November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. November 17, 2018 November 17, 2018 09:00 AM PST 11:00 AM PST Join the Quiz Saturday November 17th, 9 AM PST. The Quiz will last approximately 2 hours. Make sure you are on time or you will be at a disadvantage.
Author 
Message 
TAGS:

Hide Tags

Current Student
Joined: 29 Sep 2013
Posts: 29
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

If a and b are distinct integers and their product is not equal to
[#permalink]
Show Tags
01 Sep 2014, 11:09
Question Stats:
61% (02:33) correct 39% (03:01) wrong based on 288 sessions
HideShow timer Statistics
If a and b are distinct integers and their product is not equal to zero, is a > b? (1) \(\frac{a^3*b  b^3*a}{a^3*b + b^3*a  2a^2*b^2} < 0\) (2) b < 0
Official Answer and Stats are available only to registered users. Register/ Login.




Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488

Re: If a and b are distinct integers and their product is not equal to
[#permalink]
Show Tags
01 Sep 2014, 13:38
janani28 wrote: If a and b are distinct integers and their product is not equal to zero, is a > b?
(1) (a^(3)b  b^(3)a)/(a^(3)b + b^(3)a  2a^(2)b^(2)) < 0
(2) b < 0 Dear janani28, I'm happy to help. For questions such as this, it's a good idea to learn how to use the math notation on GC. The prompt tells us they are different integers, and neither equals zero. Statement #1: the expression can be vastly simplified. First, we can factor out (ab) from both the numerator and denominator and cancel it. \(\frac{(a^3b  b^3a)}{(a^3b + b^3a  2a^2b^2)} = \frac{(ab(a^2  b^2))}{(ab(a^2 + b^2  2ab))} = \frac{a^2  b^2}{a^2  2ab + b^2}\) That expression in the denominator is the Square of a Difference formula \(a^2  2ab + b^2 = (a  b)^2\) See: http://magoosh.com/gmat/2013/threealge ... thegmat/The denominator is always positive, we can multiply both sides by it, and it just becomes zero on the other side. We are left with the numerator: \(a^2  b^2 < 0\) \(a^2 < b^2\) This is tricky. This tells us that b has a larger absolute value than a, but either could be positive or negative, so either could be bigger. This statement, alone and by itself, is insufficient. Statement #2, by itself, is clearly insufficient. Combined: From statement #1, we know that a < b. Statement #2 tells us that b is negative. (a) if a is positive, then any positive is always bigger than any negative; then a > b, and we would answer "yes" to the prompt (b) if a is negative, then both are negative, but because b has a larger absolute value, it would be more negative  say, b = 7 and a = 4, so that a > b Either way, a has to be greater than b. With the statements combined, we can give a definitive "yes" answer to the prompt. Combined, both statements are sufficient. Answer = (C)Does all this make sense? Mike
_________________
Mike McGarry Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)




Intern
Joined: 25 Mar 2014
Posts: 31

Re: If a and b are distinct integers and their product is not equal to
[#permalink]
Show Tags
01 Sep 2014, 13:34
\(\frac{(ab(a^2b^2))}{(ab(a^22ab+b^2)}\)
\(\frac{((a+b)(ab))}{((ab)^2)}\)
\(\frac{(a+b)}{(ab)}<0\)
(a+b) AND (ab) are negative or (a+b) is positive AND (ab) are negative
1.ns 2.ns
hence C.



Manager
Status: PLAY HARD OR GO HOME
Joined: 25 Feb 2014
Posts: 150
Location: India
Concentration: General Management, Finance
GPA: 3.1

