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# If the operation @ is defined for all a and b by the equatio

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If the operation @ is defined for all a and b by the equatio [#permalink]

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18 Sep 2003, 07:58
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Question Stats:

78% (01:58) correct 22% (01:20) wrong based on 352 sessions

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If the operation @ is defined for all a and b by the equation a@b =(a^2*b)/3, then 2@(3@-1) =

A. 4
B. 2
C. -4/3
D. -2
E. -4
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Aug 2014, 00:58, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: If the operation @ is defined for all a and b by the equatio [#permalink]

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21 Sep 2003, 11:37
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Formula: a@b=((a^2)*b)/3
Question: 2@(3@-1)= ?

work within the parenthesis first so solve (3@-1) first

(3@-1)= ((3^2)*-1)/3 = (9*-1)/3= -9/3=-3
now take -3 plug back into equation and solve the rest

2@(-3)=((2^2)*-3)/3 = (4*-3)/3= -12/3= -4

so -4 is the answer....this question is merely testing order of operations

remember PEMDAS
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Re: If the operation @ is defined for all a and b by the equatio [#permalink]

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05 Aug 2014, 05:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If the operation @ is defined for all a and b by the equatio [#permalink]

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12 Aug 2014, 01:03
pete300 wrote:
If the operation @ is defined for all a and b by the equation a@b =(a^2*b)/3, then 2@(3@-1) =

A. 4
B. 2
C. -4/3
D. -2
E. -4

Check other Operations/functions defining algebraic/arithmetic expressions probelms in our Special Questions Directory.
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Re: If the operation @ is defined for all a and b by the equatio [#permalink]

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12 Aug 2014, 01:09
pete300 wrote:
If the operation @ is defined for all a and b by the equation a@b =(a^2*b)/3, then 2@(3@-1) =

A. 4
B. 2
C. -4/3
D. -2
E. -4

Hi Bunuel,

Kindly write this equation in mathematical format. It's impossible to understand otherwise

Thanks
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Re: If the operation @ is defined for all a and b by the equatio [#permalink]

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12 Aug 2014, 02:29
PareshGmat wrote:
pete300 wrote:
If the operation @ is defined for all a and b by the equation a@b =(a^2*b)/3, then 2@(3@-1) =

A. 4
B. 2
C. -4/3
D. -2
E. -4

Hi Bunuel,

Kindly write this equation in mathematical format. It's impossible to understand otherwise

Thanks

There is nothing wrong with the format.
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Re: If the operation @ is defined for all a and b by the equatio [#permalink]

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12 Aug 2014, 03:05
Bunuel wrote:
PareshGmat wrote:
pete300 wrote:
If the operation @ is defined for all a and b by the equation a@b =(a^2*b)/3, then 2@(3@-1) =

A. 4
B. 2
C. -4/3
D. -2
E. -4

Hi Bunuel,

Kindly write this equation in mathematical format. It's impossible to understand otherwise

Thanks

There is nothing wrong with the format.

a@b =(a^2*b)/3

Is it $$\frac{a^{2b}}{3}$$ or

$$\frac{ba^2}{3}$$ ??
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Re: If the operation @ is defined for all a and b by the equatio [#permalink]

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12 Aug 2014, 03:12
PareshGmat wrote:
Bunuel wrote:
PareshGmat wrote:
If the operation @ is defined for all a and b by the equation a@b =(a^2*b)/3, then 2@(3@-1) =

A. 4
B. 2
C. -4/3
D. -2
E. -4

Hi Bunuel,

Kindly write this equation in mathematical format. It's impossible to understand otherwise

Thanks

There is nothing wrong with the format.

a@b =(a^2*b)/3

Is it $$\frac{a^{2b}}{3}$$ or

$$\frac{ba^2}{3}$$ ??

If it were $$\frac{a^{2b}}{3}$$, then it would be written as (a^(2*b))/3.
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Re: If the operation @ is defined for all a and b by the equatio [#permalink]

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12 Aug 2014, 03:27
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a@b $$= \frac{ba^2}{3}$$

2@(3@-1)

Solving the bracket first

3@-1$$= \frac{-1*3^2}{3} = -3$$

2@-3$$= \frac{-3*2^2}{3} = -4$$

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Re: If the operation @ is defined for all a and b by the equatio [#permalink]

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02 Oct 2015, 08:23
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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If the operation ◙ is defined for all a and b by the equation a ◙ b = [#permalink]

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16 Mar 2016, 18:55
If the operation ◙ is defined for all a and b by the equation a ◙ b = $$a^2b$$/3
then 2 ◙ (3 ◙ – 1) =

A. 4
B. 2
C. -4/3
D. -2
E. -4
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Re: If the operation ◙ is defined for all a and b by the equation a ◙ b = [#permalink]

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16 Mar 2016, 19:07
anceer wrote:
If the operation ◙ is defined for all a and b by the equation a ◙ b = $$a^2b$$/3
then 2 ◙ (3 ◙ – 1) =

A. 4
B. 2
C. -4/3
D. -2
E. -4

Just plugging numbers into the function and following through.

First, we solve for (3 ◙ – 1).

$$((3^2)(-1)) / 3 = -3$$

Now we solve for 2 ◙ -3

$$((2^2)(-3)) / 3 = -4$$

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Re: If the operation is defined for all a and b by the equatio [#permalink]

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16 Mar 2016, 22:06
anceer wrote:
If the operation ◙ is defined for all a and b by the equation a ◙ b = $$a^2b$$/3
then 2 ◙ (3 ◙ – 1) =

A. 4
B. 2
C. -4/3
D. -2
E. -4

Merging topics. Please refer to the discussion above.
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Re: If the operation is defined for all a and b by the equatio [#permalink]

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12 Sep 2016, 14:13
I agree that the way the equation is written is kind of confusing...
Re: If the operation is defined for all a and b by the equatio   [#permalink] 12 Sep 2016, 14:13
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