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.
May 1212:00 PM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 1308:30 AM PDT
-09:30 AM PDT
In Episode 4 of our GMAT Ninja CR series, we tackle the most intimidating CR question type: Boldface & "Legalese" questions. If you've ever stared at an answer choice that reads, "The first is a consideration introduced to counter a position that...- May 14
08:30 AM PDT
-09:30 AM PDT
Most GMAT test-takers are intimidated by the hardest GMAT Verbal questions. In this session, Target Test Prep GMAT instructor Erika Tyler-John, a 100th percentile GMAT scorer, will show you how top scorers break down challenging Verbal questions..
26 Apr 2023, 03:51
Kudos
Bookmarks
If \(x + y ≠ 0\), is \(\frac{1}{x + y} < 1\)?
(1) \(x^2 = y^2\)
(2) \(\frac{1}{x} < 2\)
(1) \(x^2 = y^2\)
(2) \(\frac{1}{x} < 2\)
26 Apr 2023, 03:52
Kudos
Bookmarks
Official Solution:
If \(x + y ≠ 0\), is \(\frac{1}{x + y} < 1\)?
(1) \(x^2 = y^2\)
\(x^2 - y^2 = 0\);
\((x - y)(x + y) = 0\).
Given that \(x + y ≠ 0\), we deduce that \(x - y = 0\), which is equivalent to \(x = y\).
The question now becomes:
Is \(\frac{1}{x + y} < 1\)?
Is \(\frac{1}{2x} < 1?\)
Is \(\frac{1}{x} < 2\)?
We do not have enough information to answer this question. Hence, statement (1) is insufficient.
(2) \(\frac{1}{x} < 2\).
This statement provides no information about \(y\), so we cannot determine whether \(\frac{1}{x + y} < 1\). Insufficient.
(1)+(2) From (1), the question was simplified to "Is \(\frac{1}{x} < 2\)?" and (2) directly provides a YES answer to that. Sufficient
Answer: C
If \(x + y ≠ 0\), is \(\frac{1}{x + y} < 1\)?
(1) \(x^2 = y^2\)
\(x^2 - y^2 = 0\);
\((x - y)(x + y) = 0\).
Given that \(x + y ≠ 0\), we deduce that \(x - y = 0\), which is equivalent to \(x = y\).
The question now becomes:
Is \(\frac{1}{x + y} < 1\)?
Is \(\frac{1}{2x} < 1?\)
Is \(\frac{1}{x} < 2\)?
We do not have enough information to answer this question. Hence, statement (1) is insufficient.
(2) \(\frac{1}{x} < 2\).
This statement provides no information about \(y\), so we cannot determine whether \(\frac{1}{x + y} < 1\). Insufficient.
(1)+(2) From (1), the question was simplified to "Is \(\frac{1}{x} < 2\)?" and (2) directly provides a YES answer to that. Sufficient
Answer: C
