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Intern  Joined: 22 Dec 2009
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If the operation @ is defined for all integers a and b  [#permalink]

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Question Stats: 63% (02:04) correct 37% (02:08) wrong based on 959 sessions

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If the operation @ is defined for all integers a and b by a@b=a+b-ab, which of the following statements must be true for all integers a, b and c?

I. a@b = b@a
II. a@0 = a
III. (a@b)@c = a@(b@c)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III
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Joined: 02 Sep 2009
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If the operation is defined for all integers a and b  [#permalink]

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cmugeria wrote:
If the operation @ is defined for all integers a and b by a@b=a+b-ab, which of the following statements must be true for all integers a, b and c?

I. a@b = b@a
II. a@0 = a
III. (a@b)@c = a@(b@c)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III

We have that: $$a@b=a+b-ab$$

I. $$a@b = b@a$$ --> $$a@b=a+b-ab$$ and $$b@a=b+a-ab$$ --> $$a+b-ab=b+a-ab$$, results match;

II. $$a@0 = a$$ --> $$a@0=a+0-a*0=0$$ --> $$0=0$$, results match;

III. $$(a@b)@c = a@(b@c)$$ --> $$(a@b)@c=a@b+c-(a@b)*c=(a+b-ab)+c-(a+b-ab)c=a+b+c-ab-ac-bc+abc$$ and $$a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab-ac+abc$$, results match.

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

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jlgdr wrote:
Well, can't argue that learning the definitions is in fact quite interesting and thank you for that.
Nevertheless, I was really intereted in solving statement 3 quicker/faster/more efficient
So, being able to recognize if operations in the different order given will yield same result without having to go through all the long distribution process.

I will try to come up with a faster way but if anyone else come's up with a nice and elegant approach I'd be happy to give some nice Kudos for the collection

Cheers
J You can always test values that's an alternative

Let a=1,b=2 and c=3

Definition => a@b=a+b-ab

Option 3

III. (a@b)@c = a@(b@c)

a@b = 1 + 2 -2 = 1
1@3 = 1+3 -3 = 1

B@c = 2@3 = 5-6 = -1
1@-1 = 1 - 1 - ( -1 * 1)
=1

LHS = RHS
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Re: If the operation @ is defined for all integers a and b  [#permalink]

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1
(I) and (II) are obviously correct.

For (III)

(a+b-ab)@c = (a + b - ab)@c = a + b - ab + c - c(a + b - ab) = a + b - ab + c - ac - ab + abc

a@(b@c) = a@(b + c - bc) = a + b + c - bc - a(b + c - bc) = a + b + c - bc -ab - ac + abc

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

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Sorry for bumping up an old thread, I have a doubt: my approach for solving the question was to assume that the operation in this case was the union between two sets, a and b, and consequently the three points were the properties of the union of sets. Is that a correct approach or might it be too risky in the actual exam? (or is it even a wrong assumption, and I got it right out of luck?)
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Re: If the operation @ is defined for all integers a and b  [#permalink]

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I. a@b = b@a
$$a+b-ab=b+a-ab$$ TRUE!

II. a@0 = a
$$a+0-0 = a$$ TRUE!

III. (a@b)@c = a@(b@c)
$$a+b-ab+c-ac-bc+abc = b+c-bc + a - ab-ac+abc$$ STRIKE OUT DUPLICATES ON RHS and LHS! TRUE!

Answer: I,II, and III or (E)
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Re: If the operation @ is defined for all integers a and b  [#permalink]

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fguardini1 wrote:
Sorry for bumping up an old thread, I have a doubt: my approach for solving the question was to assume that the operation in this case was the union between two sets, a and b, and consequently the three points were the properties of the union of sets. Is that a correct approach or might it be too risky in the actual exam? (or is it even a wrong assumption, and I got it right out of luck?)

That's not correct. Stem defines some function @ for all integers a and b by a@b=a+b-ab. For example if a=1 and b=2, then a@b=1@2=1+2-1*2.

Hope it's clear.
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Re: If the operation @ is defined for all integers a and b  [#permalink]

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in equ. 3 :

I put three random numbers like (5,3,2) and tested it. however there's always a little chance of error.
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Re: If the operation @ is defined for all integers a and b  [#permalink]

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subhashghosh wrote:
(I) and (II) are obviously correct.

For (III)

(a+b-ab)@c = (a + b - ab)@c = a + b - ab + c - c(a + b - ab) = a + b - ab + c - ac - ab + abc

a@(b@c) = a@(b + c - bc) = a + b + c - bc - a(b + c - bc) = a + b + c - bc -ab - ac + abc

Been looking at this for a while and still can't figure it out.. I thought that we must ALWAYS first do the calculations in the brackets and open them, and then do the remaining calculations. as we have a@(b@c), how come do you straight come up to (a + b - ab)@c, when it's b@c in the brackets, not a@b anymore.. finding this one a bit confusing.. thanks for explaining in advance EDIT:

OK, I think I get it now.. pls, have a look at my upload and let me know if I am correct.. this is the left side of the equation in (III). With the right one, we do the exact same thing, right?
Attachments a@b.jpg [ 861.06 KiB | Viewed 88846 times ]

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Posts: 1651
Concentration: Finance
Re: If the operation @ is defined for all integers a and b  [#permalink]

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Hey there folks, sorry to bump on an old thread. Just wondering, is there a way to evaluate statement 3 faster?
I believe this questions takes around 2 minutes and evaluating statement 3 takes a lot of work and is prone to errors.

