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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]

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New post 13 Dec 2012, 10:14
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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
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Re: An operation @ is defined by the equation a@b = (a - b) / (a  [#permalink]

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New post 13 Dec 2012, 10:16
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Re: An operation @ is defined by the equation a@b = (a - b) / (a  [#permalink]

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New post 24 Nov 2013, 12:04
1
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)
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Re: An operation @ is defined by the equation a@b = (a - b) / (a  [#permalink]

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New post 24 Nov 2013, 13:47
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.

Similar questions to practice:
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the-operation-is-defined-for-all-integers-a-and-b-by-the-162119.html

Hope this helps.
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Re: An operation @ is defined by the equation a@b = (a - b) / (a  [#permalink]

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An operation θ is defined by the equation...  [#permalink]

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New post 08 Apr 2014, 20:35
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 = ?
The correct answer is that c = a

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
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Re: An operation θ is defined by the equation...  [#permalink]

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New post 08 Apr 2014, 23:11
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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 = ?
The correct answer is that c = a

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
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Re: An operation θ is defined by the equation...  [#permalink]

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New post 09 Apr 2014, 02:11
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 = ?
The correct answer is that c = a

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


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Re: An operation is defined by the equation ab = (a - b) / (a  [#permalink]

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New post 27 Jul 2015, 07:49
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
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Re: An operation is defined by the equation ab = (a - b) / (a  [#permalink]

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New post 27 Jul 2015, 07:57
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.
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An operation is defined by the equation ab = (a - b) / (a  [#permalink]

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New post 27 Jul 2015, 18:11
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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

Final Answer:

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An operation is defined by the equation ab = (a - b) / (a  [#permalink]

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New post Updated on: 28 Jul 2015, 07:08
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

Originally posted by apple08 on 27 Jul 2015, 22:27.
Last edited by apple08 on 28 Jul 2015, 07:08, edited 1 time in total.
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An operation is defined by the equation ab = (a - b) / (a  [#permalink]

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New post 27 Jul 2015, 22:53
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
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Re: An operation is defined by the equation ab = (a - b) / (a  [#permalink]

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New post 25 May 2016, 22:23
Attached is a visual that should help.
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Screen Shot 2016-05-25 at 9.42.05 PM.png
Screen Shot 2016-05-25 at 9.42.05 PM.png [ 76.69 KiB | Viewed 14510 times ]

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Re: An operation is defined by the equation ab = (a - b) / (a  [#permalink]

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New post 16 Jun 2016, 05:59
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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


We are given that operation @ is defined by a@b = (a-b)/(a+b).

We are also given that a@c = 0; thus, according to the operation, all instances of a can remain and all instances of b will be replaced with variable c. We then set that entire expression to zero.

(a-c)/(a+c) = 0

We can multiply both sides of the equation by a+c and then solve for a:

a – c = 0

a = c

The answer is E.
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Re: An operation @ is defined by the equation a@b = (a - b) / (a  [#permalink]

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Re: An operation @ is defined by the equation a@b = (a - b) / (a   [#permalink] 22 Sep 2018, 23:29
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