Re: If a and b are distinct integers and their product is not equal to
[#permalink]
Show Tags
03 Oct 2014, 11:52
mikemcgarry wrote: janani28 wrote: If a and b are distinct integers and their product is not equal to zero, is a > b?
(1) (a^(3)b  b^(3)a)/(a^(3)b + b^(3)a  2a^(2)b^(2)) < 0
(2) b < 0 Dear janani28, I'm happy to help. For questions such as this, it's a good idea to learn how to use the math notation on GC. The prompt tells us they are different integers, and neither equals zero. Statement #1: the expression can be vastly simplified. First, we can factor out (ab) from both the numerator and denominator and cancel it. \(\frac{(a^3b  b^3a)}{(a^3b + b^3a  2a^2b^2)} = \frac{(ab(a^2  b^2))}{(ab(a^2 + b^2  2ab))} = \frac{a^2  b^2}{a^2  2ab + b^2}\) That expression in the denominator is the Square of a Difference formula \(a^2  2ab + b^2 = (a  b)^2\) See: http://magoosh.com/gmat/2013/threealge ... thegmat/The denominator is always positive, we can multiply both sides by it, and it just becomes zero on the other side. We are left with the numerator: \(a^2  b^2 < 0\) \(a^2 < b^2\) This is tricky. This tells us that b has a larger absolute value than a, but either could be positive or negative, so either could be bigger. This statement, alone and by itself, is insufficient. Statement #2, by itself, is clearly insufficient. Combined: From statement #1, we know that a < b. Statement #2 tells us that b is negative. (a) if a is positive, then any positive is always bigger than any negative; then a > b, and we would answer "yes" to the prompt (b) if a is negative, then both are negative, but because b has a larger absolute value, it would be more negative  say, b = 7 and a = 4, so that a > b Either way, a has to be greater than b. With the statements combined, we can give a definitive "yes" answer to the prompt. Combined, both statements are sufficient. Answer = (C)Does all this make sense? Mike Hi there, my doubt may sound a bit off,but here it is.. when we reach at this stage= (a+b)(ab) / (ab)^2 < 0 , then after deducing,we are left with a+b/ab < 0 finally, a+b<0 = a< b ...from here,we can say that in any case, B should be greater than A.. please clear my understanding..
_________________
ITS NOT OVER , UNTIL I WIN ! I CAN, AND I WILL .PERIOD.



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488

Re: If a and b are distinct integers and their product is not equal to
[#permalink]
Show Tags
03 Oct 2014, 12:25
vards wrote: Hi there, my doubt may sound a bit off,but here it is..
when we reach at this stage= (a+b)(ab) / (ab)^2 < 0 ,
then after deducing,we are left with
a+b/ab < 0
finally,
a+b<0 = a< b ...from here,we can say that in any case, B should be greater than A.. please clear my understanding.. Dear Vards, I'm happy to respond here. The first thing I'll say, my friend, is that it's very important to distinguish the broader category of algebraic/mathematical steps that are allowed to do, that are mathematically legal to perform, with the much smaller category of those legal steps that are actually strategically beneficial in solving the problem. When we get to the stage: \(\frac{(a+b)(ab)}{(ab)^2} < 0\) You choose to cancel a factor of (a b) in the numerator and denominator. That's definitely a legal move to do, but it's doubtful whether that is a particularly strategic thing to do. In fact, in terms of strategy, in terms of getting us closer to the answer, I think it's a not very good move. By the way, canceling is not called " deducing," at least in American mathematical terms. You see, that act of canceling leaves us with: \(\frac{a+b}{ab} < 0\) But the trouble is, with just statement #1 for the expressions a, b, (a + b), and (a  b): we don't know whether any of them are positive or negative or zero. That means if we multiply both sides by the denominator (a  b), we don't know whether we are multiplying by a positive or a negative number, so we don't know whether we reverse the inequality sign. Technically, at that stage, we would have to create two different scenarios, one in which we specified that (a  b) > 0, and the other in which we specified that (a  b) < 0, and the follow out the consequences of each case. We could do all that, but that's a very long, very complicated way to get to answer. That's precisely why that initial canceling move was a notverystrategic thing to do. What I did was considerably more elegant in terms of strategy. Noting that the original denominator, (a  b)^2, is a square and therefore always positive, we can just multiply both sides by that, we are left with the fact that a^2  b^2 < 0. That immediate leads to the important statement a^2 < b^2, which is very helpful in answering the question. Don't assume, because you can do something, because a move is allowed by the laws of algebra, that just because of this, the move is a particularly useful, a particularly strategic move. There are always dozens of moves that are perfectly legal and disastrous in terms of strategy. Does all this make sense? Mike
_________________
Mike McGarry Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)



