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In the expression a $ b, the $ symbol represents one of the

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In the expression a $ b, the $ symbol represents one of the following arithmetic operations on a and b (in the order the variables are shown): addition, subtraction, multiplication, and division. Given that it is not true that a $ b = b $ a for all possible values of a and b, a pair of nonzero, non-identical values for a and b is chosen such that a $ b produces the same result, no matter which of the operations (under the given constraints) that $ represents. The nonzero value of b that cannot be chosen, no matter the value of a, is

A. -2
B. -1
C. -1/2
D. 1
E. 1/2

[Reveal] Spoiler:
I would like to have some discussion before posting the correct answer
[Reveal] Spoiler: OA

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Last edited by Bunuel on 19 Mar 2013, 11:05, edited 1 time in total.
OA added.

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emmak wrote:
In the expression a $ b, the $ symbol represents one of the following arithmetic operations on a and b (in the order the variables are shown): addition, subtraction, multiplication, and division. Given that it is not true that a $ b = b $ a for all possible values of a and b, a pair of nonzero, non-identical values for a and b is chosen such that a $ b produces the same result, no matter which of the operations (under the given constraints) that $ represents. The nonzero value of b that cannot be chosen, no matter the value of a, is

A. -2
B. -1
C. -1/2
D. 1
E. 1/2

[Reveal] Spoiler:
I would like to have some discussion before posting the correct answer


Since it's NOT true that a$b=b$a for all possible values of a and b, then $ is neither addition not multiplication (because \(a+b=b+a\) and \(ab=ba\) for all possible values of a and b).

So, we have that $ is either subtraction or division.

Next, we are told that a$b produces the same result, no matter which of the operations (under the given constraints) that $ represents so no matter whether $ is subtraction or division a$b will produce the same result, so \(a-b=\frac{a}{b}\) --> \(ab-b^2=a\) --> \(a=\frac{b^2}{b-1}\) --> b cannot be 1, because in this case \(b-1=0\) and we cannot divide by zero.

Answer: D.

Hope it's clear.
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Re: In the expression a $ b, the $ symbol represents one of the [#permalink]

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"Given that it is not true that a $ b = b $ a for all possible values of a and b"

so $ cannot be + or *
because
\(a * b = b * a\)

\(a + b = b + a\)

for every a,b

we know that
\(a/b = a - b\)

\(a = ab - b^2\)

\(a - ab = -b^2\)

\(a(1 - b) = -b^2\)

\(a = -b^2/(1 - b)\)

B cannot be 1
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Re: In the expression a $ b, the $ symbol represents one of the [#permalink]

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Great explainations. I have 1 question. In the equation, why do you solve for a instead of solving for b? Is there something in the wording that tells you to solve "a=" instead of "b=?"


a/b=a-b
ab-b^2=a After this step, how do I know whether I try to simplify a or b?

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DoItRight wrote:
Great explainations. I have 1 question. In the equation, why do you solve for a instead of solving for b? Is there something in the wording that tells you to solve "a=" instead of "b=?"


a/b=a-b
ab-b^2=a After this step, how do I know whether I try to simplify a or b?


I tried to get all "a" and "b" terms on different sides, that's the main point.

However if you try to simplify b you get:

\(b(a-b)=a\)

\(b=\frac{a}{(a-b)}\)

and you go nowhere.
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New post 23 Jul 2013, 18:42
DoItRight wrote:
Great explainations. I have 1 question. In the equation, why do you solve for a instead of solving for b? Is there something in the wording that tells you to solve "a=" instead of "b=?"


a/b=a-b
ab-b^2=a After this step, how do I know whether I try to simplify a or b?



Simplifying b does not work. i.e. it wont give you any result:

You will have:

b^2 = ab - a
b = a - a/b

The value of b cannot be 1 because substituting b = 1 give you: 1 = a-a --> 1 = 0?? Makes no sense

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In the expression a $ b, the $ symbol represents one of the [#permalink]

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New post 21 Aug 2017, 21:42
Need some help
let b=-2
a=1/2
a/b=1/2/-2=1/-4
a-b=1/2-(-2)=5/2
hence b cannot be -2?

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Re: In the expression a $ b, the $ symbol represents one of the   [#permalink] 21 Aug 2017, 22:06
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