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# Is a > c?

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Updated on: 14 Aug 2012, 01:14
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Question Stats:

69% (01:34) correct 31% (01:50) wrong based on 547 sessions

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Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

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Thanks Rphardu

Originally posted by rphardu on 13 Aug 2012, 12:06.
Last edited by Bunuel on 14 Aug 2012, 01:14, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Is a > c?  [#permalink]

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14 Aug 2012, 01:16
7
8
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.
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13 Aug 2012, 12:33
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1
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

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Re: Is a > c?  [#permalink]

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14 Aug 2012, 07:45
2
Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

When we subtract two inequalities with their signs in opposite directions, we are in fact using addition of two inequalities in the same direction:

$$a>b$$

$$C<D$$ -> this can be rewritten as

$$-C>-D$$

Now we can add the first and the third inequality, because they have the same direction and get $$a-C>b-D.$$
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12 Aug 2013, 08:05
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?
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12 Aug 2013, 10:22
1
nikhil007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?

From (1) and (2) we see $$b^2(a-c)>0$$

Since LHS >0 we must have $$b =! 0$$ and$$a > c$$
as if $$b = 0$$ then LHS = 0 .
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29 Aug 2013, 13:32
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.
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30 Aug 2013, 05:49
swati007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.

We have $$b^2(a-c)>0$$ ($$b\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.

Hope it's clear.
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Re: Is a > c?  [#permalink]

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14 Dec 2013, 14:47
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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17 Feb 2015, 08:22
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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17 Feb 2015, 11:49
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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Re: Is a > c?  [#permalink]

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16 Jun 2015, 13:39
Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

What if b=0 and d=-1?

In that situation, wouldn't the 2nd equation become:

a(0) > (0)c – d
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Re: Is a > c?  [#permalink]

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16 Jun 2015, 13:50
metskj127 wrote:
Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

What if b=0 and d=-1?

In that situation, wouldn't the 2nd equation become:

a(0) > (0)c – d

If b = 0 and d = -1, then ab^2 – b = 0 and b^2c – d = 1. Anyway, what are you trying to say?
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Re: Is a > c?  [#permalink]

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21 Jan 2018, 22:48
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

Question asks whether a > c or Is a-c > 0

(1) First statement is obviously not sufficient. Just b>d or b-d > 0 but nothing about a and c is mentioned.

(2) Second statement can be rewritten as:
ab^2 – b^2c > b – d
b^2*(a-c) > (b-d)
Here b^2 cannot be negative but whether a-c is positive or not depends on b-d also (on right hand side). Nothing is mentioned about that so insufficient.

Combining the two statements, from first we know that b-d is positive and since left hand side: b^2*(a-c) is greater than b-d so b^2*(a-c) also must be positive.
Now b^2 cannot be negative. So for b^2*(a-c) to be positive, a-c also must be positive. Which means a-c > 0 or a > c. Sufficient.

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Re: Is a > c?  [#permalink]

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23 Jan 2018, 23:47
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 4 variables (a, b, c and d) and 0 equations, E is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):

b > d
ab^2 - b > b^2c - d
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 > 0 since b > d and b^2 > 0 if b≠0.
⇔ a > c

If b = 0, we have d < 0
ab^2 - b > b^2c - d
⇔ 0 > -d which contradicts b > d
Thus b≠0.

Both conditions together are sufficient.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1):
We don't have any information about a and c from the condition 1) only.
The condition 1) only is not sufficient.

Condition 2):
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 since b^2 > 0
⇔ a - c > b - d
a = 2, c = 1, b = 1, d = 1 : Yes
a = 1, c = 2, b = 0, d = 1 : No
The condition 2) only is not sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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Re: Is a > c?  [#permalink]

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26 Jan 2018, 04:29
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