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

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24 Nov 2013, 12:04

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)

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]

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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 = ?

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
_________________

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]

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11 May 2015, 04:07

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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.
_________________

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]

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27 Jul 2015, 22:27

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, 07:08, edited 1 time in total.

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

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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

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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