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Re: If the operation @ is defined for all integers a and b by a@b = a+bab
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02 Aug 2015, 02:18
visram04 wrote: If operation # for all integers is defined as a#b=a+bab, 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+bab(i) a#b=b#a a#b = a+bab b#a = b+aba > we can rewrite this as a+bab 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+bab =x. Sub in the above equation a+bab+c[(a+bab)*c) = a+b+cab [ac+bcabc] = a+b+cabacbc+abc(1) a#(b#c) Consider b#c =y. Thus a#(b#c) > a#y > a+yay Now b#c=y > b+cbc =y. Sub in the above equation a+b+cbc[a*(b+cbc)] > a+b+cbc[ab+acabc] > a+b+cbcabac+abc (2) (1) =(2) Thus (a#b)#c=a#(b#c) is true. Hence all are true



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Re: If the operation @ is defined for all integers a and b by a@b = a+bab
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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?



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Re: If the operation @ is defined for all integers a and b by a@b = a+bab
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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



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Re: If the operation @ is defined for all integers a and b by a@b = a+bab
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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.
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Re: If the operation @ is defined for all integers a and b by a@b = a+bab
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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+bab. For example if a=1 and b=2, then a@b=1@2=1+21*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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Re: If the operation @ is defined for all integers a and b by a@b = a+bab
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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+bab, 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: \(a@b=a+bab\) I. \(a@b = b@a\) > \(a@b=a+bab\) and \(b@a=b+aab\) > \(a+bab=b+aab\), results match; II. \(a@0 = a\) > \(a@0=a+0a*0=0\) > \(0=0\), results match; III. \((a@b)@c = a@(b@c)\) > \((a@b)@c=a@b+c(a@b)*c=(a+bab)+c(a+bab)c=a+b+cabacbc+abc\) and \(a@(b@c)=a+b@ca*(b@c)=a+(b+cbc)(b+cbc)a=a+b+cbcabac+abc\), results match. Answer: E. Hi, FOr this particular questions, can we substitute numbers? WIll that apprach always work? Please advise



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Re: If the operation @ is defined for all integers a and b by a@b = a+bab
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Updated on: 24 Nov 2018, 18:26
A video explanation can be found here: https://www.youtube.com/watch?v=gCL6DmaIqK4The 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 timeconsuming, but statement 3 turns out to be true. Answer E
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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.



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Re: If the operation @ is defined for all integers a and b by a@b = a+bab
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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.
Thanks in advance.



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Re: If the operation @ is defined for all integers a and b by a@b = a+bab
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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.



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If the operation @ is defined for all integers a and b by a@b = a+bab
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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 From I we already computed 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)= =1124=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+bab
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11 Apr 2020, 07:57



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