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# If a and b are distinct integers and their product is not equal to

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If a and b are distinct integers and their product is not equal to  [#permalink]

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01 Sep 2014, 12:09
1
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60% (01:50) correct 40% (02:15) wrong based on 217 sessions

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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
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Joined: 28 Dec 2011
Posts: 4666
Re: If a and b are distinct integers and their product is not equal to  [#permalink]

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01 Sep 2014, 14:38
9
3
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/three-alge ... -the-gmat/
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.

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)

##### General Discussion
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Joined: 25 Mar 2014
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Re: If a and b are distinct integers and their product is not equal to  [#permalink]

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01 Sep 2014, 14:34
1
1
$$\frac{(ab(a^2-b^2))}{(ab(a^2-2ab+b^2)}$$

$$\frac{((a+b)(a-b))}{((a-b)^2)}$$

$$\frac{(a+b)}{(a-b)}<0$$

(a+b) AND (a-b) are negative
or
(a+b) is positive AND (a-b) are negative

1.ns
2.ns

hence C.
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Re: If a and b are distinct integers and their product is not equal to  [#permalink]

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03 Oct 2014, 12:52
2
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/three-alge ... -the-gmat/
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.

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)(a-b) / (a-b)^2 < 0 ,

then after deducing,we are left with-

a+b/a-b < 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..
_________________

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Joined: 28 Dec 2011
Posts: 4666
Re: If a and b are distinct integers and their product is not equal to  [#permalink]

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03 Oct 2014, 13:25
3
vards wrote:
Hi there,
my doubt may sound a bit off,but here it is..

when we reach at this stage= (a+b)(a-b) / (a-b)^2 < 0 ,

then after deducing,we are left with-

a+b/a-b < 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:
$$\frac{(a+b)(a-b)}{(a-b)^2} < 0$$
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:
$$\frac{a+b}{a-b} < 0$$
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 not-very-strategic 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
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Re: If a and b are distinct integers and their product is not equal to  [#permalink]

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03 Oct 2014, 13:51
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)(a-b) / (a-b)^2 < 0 ,

then after deducing,we are left with-

a+b/a-b < 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:
$$\frac{(a+b)(a-b)}{(a-b)^2} < 0$$
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:
$$\frac{a+b}{a-b} < 0$$
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 not-very-strategic 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 ..!
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ITS NOT OVER , UNTIL I WIN ! I CAN, AND I WILL .PERIOD.

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Re: If a and b are distinct integers and their product is not equal to  [#permalink]

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01 Nov 2015, 10:28
4
1
(a+b) AND (a-b) are negative
or
(a+b) is positive AND (a-b) are negative

1.ns
2.ns

hence C.
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Thanks

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Re: If a and b are distinct integers and their product is not equal to  [#permalink]

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22 Jul 2018, 22:40
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