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
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: An operation @ is defined by the equation a@b = (a - b) / (a [#permalink]
24 Nov 2013, 11:04
Bunuel wrote:
Walkabout wrote:
An operation @ is defined by the equation a@b = (a - b) / (a + b), for all numbers a and b such that a ≠ -b. If a ≠ -c and a@c = 0, then c =
(A) -a (B) -1/a (C) 0 (D) 1/a (E) a
Given that \(a@b = \frac{a - b}{a + b}\), thus \(a@c = \frac{a - c}{a + c}\).
Also given that \(a@c = \frac{a - c}{a + c}=0\) --> \(a-c=0\) --> \(a=c\).
Answer: E.
Hi Bunnuel,
I've never senn such a task before in my life. Could you please explain what the @ means or what the whole expression "operation @" means? or do you recommend any books/links for this ? (I already worked through MGMAT Foundations of Math but really NEVER seen this before)
Re: An operation @ is defined by the equation a@b = (a - b) / (a [#permalink]
24 Nov 2013, 12:47
Expert's post
1
This post was BOOKMARKED
unceldolan wrote:
Bunuel wrote:
Walkabout wrote:
An operation @ is defined by the equation a@b = (a - b) / (a + b), for all numbers a and b such that a ≠ -b. If a ≠ -c and a@c = 0, then c =
(A) -a (B) -1/a (C) 0 (D) 1/a (E) a
Given that \(a@b = \frac{a - b}{a + b}\), thus \(a@c = \frac{a - c}{a + c}\).
Also given that \(a@c = \frac{a - c}{a + c}=0\) --> \(a-c=0\) --> \(a=c\).
Answer: E.
Hi Bunnuel,
I've never senn such a task before in my life. Could you please explain what the @ means or what the whole expression "operation @" means? or do you recommend any books/links for this ? (I already worked through MGMAT Foundations of Math but really NEVER seen this before)
@ is a made up operation (function), defined by the equation a@b = (a - b)/(a + b). For example 2@3=(2-3)/(2+3)=-1/5.
An operation θ is defined by the equation... [#permalink]
08 Apr 2014, 19:35
Hello and thank you for your help! I don't understand the answer provided by my book.
Question: An operation θ is defined by the equation a θ b = a-b/a+b, for all numbers a and b such that a does not equal -b. If a does not equal -c and a θ c = 0, then c = ?
Re: An operation θ is defined by the equation... [#permalink]
08 Apr 2014, 22:11
1
This post received KUDOS
Expert's post
raharu wrote:
Hello and thank you for your help! I don't understand the answer provided by my book.
Question: An operation θ is defined by the equation a θ b = a-b/a+b, for all numbers a and b such that a does not equal -b. If a does not equal -c and a θ c = 0, then c = ?
Answer: Substitute c for b and 0 for a θ c in the given equation and solve for c.
So 0 = a - c / a + c
Multiply each side by a + c
So 0 = a - c So c = a
My question is: why can they substitute c for b?
Source: 12th edition Gmat Review
θ has been defined as an operator which takes input values of two variables and gives the answer by calculating (First variable - Second variable)/(First variable + Second variable). Here a and b are just taken as an example.
a θ b = (a - b)/(a + b) x θ y = (x - y)/(x + y) a θ c = (a - c)/(a + c) It doesn't what the two variables are.
Given a θ c = 0 = (a - c)/(a + c) Then a - c = 0 a = c _________________
Re: An operation θ is defined by the equation... [#permalink]
09 Apr 2014, 01:11
Expert's post
raharu wrote:
Hello and thank you for your help! I don't understand the answer provided by my book.
Question: An operation θ is defined by the equation a θ b = a-b/a+b, for all numbers a and b such that a does not equal -b. If a does not equal -c and a θ c = 0, then c = ?
Re: An operation @ is defined by the equation a@b = (a - b) / (a [#permalink]
11 May 2015, 03:07
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: An operation is defined by the equation ab = (a - b) / (a [#permalink]
27 Jul 2015, 06:57
apple08 wrote:
Need great help How can I know that b can be substituted with c in the new equation ? Hope to hear from you Many thanks
The trick here is to recognize 2 things:
1. Unless you can substitute some variable by c in a@b , how are you going to get the desired relation between a and c? 2. The question stem mentions that the relation for a@b is true for "all numbers a and b". Thus you can substitute c for b and get the desired result. _________________
An operation is defined by the equation ab = (a - b) / (a [#permalink]
27 Jul 2015, 17:11
Expert's post
Hi All,
This question is an example of a 'Symbolism' question; in these types of prompt, the GMAT 'makes up' a math symbol, tells you how to use it, then asks you to use it to perform a calculation.
Here, we're given a made-up calculation that uses the @ symbol.... A@B = (A-B)/(A+B)
eg. 1@2 = (1-2)/(1+2) = -1/3
We're told that A@C = 0 and A ≠ -C. We're asked for the value of C....
We can TEST VALUES to answer this question, but we have to start by TESTing a VALUE for A, then figure out what C would have to equal....
IF.... A = 2 A@C = 2@C = (2-C)/(2+C) = 0
So, what would C have to equal to make this equation equal 0?
(2-C)/(2+C) = 0
Since we're dealing with a fraction, we need the NUMERATOR to equal 0. In this example, that would ONLY happen when C = 2. So we're looking for an answer that equals 2 when A=2.
Answer A: - A = -2 NOT a match Answer B: -1/A = -1/2 NOT a match Answer C: 0 NOT a match Answer D: 1/A = 1/2 NOT a match Answer E: A = 2 This IS a MATCH
An operation is defined by the equation ab = (a - b) / (a [#permalink]
27 Jul 2015, 21:27
Dear All, Many thanks , I'm sorry,i need help to understand it
Can I also know b can be replaced by c based on the following statements: a not equal -b a not equal -c As it indicate a-b and a-c has same relationship , since it is the same relationship,I can substitute b with c. Appreciate your comments Hope to hear from you many thanks for the great help
Last edited by apple08 on 28 Jul 2015, 06:08, edited 1 time in total.
An operation is defined by the equation ab = (a - b) / (a [#permalink]
27 Jul 2015, 21:53
Or since a@c = 0, I can plug c into b since @ is common in both a@b and a@c. How can I make use of "a not equal to -b" or "a not equal to -c"? Is it from "a not equal to -c" I know c is not equal to -a, hence eliminate answer (a) -a , really appreciate your great help, many many thanks
gmatclubot
An operation is defined by the equation ab = (a - b) / (a
[#permalink]
27 Jul 2015, 21:53
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...