Manager
Status: PLAY HARD OR GO HOME
Joined: 25 Feb 2014
Posts: 150
Location: India
Concentration: General Management, Finance
GPA: 3.1

Re: If a and b are distinct integers and their product is not equal to
[#permalink]
Show Tags
03 Oct 2014, 12:51
mikemcgarry wrote: vards wrote: Hi there, my doubt may sound a bit off,but here it is..
when we reach at this stage= (a+b)(ab) / (ab)^2 < 0 ,
then after deducing,we are left with
a+b/ab < 0
finally,
a+b<0 = a< b ...from here,we can say that in any case, B should be greater than A.. please clear my understanding.. Dear Vards, I'm happy to respond here. The first thing I'll say, my friend, is that it's very important to distinguish the broader category of algebraic/mathematical steps that are allowed to do, that are mathematically legal to perform, with the much smaller category of those legal steps that are actually strategically beneficial in solving the problem. When we get to the stage: \(\frac{(a+b)(ab)}{(ab)^2} < 0\) You choose to cancel a factor of (a b) in the numerator and denominator. That's definitely a legal move to do, but it's doubtful whether that is a particularly strategic thing to do. In fact, in terms of strategy, in terms of getting us closer to the answer, I think it's a not very good move. By the way, canceling is not called " deducing," at least in American mathematical terms. You see, that act of canceling leaves us with: \(\frac{a+b}{ab} < 0\) But the trouble is, with just statement #1 for the expressions a, b, (a + b), and (a  b): we don't know whether any of them are positive or negative or zero. That means if we multiply both sides by the denominator (a  b), we don't know whether we are multiplying by a positive or a negative number, so we don't know whether we reverse the inequality sign. Technically, at that stage, we would have to create two different scenarios, one in which we specified that (a  b) > 0, and the other in which we specified that (a  b) < 0, and the follow out the consequences of each case. We could do all that, but that's a very long, very complicated way to get to answer. That's precisely why that initial canceling move was a notverystrategic thing to do. What I did was considerably more elegant in terms of strategy. Noting that the original denominator, (a  b)^2, is a square and therefore always positive, we can just multiply both sides by that, we are left with the fact that a^2  b^2 < 0. That immediate leads to the important statement a^2 < b^2, which is very helpful in answering the question. Don't assume, because you can do something, because a move is allowed by the laws of algebra, that just because of this, the move is a particularly useful, a particularly strategic move. There are always dozens of moves that are perfectly legal and disastrous in terms of strategy. Does all this make sense? Mike Ohk..! Great insights..! Thanks ..!
_________________
ITS NOT OVER , UNTIL I WIN ! I CAN, AND I WILL .PERIOD.



Intern
Joined: 30 Aug 2015
Posts: 31
Concentration: Marketing, Finance
WE: Brand Management (Manufacturing)

Re: If a and b are distinct integers and their product is not equal to
[#permalink]
Show Tags
01 Nov 2015, 09:28
(a+b) AND (ab) are negative or (a+b) is positive AND (ab) are negative 1.ns 2.ns hence C.
_________________
Please award kudos if you like my explanation. Thanks



Manager
Joined: 02 Aug 2017
Posts: 67
Concentration: Strategy, Nonprofit
GPA: 3.71

Re: If a and b are distinct integers and their product is not equal to
[#permalink]
Show Tags
30 Aug 2018, 18:01
Kudos please, if it helps Posted from my mobile device
Attachments
15356808401331381215439420859325.jpg [ 1.93 MiB  Viewed 470 times ]
_________________
Everything is in flux, nothing stays still
MGMAT1 :590 Q42 V30 (07/07/18) VERITAS :660 Q48 V33 (16/07/18) GMATPREP1 :690 Q46 V36 (22/07/18) GMATPREP2 :740 Q51 V39 (06/08/18) ECONOMIST :740 Q49 V44 (11/08/18) KAPLAN :690 Q49 V36 (17/08/18) PRINCETON :690 Q48 V38 (26/08/18) MGMAT2 :720 Q43 V45 (02/09/18)




Re: If a and b are distinct integers and their product is not equal to &nbs
[#permalink]
30 Aug 2018, 18:01






