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# An operation @ is defined by the equation a@b = (a - b) / (a

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An operation @ is defined by the equation a@b = (a - b) / (a [#permalink]  13 Dec 2012, 09:14
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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
[Reveal] Spoiler: OA
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Re: An operation @ is defined by the equation a@b = (a - b) / (a [#permalink]  13 Dec 2012, 09:16
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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 $$a@c = \frac{a - c}{a + c}=0$$ --> $$a-c=0$$ --> $$a=c$$.

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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:
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 $$a@c = \frac{a - c}{a + c}=0$$ --> $$a-c=0$$ --> $$a=c$$.

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)
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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:
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 $$a@c = \frac{a - c}{a + c}=0$$ --> $$a-c=0$$ --> $$a=c$$.

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.

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Hope this helps.
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Re: An operation @ is defined by the equation a@b = (a - b) / (a [#permalink]  24 Nov 2013, 23:40
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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 = ?
[Reveal] Spoiler:
The correct answer is that c = a

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
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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 = ?
[Reveal] Spoiler:
The correct answer is that c = a

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
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Math Expert
Joined: 02 Sep 2009
Posts: 27512
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Kudos [?]: 42453 [0], given: 6029

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 = ?
[Reveal] Spoiler:
The correct answer is that c = a

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

Merging similar topics. Please refer to the discussion above.

All OG13 questions are here: the-official-guide-quantitative-question-directory-143450.html

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Re: An operation @ is defined by the equation a@b = (a - b) / (a [#permalink]  11 May 2015, 03:07
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Re: An operation @ is defined by the equation a@b = (a - b) / (a   [#permalink] 11 May 2015, 03:07
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