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

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Intern
Joined: 17 Sep 2003
Posts: 3
Location: NY
If the operation @ is defined for all a and b by the equatio  [#permalink]

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Updated on: 11 Aug 2014, 23:58
2
6
00:00

Difficulty:

5% (low)

Question Stats:

81% (01:00) correct 19% (01:16) wrong based on 360 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

Originally posted by pete300 on 18 Sep 2003, 06:58.
Last edited by Bunuel on 11 Aug 2014, 23:58, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Joined: 15 Sep 2003
Posts: 70
Location: california
Re: If the operation @ is defined for all a and b by the equatio  [#permalink]

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21 Sep 2003, 10:37
2
1
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
Math Expert
Joined: 02 Sep 2009
Posts: 52285
Re: If the operation @ is defined for all a and b by the equatio  [#permalink]

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12 Aug 2014, 00:03
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, 00:09
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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Math Expert
Joined: 02 Sep 2009
Posts: 52285
Re: If the operation @ is defined for all a and b by the equatio  [#permalink]

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12 Aug 2014, 01:29
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, 02:05
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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Math Expert
Joined: 02 Sep 2009
Posts: 52285
Re: If the operation @ is defined for all a and b by the equatio  [#permalink]

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12 Aug 2014, 02:12
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, 02:27
1
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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Posts: 96
Location: India
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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, 17:55
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
Intern
Joined: 10 Nov 2014
Posts: 8
Re: If the operation ◙ is defined for all a and b by the equation a ◙ b =  [#permalink]

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16 Mar 2016, 18:07
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

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

Math Expert
Joined: 02 Sep 2009
Posts: 52285
Re: If the operation is defined for all a and b by the equatio  [#permalink]

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16 Mar 2016, 21:06
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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Intern
Joined: 01 Sep 2016
Posts: 8
Re: If the operation is defined for all a and b by the equatio  [#permalink]

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

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22 Dec 2018, 18:44
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

$$a@b =\frac{(a^2*b)}{3}$$

$$2@(3@-1)$$ ----- $$(i)$$

$$3@-1 = \frac{3^2*(-1)}{3} = \frac{-9}{3} = -3$$

Hence $$(i)$$ becomes $$=> 2@-3$$

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

Re: If the operation @ is defined for all a and b by the equatio &nbs [#permalink] 22 Dec 2018, 18:44
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