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Is a > c?

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Is a > c? [#permalink] New post 13 Aug 2012, 11:06
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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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Last edited by Bunuel on 14 Aug 2012, 00:14, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Can we add inequalities [#permalink] New post 13 Aug 2012, 11:33
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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.

Answer C
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Re: Is a > c? [#permalink] New post 14 Aug 2012, 00:16
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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] New post 14 Aug 2012, 06:45
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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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Re: Is a > c? [#permalink] New post 15 Aug 2012, 09:29
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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Re: Can we add inequalities [#permalink] New post 12 Aug 2013, 07:05
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.

Answer C


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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Re: Can we add inequalities [#permalink] New post 12 Aug 2013, 09:22
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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.

Answer C


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 anda > c
as if b = 0 then LHS = 0 .
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Re: Can we add inequalities [#permalink] New post 29 Aug 2013, 12:32
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.

Answer C




I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.
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Re: Can we add inequalities [#permalink] New post 30 Aug 2013, 04:49
Expert's post
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.

Answer C




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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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Can we add inequalities [#permalink] New post 30 Aug 2013, 23:14
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] New post 15 Oct 2014, 14:49
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Re: Is a > c?   [#permalink] 15 Oct 2014, 14:49
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