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# If the operation @ is defined for all integers a and b by a@b = a+b-ab

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Manager
Joined: 14 Mar 2014
Posts: 139
GMAT 1: 710 Q50 V34
Re: If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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02 Aug 2015, 02:18
1
visram04 wrote:
If operation # for all integers is defined as a#b=a+b-ab, which of the following is must be ture for all integers a,b and c?
i) a#b=b#a
ii) a#0=a
iii) (a#b)#c=a#(b#c)

Answer is all 3. Can someone help me understand i) and iii)?

Given: a#b=a+b-ab

(i) a#b=b#a
a#b = a+b-ab
b#a = b+a-ba --> we can rewrite this as a+b-ab
which is same as a#b. Thus it is true

(ii) a#0=a
a#0 = a+0-(a*0) = a. Thus true

(iii) (a#b)#c=a#(b#c)
(a#b)#c
Consider a#b = x. Thus (a#b)#c ==> x#c => x+c-(x*c)
Now a#b = x => a+b-ab =x. Sub in the above equation
a+b-ab+c-[(a+b-ab)*c) = a+b+c-ab -[ac+bc-abc] = a+b+c-ab-ac-bc+abc--(1)

a#(b#c)
Consider b#c =y. Thus a#(b#c) --> a#y -> a+y-ay
Now b#c=y --> b+c-bc =y. Sub in the above equation
a+b+c-bc-[a*(b+c-bc)] --> a+b+c-bc-[ab+ac-abc] --> a+b+c-bc-ab-ac+abc --(2)
(1) =(2)
Thus (a#b)#c=a#(b#c) is true.

Hence all are true
Intern
Joined: 14 Oct 2014
Posts: 3
Re: If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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25 Dec 2015, 13:32
I apologize for the rudimentary questions here, but can you break down the strategy for solving this problem? I always get confused with custom character problems. What are we trying to prove here?
Manager
Joined: 03 Jan 2017
Posts: 132
Re: If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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28 Mar 2017, 09:20
I just picked random positive integers and started testing right away, all 3 worked for me

Though time consuming, but one can arrive pretty quickly at 1,2. 3 took much more time
Manager
Joined: 31 Dec 2016
Posts: 65
Re: If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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16 Jul 2017, 09:53
While Brunnel provides terrific answers the issues are that he sometimes assumes we have his level of knowledge of math, which we don't. Here is a more simple version of what he did in answer choice (C) which explains the steps.

Essentially, for operator questions one variable becomes another expression. So (A operator B) Operator C means A operator B becomes A and C becomes B. Hopefully, the logic flows through below.
Attachments

gmat answer.png [ 297.08 KiB | Viewed 2426 times ]

Intern
Joined: 08 Jul 2016
Posts: 34
Location: Singapore
GMAT 1: 570 Q43 V25
GMAT 2: 640 Q42 V36
WE: Underwriter (Insurance)
Re: If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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07 Feb 2018, 07:51
Bunuel wrote:
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.

Hey Bunuel,

Since I always tend to mix up the variables by the end of the problem, please tell if it would be wrong to test the function with random numbers (positive integers)?
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Joined: 22 Nov 2016
Posts: 7
Re: If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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08 Apr 2018, 17:59
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 $$a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab-ac+abc$$, results match.

Hi,
FOr this particular questions, can we substitute numbers? WIll that apprach always work?
Intern
Joined: 20 Aug 2018
Posts: 25
Re: If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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Updated on: 24 Nov 2018, 18:26
A video explanation can be found here:

The fastest way to solve the problem is to pick numbers for a, b, and c that are small and easy to work with - for example, 1, 2 and 3

You can see pretty quickly that statements I and II must be true

Testing statement 3 is more time-consuming, but statement 3 turns out to be true.

_________________

Originally posted by CrushGMAT on 24 Nov 2018, 14:28.
Last edited by CrushGMAT on 24 Nov 2018, 18:26, edited 1 time in total.
Intern
Joined: 04 Sep 2018
Posts: 25
GPA: 3.33
Re: If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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08 Jan 2019, 12:45
Can someone please comment if picking numbers, and checking if the Left hand side equals Right Hand side for the equations is a good way to approach the problem?

I think that approach makes it very easy for me personally to attempt the question.

Intern
Joined: 03 Jun 2018
Posts: 1
Re: If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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04 May 2019, 07:36
Hi All,
Just wondering if this approach will work.. Not sure if it has been pointed out or if I am correct. Please correct me if I am wrong.
We know that statement II is true, i.e- a@0 = a,
So, for statement 3, (a@b)@c = a@(b@c), if we consider b = 0, the statement reduces to
(a@0)@c = a@(0@c)
and from statement II, we know a@0 = a and 0@c = c,
Hence, a@c = a@c.

Therefore, correct answer is option E.
Intern
Joined: 20 Mar 2019
Posts: 23
If the operation @ is defined for all integers a and b by a@b = a+b-ab  [#permalink]

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11 Apr 2020, 07:57
PICKING NUMBER WHILE MINIMIZING CALCULATIONS
(where calculations were performed there are italic characters)
Take following values to test the three statements:

Choose:
a=3
b=8

I.
a + b - ab = b + a - ba
3 + 8 - 24 = 8 + 3 - 24
-13 = -13

works
I is true

II.
a + 0 - (a*0) = a
3 + 0 - 0 = 3
3 = 3

works
II is true

Take advantage of II being true.
Choose:
c=0

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

1st member
( a @ 'b) @ c
a @ 'b = -13
so, since II is true and c=0
( a @ 'b ) @ c = ( a @ 'b )
( a @ 'b) @ 0 = -13

2nd member
a @ ( b @ c )
From II we know that for c=0 we get
b @ c = b
8 @ 0 = 8
so,
a @ ( b @ c )=
=3 + 8 - (3*8)=
=11-24=-13

In conclusion
we got
1st member=2nd member
-13=-13
III is true

CORRECT ANS: E
If the operation @ is defined for all integers a and b by a@b = a+b-ab   [#permalink] 11 Apr 2020, 07:57

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