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Is a > c?

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Is a > c?  [#permalink]

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New post 14 Dec 2013, 13:13
2
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

68% (01:42) correct 32% (01:44) wrong based on 195 sessions

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Is a > c?

(1) b > d
(2) a*b^2 - b > (b^2)*c - d
Magoosh GMAT Instructor
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Re: Is a > c?  [#permalink]

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New post 14 Dec 2013, 13:47
jjack0310 wrote:
Is a > c?

(1) b > d
(2) a*(b^2) - b > (b^2)*c - d

The correct answer that I have is supposed to be C, but I don't know how. Thanks
Sorry if this has been posted before. I tried searching and could not find anything.

Dear jjack0310,
This is a great question and I am happy to help. :-)

First of all, you may find this post on DS & inequalities helpful:
http://magoosh.com/gmat/2013/gmat-quant ... qualities/

I will take it for granted that each of the statements individually is not sufficient. Let's focus on what happens when we combine the statements.

In statement #2, I will add b to both sides, and subtract (b^2)*c. It just seems like a good idea to get both (b^2) terms on the same side.
a*(b^2) - (b^2)*c > b - d
Now, notice that since b > d, we know that (b - d) > 0
a*(b^2) - (b^2)*c > b - d > 0
Then, notice, we can factor out (b^2) from the far left expression:
(b^2)*(a - c) > b - d > 0
This means
(b^2)*(a - c) > 0
Now, we can divide both sides by (b^2). Ordinarily, dividing both side of an inequality by a variable is a dicey business, because in general, a variable could be positive or negative, but here, we are guaranteed that, whether b is positive or negative, (b^2) must be a positive number, and therefore we can divide by it and definitely not alter the order of the inequality.
a - c > 0
a > c

Thus, from the combined information, we were able to deduce the prompt statement. Combined, the statements are sufficient. OA = (C).

That is a really tricky bit of algebra this question demands, but that's what the harder problems on the GMAT could demand. Please let me know if you have any further questions.

Mike :-)
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Re: Is a > c?  [#permalink]

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New post 17 Feb 2015, 07:22
In this question, how do we know b is not equal to zero? Statement 1 tells us b>d, but it could be that b = 0 and d = -1. If that were the case and b=0, then we can't divide by b^2 when combining the statements.
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Re: Is a > c?  [#permalink]

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New post 17 Feb 2015, 10:49
ericmshields wrote:
In this question, how do we know b is not equal to zero? Statement 1 tells us b > d, but it could be that b = 0 and d = -1. If that were the case and b=0, then we can't divide by b^2 when combining the statements.

Dear ericmshields,
That's a great question, and I am happy to help. :-)

In the calculations above, obviously, I didn't not include the possibility that b = 0. What if this is true?

Let's assume b = 0. Then d would have to be some negative number, say, d = -1. Let's assume both statements are true. Look at statement #2: this becomes ---
a*(0^2) - 0 > (0^2)*c - d
0 - 0 > 0 - d
0 > -d
Now, if we multiply both sides by a negative sign, this reverses the order of the inequality:
0 < d
This is a direct contradiction of our assumptions. In other words, if we assume b = 0, it leads directly to a contradiction between the two statements. Because of this, if we assume that both statements are true, it automatically means that b can't possibly equal zero. Therefore, we are 100% free to divide by (b^2), because we are guaranteed that it can't possibly be zero.

In general, considering what happens when a variable equals zero is very important. Technically, we can never simply assume that a variable isn't zero, and we can never simply assume that a variable could equal zero. Both assumptions are things we need to investigate.

Does all this make sense?
Mike :-)
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Mike McGarry
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Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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Re: Is a > c?  [#permalink]

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New post 20 Oct 2016, 11:30
(1) b>d, not sufficient

(2) ab² - b > b²c - d, not sufficient

(combined), sufficient
from (1) : b-d>0
from (2) : b² (a-c)>b-d
so, b²(a-c)>0
b² is >0, so (a-c) is necessarily >0
a-c>0 => a>c
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Re: Is a > c?  [#permalink]

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New post 03 Feb 2019, 10:12
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Re: Is a > c?   [#permalink] 03 Feb 2019, 10:12
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