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Is b < 0 ?

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Is b < 0 ?  [#permalink]

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New post 08 Oct 2016, 05:55
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Is b < 0 ?

(1) b^3 < b
(2) b^2 > b
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Re: Is b < 0 ?  [#permalink]

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New post 08 Oct 2016, 06:03
1
Manonamission wrote:
Is b<0 ?

(1) b^3 < b
(2) b^2 > b

Please help in understanding the solution?


Statement 1 : \(b^3 < b\), here b could be a fraction less than 1 or it could be a negative integer. Hence, insufficient.

Statement 2 : \(b^2 > b\), here b could be negative integer or a positive integer but not a fraction less than 1. Hence, insufficient.

Combining , From 1 and 2 together, we can say b canot be a fraction or a positive integer, hence, it MUST be -ve. Sufficient. Answer C.
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Re: Is b < 0 ?  [#permalink]

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New post 08 Oct 2016, 06:55
abhimahna wrote:
Manonamission wrote:
Is b<0 ?

(1) b^3 < b
(2) b^2 > b

Please help in understanding the solution?


Statement 1 : \(b^3 < b\), here b could be a fraction less than 1 or it could be a negative integer. Hence, insufficient.

Statement 2 : \(b^2 > b\), here b could be negative integer or a positive integer but not a fraction less than 1. Hence, insufficient.

Combining , From 1 and 2 together, we can say b canot be a fraction or a positive integer, hence, it MUST be -ve. Sufficient. Answer C.
ou

Can you please specify how b could be a fraction < 1 or could be a negative integer via mathematical calculation?
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Re: Is b < 0 ?  [#permalink]

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New post 08 Oct 2016, 07:20
Manonamission wrote:
Is b < 0 ?

(1) b^3 < b
(2) b^2 > b


Hi,

(1) \(b^3<b.......b^3-b<0......b(b^2-1)<0\)
Two cases..
b is positive then b^2-1<0 or b^2<1.... b is between 0and1.. possible
b is negative then b^2-1>0 or b^2>1...b is less than -1... possible
So b can be >0 or <0..
Insufficient..

(2) \(b^2>b.....b^2-b>0.......b(b-1)>0\)
Again two cases..
b is positive, then b-1also has to be >0....b-1>0....b>1.... possible
b is negative, then b-1<0....b<1....... possible
Insufficient..

Combined..
I) if b is positive, FROM statement 1, b should be between 0&1 AND FROM statement 2, b>1.... not possible
II) if b is negative, FROM statement 1, b should be less than -1 AND FROM statement 2, b is less than 1.... possible
Means b is less than -1

Sufficient... b<0....
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Re: Is b < 0 ?  [#permalink]

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New post 08 Oct 2016, 07:46
1
Manonamission wrote:
abhimahna wrote:
Manonamission wrote:
Is b<0 ?

(1) b^3 < b
(2) b^2 > b

Please help in understanding the solution?


Statement 1 : \(b^3 < b\), here b could be a fraction less than 1 or it could be a negative integer. Hence, insufficient.

Statement 2 : \(b^2 > b\), here b could be negative integer or a positive integer but not a fraction less than 1. Hence, insufficient.

Combining , From 1 and 2 together, we can say b canot be a fraction or a positive integer, hence, it MUST be -ve. Sufficient. Answer C.
ou

Can you please specify how b could be a fraction < 1 or could be a negative integer via mathematical calculation?


\(b^3 < b\) can be written as \(b^3 - b < 0\) or \(b(b^2 -1)<0\)

This would mean b could be anything out of b<0 or b<1 or b<-1. That means b could be 1/2 also. In all the cases the equation will be satisfied.

Similarly, \(b^2 > b\) can be written as \(b^2 - b> 0\) or \(b(b-1)>0\), which means either b>1 or b<0.

Combining both the statements, I can say b < 0. Hence, Sufficient.
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Re: Is b < 0 ?  [#permalink]

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New post 10 Oct 2016, 00:34
Manonamission wrote:
Is b < 0 ?

(1) b^3 < b
(2) b^2 > b


from 1

b is a +ve fraction or b is -ve..insuff

from 2

1<b or b -ve

from both together the common denominator of b is b<0... suff
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Is b<0?  [#permalink]

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New post 21 Oct 2016, 01:18
(1) b(^3)<b
(2) b(^2)>b

that is a question from manhattan Algebra book. The answer the manual suggests is C, but I have a doubt. It's true that the first statement is insufficient, but what about the second one?

In my opinion, If b(^2)>b, and I divide both the left term and the right term for b, the result is b>0, so that's sufficient.

The book solve this question testing it with real numbers:
first using b=2, so 4>2.
then using b=-2, so 4>-2.
Because both a positive and a negative number are possible solutions, that is not sufficient.

So i'm asking why simplifying the inequality dividing for b is not the right method.

Thank you.
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Re: Is b < 0 ?  [#permalink]

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New post 21 Oct 2016, 01:25
Aletoz wrote:
(1) b(^3)<b
(2) b(^2)>b

that is a question from manhattan Algebra book. The answer the manual suggests is C, but I have a doubt. It's true that the first statement is insufficient, but what about the second one?