Just wondering if I'm doing this the correct/most efficient way

Thanks guys
Cheers!

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

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Would it be smart to pick numbers for each of the variables? I selected 1&3 only since I got mixed up with the letter variables. Is picking number the most efficient way to approach this problem?
Intern  Joined: 03 Oct 2013
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Re: If the operation @ is defined for all integers a and b  [#permalink]

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jlgdr wrote:
Hey there folks, sorry to bump on an old thread. Just wondering, is there a way to evaluate statement 3 faster?
I believe this questions takes around 2 minutes and evaluating statement 3 takes a lot of work and is prone to errors.

Just wondering if I'm doing this the correct/most efficient way

Thanks guys
Cheers!

J In this problem we have been asked to check the commutative and associative property of the given function. These properties are defined as below:

Commutative: In mathematics, a binary operation is commutative if changing the order of the operands does not change the result.

Associative: Within an expression containing two or more occurrences in a row of the same associative operator, the order in which the operations are performed does not matter as long as the sequence of the operands is not changed. That is, rearranging the parentheses in such an expression will not change its value.

If you're wondering if commutativity implies associativity in mathematics then the answer is NO. However, for simple addition and multiplication functions commutativity does imply associativity and hence in such cases option 3 need not be tested if option 1 is true. However, the only way to solve such problems which involve functions other than simple addition and multiplication would be to solve the expression completely as stated above.
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Re: If the operation @ is defined for all integers a and b  [#permalink]

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Well, can't argue that learning the definitions is in fact quite interesting and thank you for that.
Nevertheless, I was really intereted in solving statement 3 quicker/faster/more efficient
So, being able to recognize if operations in the different order given will yield same result without having to go through all the long distribution process.

I will try to come up with a faster way but if anyone else come's up with a nice and elegant approach I'd be happy to give some nice Kudos for the collection

Cheers
J Manager  Joined: 10 Mar 2013
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GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Re: If the operation @ is defined for all integers a and b  [#permalink]

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Just took this question today, and I was also wondering if there were a way to solve it more quickly than performing the heavy manipulations that are in III or guessing numbers.
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Joined: 10 Mar 2014
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Re: If the operation @ is defined for all integers a and b  [#permalink]

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Bunuel wrote:
cmugeria wrote:
If the operation @ is defined for all integers a and b by a@b=a+b-ab, which of the following statements must be true for all integers a, b and c?

I. a@b = b@a
II. a@0 = a
III. (a@b)@c = a@(b@c)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III

We have that: $$a@b=a+b-ab$$

I. $$a@b = b@a$$ --> $$a@b=a+b-ab$$ and $$b@a=b+a-ab$$ --> $$a+b-ab=b+a-ab$$, results match;

II. $$a@0 = a$$ --> $$a@0=a+0-a*0=0$$ --> $$0=0$$, results match;

III. $$(a@b)@c = a@(b@c)$$ --> $$(a@b)@c=a@b+c-(a@b)*c=(a+b-ab)+c-(a+b-ab)c=a+b+c-ab-ac-bc+abc$$ and $$a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab+abc$$, results match.

Hi Bunuel,

Here in second statement how you are getting 0=0?

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

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PathFinder007 wrote:
Bunuel wrote:
cmugeria wrote:
If the operation @ is defined for all integers a and b by a@b=a+b-ab, which of the following statements must be true for all integers a, b and c?

I. a@b = b@a
II. a@0 = a
III. (a@b)@c = a@(b@c)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III

We have that: $$a@b=a+b-ab$$

I. $$a@b = b@a$$ --> $$a@b=a+b-ab$$ and $$b@a=b+a-ab$$ --> $$a+b-ab=b+a-ab$$, results match;

II. $$a@0 = a$$ --> $$a@0=a+0-a*0=0$$ --> $$0=0$$, results match;

III. $$(a@b)@c = a@(b@c)$$ --> $$(a@b)@c=a@b+c-(a@b)*c=(a+b-ab)+c-(a+b-ab)c=a+b+c-ab-ac-bc+abc$$ and $$a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab+abc$$, results match.

Hi Bunuel,

Here in second statement how you are getting 0=0?

Thanks.

II says: $$a@0 = a$$

LHS = $$a@0=a+0-a*0=a$$.
RHS = a

a=a.
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Re: If the operation is defined for all integers a and b  [#permalink]

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I have a question regarding this problem. Don't we have to test for negative numbers while expanding the equation? This is where I got stuck and it became time consuming for me at which point I guessed and moved on.
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Re: If the operation is defined for all integers a and b  [#permalink]

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Yes, tested for negative values and got stuck. For a@0 = a. If we put a is negative then it doesn't hold. Any thoughts on this? Where am I going wrong?
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Re: If the operation is defined for all integers a and b  [#permalink]

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Yoshit wrote:
Yes, tested for negative values and got stuck. For a@0 = a. If we put a is negative then it doesn't hold. Any thoughts on this? Where am I going wrong?

It does: lets say a=-4 ---> a@0=-4@0 = -4+0-0*-4 = -4 =a. This HAS to be true even by the underlying algebra!

Thus satisfies the given equation.

Can you show your steps for calculations?
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Re: If the operation is defined for all integers a and b  [#permalink]

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You are right, taking negative values I jumbled up equations. Thank you for your help. Re: If the operation is defined for all integers a and b   [#permalink] 26 Jul 2015, 08:17

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