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.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
Thu Dec 13, 2018 6:27 am
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
82% (01:09) correct
18%
(01:21)
wrong
based on 1353
sessions
History
Date
Time
Result
Not Attempted Yet
If \(r☉s = rs + r + s\), then for what value of \(s\) is \(r☉s\) equal to \(r\) for all values of \(r\)?
(A) –1
(B) 0
(C) 1
(D) \(\frac{1}{r+1}\)
(E) \(r\)
(A) –1
(B) 0
(C) 1
(D) \(\frac{1}{r+1}\)
(E) \(r\)
Every tricky question should trigger a follow up quiz.
Build that habit with
GMAT Club Forum Quiz →.
Thu Dec 13, 2018 6:36 am
Kudos
Bookmarks
HKD1710
We want: \(r☉s = r\)
In other words: \(rs + r + s = r\)
Subtract r from both sides: \(rs + s = 0\)
Factor the left side: \(s(r + 1) = 0\)
So, either s = 0, or r = -1
So, \(r☉s = r\) when s = 0, or r = -1
The question asks about the value of s, so the correct answer is B.
Cheers,
Brent
General Discussion
Thu Dec 13, 2018 9:00 pm
Kudos
Bookmarks
We want: r☉s=r
Hence rs+r+s=r
Subtract r from both sides: rs+s=0
Taking s common: s(r+1)=0
So, either s = 0, or r = -1
But in the question, find s for all values of r .
Hence r can be any real number not only -1 . So s=0 .
IMO B.
Hence rs+r+s=r
Subtract r from both sides: rs+s=0
Taking s common: s(r+1)=0
So, either s = 0, or r = -1
But in the question, find s for all values of r .
Hence r can be any real number not only -1 . So s=0 .
IMO B.
