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..
16 Sep 2014, 00:54
Kudos
Bookmarks
Is \(2(a + b - c)\) an odd integer?
(1) \(a\), \(b\), and \(c\) are consecutive integers
(2) \(b = a + c\)
(1) \(a\), \(b\), and \(c\) are consecutive integers
(2) \(b = a + c\)
16 Sep 2014, 00:54
Kudos
Bookmarks
Official Solution:
Is \(2(a + b - c)\) an odd integer?
(1) \(a\), \(b\), and \(c\) are consecutive integers.
The crucial piece of information above is that \(a\), \(b\), and \(c\) are integers. Given that \(a\), \(b\), and \(c\) are integers, \(2(a + b - c) = 2*integer = even\). Therefore, the answer to the question is NO. Sufficient.
(2) \(b = a + c\)
If \(a\), \(b\), and \(c\) are integers, for example, \(a = b = c = 0\), then \(2(a + b - c) = 2*integer = even\). However, if \(a\), \(b\), and \(c\) are NOT integers, it's possible for \(2(a + b - c)\) to be an odd integer in cases where \((a + b - c)=\frac{odd}{2}\). For example, consider \(a = \frac{1}{4}\), \(b = 0\) and \(c = -\frac{1}{4}\); in this case, \(2(a + b - c) = 2*\frac{1}{2} = 1 = odd\). Not sufficient.
Alternatively, substituting \(b = a + c\) into \(2(a + b - c)\) gives \(2(a + b - c) = 2(a + a + c - c) = 4a\). Hence, the question becomes whether \(4a\) is an odd integer. If \(a = \frac{odd}{4}\), the answer will be YES, because in this case \(4a = 4*\frac{odd}{4} = odd\). However, if \(a \neq \frac{odd}{4}\), the answer will be NO. Not sufficient.
Answer: A
Is \(2(a + b - c)\) an odd integer?
(1) \(a\), \(b\), and \(c\) are consecutive integers.
The crucial piece of information above is that \(a\), \(b\), and \(c\) are integers. Given that \(a\), \(b\), and \(c\) are integers, \(2(a + b - c) = 2*integer = even\). Therefore, the answer to the question is NO. Sufficient.
(2) \(b = a + c\)
If \(a\), \(b\), and \(c\) are integers, for example, \(a = b = c = 0\), then \(2(a + b - c) = 2*integer = even\). However, if \(a\), \(b\), and \(c\) are NOT integers, it's possible for \(2(a + b - c)\) to be an odd integer in cases where \((a + b - c)=\frac{odd}{2}\). For example, consider \(a = \frac{1}{4}\), \(b = 0\) and \(c = -\frac{1}{4}\); in this case, \(2(a + b - c) = 2*\frac{1}{2} = 1 = odd\). Not sufficient.
Alternatively, substituting \(b = a + c\) into \(2(a + b - c)\) gives \(2(a + b - c) = 2(a + a + c - c) = 4a\). Hence, the question becomes whether \(4a\) is an odd integer. If \(a = \frac{odd}{4}\), the answer will be YES, because in this case \(4a = 4*\frac{odd}{4} = odd\). However, if \(a \neq \frac{odd}{4}\), the answer will be NO. Not sufficient.
Answer: A
30 Jan 2015, 08:01
Kudos
Bookmarks
Wont 2(a+b-c) result into an even integer irrespective of the statements below? Thus, making either statement sufficient.
