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If the operation @ is defined for all integers a and b [#permalink]
 18 Sep 2010, 20:16
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46% (01:15) wrong based on 17 sessions
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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Re: GMAT PREP QUESTION [#permalink]
18 Sep 2010, 20:29
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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-abI. 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. Answer: E.
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Re: GMAT PREP QUESTION [#permalink]
19 Apr 2011, 19:25
(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 Answer - E
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Re: GMATPrep#1 Question: If the operation (*) is defined for... [#permalink]
15 Jan 2012, 17:17
I was careless in writing the equation for option 3, else it was easy. I got it wrong but once you write it you clearly see that C also satifies the equation.
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Re: GMAT PREP QUESTION [#permalink]
22 May 2012, 02:25
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
Answer - E small typo : ab should be bc , in the above post
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Re: If the operation @ is defined for all integers a and b [#permalink]
22 May 2012, 03:09
E is the answer. It took some time to solve this but was able to slove faster when assuming values for a b and c. Posted from my mobile device  Posted from my mobile device
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Re: If the operation @ is defined for all integers a and b [#permalink]
18 Dec 2012, 08:45
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]
18 Dec 2012, 21:29
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]
19 Dec 2012, 03:24
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]
04 Jan 2013, 11:47
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: GMAT PREP QUESTION [#permalink]
28 Jan 2013, 22:58
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
Answer - E 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 1425 times ]
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Re: GMAT PREP QUESTION [#permalink]
07 May 2013, 12:22
Bunuel wrote: 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.
Answer: E. Bunuel, is there a faster method than solving it with algebra or picking numbers? Thanks!
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Re: GMAT PREP QUESTION
[#permalink]
07 May 2013, 12:22
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