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# Is a > c?

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Is a > c? [#permalink]

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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:
Can we add (1) and (2) to get the answers.
[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:
Can we add (1) and (2) to get the answers.

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:
Can we add (1) and (2) to get the answers.

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
2
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Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

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, 10:22
1
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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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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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29 Aug 2013, 13:32
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

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
1
This post was
BOOKMARKED
swati007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

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

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

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

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

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.

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Re: Is a > c? [#permalink]

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21 Nov 2016, 23:51
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Re: Is a > c?   [#permalink] 21 Nov 2016, 23:51
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