In my opinion, If b(^2)>b, and I divide both the left term and the right term for b, the result is b>0, so that's sufficient.

The book solve this question testing it with real numbers:
first using b=2, so 4>2.
then using b=-2, so 4>-2.
Because both a positive and a negative number are possible solutions, that is not sufficient.

So i'm asking why simplifying the inequality dividing for b is not the right method.

Thank you.


Merging topics. Please refer to the discussion above.
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Re: Is b < 0 ?  [#permalink]

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New post 21 Oct 2016, 01:30
Aletoz wrote:
(1) b(^3)<b
(2) b(^2)>b

that is a question from manhattan Algebra book. The answer the manual suggests is C, but I have a doubt. It's true that the first statement is insufficient, but what about the second one?

In my opinion, If b(^2)>b, and I divide both the left term and the right term for b, the result is b>0, so that's sufficient.

The book solve this question testing it with real numbers:
first using b=2, so 4>2.
then using b=-2, so 4>-2.
Because both a positive and a negative number are possible solutions, that is not sufficient.

So i'm asking why simplifying the inequality dividing for b is not the right method.

Thank you.


As for your question: we cannot divide b^2 > b by b because we don't know the sign of b. If b > 0, then when dividing we get b > 1 BUT if b < 0, then when dividing we get b < 1 (recall that we should flip the sign of an inequality if we multiply/divide it by negative value).

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.

For more check the links below:
Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach
Inequality tips
Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

Hope it helps.
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Re: Is b < 0 ?  [#permalink]

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New post 18 Jul 2018, 00:54
Manonamission wrote:
Is b < 0 ?

(1) b^3 < b
(2) b^2 > b


Dear Moderator,
I have another question , why is another rule of inequality not working here :

"When two inequalities have different signs , they can be subtracted , and we take the sign of the inequality from which we subtract".

hence \(b^3- b^2 < 0\)

\(b^2(b-1) <0\)
here b can be 1/2 or any negative number , hence this way after combining 1+2 we get that b can be >0 or b<0 hence insuff.

Where/what is the flaw in the above thank you.
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Re: Is b < 0 ?  [#permalink]

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New post 18 Jul 2018, 01:07
1
stne wrote:
Manonamission wrote:
Is b < 0 ?

(1) b^3 < b
(2) b^2 > b


Dear Moderator,
I have another question , why is another rule of inequality not working here :

"When two inequalities have different signs , they can be subtracted , and we take the sign of the inequality from which we subtract".

hence \(b^3- b^2 < 0\)

\(b^2(b-1) <0\)
here b can be 1/2 or any negative number , hence this way after combining 1+2 we get that b can be >0 or b<0 hence insuff.

Where/what is the flaw in the above thank you.


You subtracted correctly.

\(b^2(b - 1) < 0\) gives b < 1 (b ≠ 0). But we have two other inequalities and we cannot ignore them. I'll let you find what ranges each of them gives and what is the intersection of these ranges.
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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

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Re: Is b < 0 ?  [#permalink]

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New post 18 Jul 2018, 02:08
Bunuel wrote:
stne wrote:
Manonamission wrote:
Is b < 0 ?

(1) b^3 < b
(2) b^2 > b


Dear Moderator,
I have another question , why is another rule of inequality not working here :

"When two inequalities have different signs , they can be subtracted , and we take the sign of the inequality from which we subtract".

hence \(b^3- b^2 < 0\)

\(b^2(b-1) <0\)
here b can be 1/2 or any negative number , hence this way after combining 1+2 we get that b can be >0 or b<0 hence insuff.

Where/what is the flaw in the above thank you.


You subtracted correctly.

\(b^2(b - 1) < 0\) gives b < 1 (b ≠ 0). But we have two other inequalities and we cannot ignore them. I'll let you find what ranges each of them gives and what is the intersection of these ranges.



Right , after reading your response , the answer just hit me .
I should have brought all the variables to one side and then subtracted to get the correct range of b.

\(b^3 < b\)
\(b^3 - b <0\)
\(b(b^2 -1)<0\)
here b can be b <-1 and 0<b<1 so Insuff.

\(b^2 > b\)
\(b^2 - b >0\)
b(b-1)>0
here b<0 or b >1 hence insuff.

so clearly intersection of these ranges is b<-1

My Flaw : while combining both equations instead of bringing all the variables to one side and subtracting as below:
( Subtraction statement 2 from 1 after bringing all the variables to left side )
\(b(b^2 -1) - (b^2 - b) < 0\)
after simplification, the above eqn. becomes \(b(b^2-1-b)+1<0\) we find that only for b< -1 this eqn is satisfied.

I was simply doing \(b^3- b^2 <b-b\)
\(b^3- b^2 <0\)
\(b^2(b-1) <0\) so this eqn. does not give the true values
rather \(b(b^2-1-b)+1<0\) gives the true values of b.
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Re: Is b < 0 ? &nbs [#permalink] 18 Jul 2018, 02:08
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