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.
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 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 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.
Jul 1911:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment
BillyZ
Current Student
Joined: 14 Nov 2016
Last visit: 03 May 2025
Posts: 1,147
Given Kudos: 926
Location: Malaysia
Concentration: General Management, Strategy
Schools: Sloan '23 (D) Tuck '23 (D) Darden '23 (D) Ross '23 (D) Kellogg '23 (D) Wharton '23 (D) Booth '23 (D) Fuqua '24 (A) Harvard '24 (D) Stanford '24 (D)
GMAT 1: 750 Q51 V40 (Online)
GPA: 3.53
Products:
Schools: Sloan '23 (D) Tuck '23 (D) Darden '23 (D) Ross '23 (D) Kellogg '23 (D) Wharton '23 (D) Booth '23 (D) Fuqua '24 (A) Harvard '24 (D) Stanford '24 (D)
GMAT 1: 750 Q51 V40 (Online)
Posts: 1,147
22 Feb 2017, 22:58
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
60% (02:03) correct
40%
(01:54)
wrong
based on 469
sessions
History
Date
Time
Result
Not Attempted Yet
Assume that all positive integers are colored either red or blue. If a set S consisting entirely of positive integers is chosen, where R is the number of red integers chosen and B is the number of blue integers chosen, the discrepancy of set S is \(|R-B|\). If set Q is such a set, is its discrepancy greater than \(\frac{1}{2}\) ?
(1) \(\frac{(R)}{(R+B)}\) < \(\frac{(B)}{(R+B)}\)
(2) \(R<B\)
(1) \(\frac{(R)}{(R+B)}\) < \(\frac{(B)}{(R+B)}\)
(2) \(R<B\)
23 Feb 2017, 12:15
Kudos
Bookmarks
AustinKL
Assume that all positive integers are colored either red or blue. If a set S consisting entirely of positive integers is chosen, where R is the number of red integers chosen and B is the number of blue integers chosen, the discrepancy of set S is \(|R-B|\). If set Q is such a set, is its discrepancy greater than \(\frac{1}{2}\) ?
(1) \(\frac{(R)}{(R+B)}\) < \(\frac{(B)}{(R+B)}\)
(2) \(R<B\)
Dear AustinKL, (1) \(\frac{(R)}{(R+B)}\) < \(\frac{(B)}{(R+B)}\)
(2) \(R<B\)
I'm happy to respond.
Like many Veritas questions, this is a great question. It has the feel of a GMAT Quant question: it looks as if it would require extensive complicated calculations, but when we unpack what it is really asking, it's very simple. Let's think about this. R & B are integers, because they are counts. The difference (R - B) could be any integer, positive or negative or zero. The absolute value |R - B| could be any positive integer or zero. The only way this absolute value difference would be less than 1/2 is if it is zero. If it is zero, then it must be true that R = B, the number of reds equals the number of blues. That's what the prompt question is really asking. If we know that R = B, then we can give a definitive "yes" answer. If we know for a fact that R and B are not equal, then we can give a definitive no answer.
Statement #1: \(\frac{R}{(R + B)}\) < \(\frac{B}{(R + B)}\)
Multiply both side by (R + B), and we get R < B. They are not equal, so we can give a definitive "no" to the prompt question. Because we can give a definitive answer, this statement is sufficient.
Forget statement #1 now.
Statement #2: \(R < B\)
This is a direct statement that R and B are not equal. Again, we can give a definitive "no" to the prompt question. Again, because we can give a definitive answer, this statement is sufficient.
Each statement, on its own, is sufficient. OA = (D)
Does all this make sense?
Mike
General Discussion
31 Jul 2017, 10:30
Kudos
Bookmarks
Can you please explain this
"The only way this absolute value difference would be less than 1/2 is if it is zero"
I'm happy to respond. Like many Veritas questions, this is a great question. It has the feel of a GMAT Quant question: it looks as if it would require extensive complicated calculations, but when we unpack what it is really asking, it's very simple.
Let's think about this. R & B are integers, because they are counts. The difference (R - B) could be any integer, positive or negative or zero. The absolutely value |R - B| could be any positive integer or zero. The only way this absolute value difference would be less than 1/2 is if it is zero. If it is zero, then it must be true that R = B, the number of reds equals the number of blues. That's what the prompt question is really asking. If we know that R = B, then we can give a definitive "yes" answer. If we know for a fact that R and B are not equal, then we can give a definitive no answer.
Statement #1: \(\frac{R}{(R + B)}\) < \(\frac{B}{(R + B)}\)
Multiply both side by (R + B), and we get R < B. They are not equal, so we can give a definitive "no" to the prompt question. Because we can give a definitive answer, this statement is sufficient.
Forget statement #1 now.
Statement #2: \(R < B\)
This is a direct statement that R and B are not equal. Again, we can give a definitive "no" to the prompt question. Again, because we can give a definitive answer, this statement is sufficient.
Each statement, on its own, is sufficient. OA = (D)
Does all this make sense?
Mike
"The only way this absolute value difference would be less than 1/2 is if it is zero"
mikemcgarry
AustinKL
Assume that all positive integers are colored either red or blue. If a set S consisting entirely of positive integers is chosen, where R is the number of red integers chosen and B is the number of blue integers chosen, the discrepancy of set S is \(|R-B|\). If set Q is such a set, is its discrepancy greater than \(\frac{1}{2}\) ?
(1) \(\frac{(R)}{(R+B)}\) < \(\frac{(B)}{(R+B)}\)
(2) \(R<B\)
Dear AustinKL, (1) \(\frac{(R)}{(R+B)}\) < \(\frac{(B)}{(R+B)}\)
(2) \(R<B\)
I'm happy to respond.
Like many Veritas questions, this is a great question. It has the feel of a GMAT Quant question: it looks as if it would require extensive complicated calculations, but when we unpack what it is really asking, it's very simple. Let's think about this. R & B are integers, because they are counts. The difference (R - B) could be any integer, positive or negative or zero. The absolutely value |R - B| could be any positive integer or zero. The only way this absolute value difference would be less than 1/2 is if it is zero. If it is zero, then it must be true that R = B, the number of reds equals the number of blues. That's what the prompt question is really asking. If we know that R = B, then we can give a definitive "yes" answer. If we know for a fact that R and B are not equal, then we can give a definitive no answer.
Statement #1: \(\frac{R}{(R + B)}\) < \(\frac{B}{(R + B)}\)
Multiply both side by (R + B), and we get R < B. They are not equal, so we can give a definitive "no" to the prompt question. Because we can give a definitive answer, this statement is sufficient.
Forget statement #1 now.
Statement #2: \(R < B\)
This is a direct statement that R and B are not equal. Again, we can give a definitive "no" to the prompt question. Again, because we can give a definitive answer, this statement is sufficient.
Each statement, on its own, is sufficient. OA = (D)
Does all this make sense?
Mike
