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# Is a > c?

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13 Aug 2012, 12:06
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Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
[Reveal] Spoiler: OA

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Thanks Rphardu

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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13 Aug 2012, 12:33
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rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:

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, 01:16
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rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:

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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Re: Is a > c? [#permalink]

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

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:

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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Re: Is a > c? [#permalink]

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15 Aug 2012, 10:29
1 &2 combo-

a(b^2)-b-(b^2)c+d>0
(b^2)(a-c)-(b-d)>0

note, that 1 states that b>d. in order to make the expression above positive a must be > c
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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
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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

[Reveal] Spoiler:

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
Expert's post
swati007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:

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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31 Aug 2013, 00:14
Quote:

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.

Wonderful explanation!!! Thanks Bunuel
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Re: Is a > c? [#permalink]

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15 Oct 2014, 15:49
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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

[Reveal] Spoiler:

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
Expert's post
metskj127 wrote:
Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:

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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16 Jun 2015, 13:58
Never mind- I see my mistake now. Thank you for the help.
Re: Is a > c?   [#permalink] 16 Jun 2015, 13:58
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