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 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 12
07:30 AM PDT
-08:30 AM PDT
Join the Live and learn more about Picking the Right Schools 
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 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.. - Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
10 Nov 2021, 03:07
Kudos
Bookmarks
Is \(k\) an even integer?
(1) \(5k + 2\) is an even integer.
(2) \(k^2 + 1\) is an odd integer.
(1) \(5k + 2\) is an even integer.
(2) \(k^2 + 1\) is an odd integer.
10 Nov 2021, 03:07
Kudos
Bookmarks
Official Solution:
Is \(k\) an even integer?
(1) \(5k + 2\) is an even integer.
If \(k\) is an integer, then from \(5k + 2=even\) it follows that \(5k=even\) and in this case \(k\) must be even.
If \(k\) is NOT an integer, then from \(5k + 2=even\) it follows that \(k =\frac{even}{5}\) and in this case \(k\) must be some fraction like: \(..., \ -\frac{4}{5}, \ -\frac{2}{5}; \ \frac{2}{5},\ \frac{4}{5}, ...\), so NOT an even intger.
Not sufficient.
(2) \(k^2 + 1\) is an odd integer.
If \(k\) is an integer, then from \(k^2+1=odd\) it follows that \(k^2=even\) and in this case \(k\) must be even.
If \(k\) is NOT an integer, then from \(k^2+1=odd\) it follows that \(k =\sqrt{even}\) and in this case \(k\) must some irrational number like: \(..., \ -\sqrt{6}, \ -\sqrt{2}; \ \sqrt{2},\ \sqrt{6}, ...\), so NOT an even intger.
Not sufficient.
(1)+(2) No irrational number satisfices (1): \(5k + 2=5*irrational + integer=irrational +integer=irrational\). So, we are left with the first case from (2): \(k\) is an even integer. Sufficient.
Answer: C
Is \(k\) an even integer?
(1) \(5k + 2\) is an even integer.
If \(k\) is an integer, then from \(5k + 2=even\) it follows that \(5k=even\) and in this case \(k\) must be even.
If \(k\) is NOT an integer, then from \(5k + 2=even\) it follows that \(k =\frac{even}{5}\) and in this case \(k\) must be some fraction like: \(..., \ -\frac{4}{5}, \ -\frac{2}{5}; \ \frac{2}{5},\ \frac{4}{5}, ...\), so NOT an even intger.
Not sufficient.
(2) \(k^2 + 1\) is an odd integer.
If \(k\) is an integer, then from \(k^2+1=odd\) it follows that \(k^2=even\) and in this case \(k\) must be even.
If \(k\) is NOT an integer, then from \(k^2+1=odd\) it follows that \(k =\sqrt{even}\) and in this case \(k\) must some irrational number like: \(..., \ -\sqrt{6}, \ -\sqrt{2}; \ \sqrt{2},\ \sqrt{6}, ...\), so NOT an even intger.
Not sufficient.
(1)+(2) No irrational number satisfices (1): \(5k + 2=5*irrational + integer=irrational +integer=irrational\). So, we are left with the first case from (2): \(k\) is an even integer. Sufficient.
Answer: C
