfied. Now lets check for values of x between -1 and 1
x = -0.5
\(\frac{-0.5}{(0.5) }> \frac{-(-0.5)}{[1- (- 0.5)]}\)
= -1 > \(\frac{1}{3 }-\) Not true
x = 0.4
\(\frac{0.4}{1.4} > \frac{-0.4 }{ 0.6 }\)
\(\frac{2}{7}> - 2/3\) - True
therefore statement 2 is Not sufficient
Answer A
|x|≠1 implies x ≠ 1 or -1
(1) x > 1
We can check the above inequality for different values of x(say) = 1.5 or 2 or 10.
x = 1.5 . Inequality is \(\frac{1.5}{2.5} > \frac{-1.5}{(-0.5)}\) i.e., 0.6 > 3 which is not true. Similarly we can prove that for any value of x >1, the above inequality is false.
Hence statement 1 is sufficient
(2) \(\frac{x}{(x+1)}\)<1
this inequality is true for all values of x>-1
For ex - \(\frac{-0.5}{0.5 }< 1\) is true; 0/1 < 1 is true ; \(\frac{5}{6 }< 1\) is true.
we have already seen that for x >1 our main inequality is not satisfied. Now lets check for values of x between -1 and 1
x = -0.5
\(\frac{-0.5}{(0.5) }> \frac{-(-0.5)}{[1- (- 0.5)]}\)
= -1 > \(\frac{1}{3 }-\) Not true
x = 0.4
\(\frac{0.4}{1.4} > \frac{-0.4 }{ 0.6 }\)
\(\frac{2}{7}> - 2/3\) - True
therefore statement 2 is Not sufficient
Answer A
12 Feb 2023, 05:02
Kudos
Bookmarks
Bunuel
\(\frac{x}{x + 1} > \frac{-x}{1 - x}\)
\(\frac{x}{x + 1} > \frac{-x}{-(x+1)}\)
\(\frac{x}{x + 1} > \frac{x}{(x+1)}\)
This is true for all values of x except for x = 0.
Hence, we can translate the question to "Is x = 0?"
Statement 1
(1) \(x > 1\)
\(x \neq 0,\) hence the statement is sufficient to answer the question.
Sufficient.
Statement 2
(2) \(\frac{x}{x + 1} < 1\)
\(\frac{x}{x + 1} < 1\)
\(\frac{x}{x + 1} - 1 < 0\)
\(\frac{x - x - 1}{x + 1} < 0\)
\(\frac{- 1}{x + 1} < 0\)
Therefore x + 1 > 0
x > -1.
x can be 0 or it can be any other number. Hence, the statement is not sufficient.
Option A
