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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

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

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

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