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Manager  Joined: 11 Jul 2016
Posts: 78
Is b < 0 ?  [#permalink]

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5 00:00

Difficulty:   35% (medium)

Question Stats: 64% (01:26) correct 36% (01:21) wrong based on 123 sessions

### HideShow timer Statistics Is b < 0 ?

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

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Manonamission wrote:
Is b<0 ?

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

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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Manager  Joined: 11 Jul 2016
Posts: 78
Re: Is b < 0 ?  [#permalink]

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abhimahna wrote:
Manonamission wrote:
Is b<0 ?

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

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?
Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Re: Is b < 0 ?  [#permalink]

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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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Manonamission wrote:
abhimahna wrote:
Manonamission wrote:
Is b<0 ?

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

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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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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Joined: 20 Oct 2016
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(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.
Math Expert V
Joined: 02 Sep 2009
Posts: 56260
Re: Is b < 0 ?  [#permalink]

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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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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:

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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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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Joined: 02 Sep 2009
Posts: 56260
Re: Is b < 0 ?  [#permalink]

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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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Re: Is b < 0 ?  [#permalink]

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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.

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 ?  [#permalink]

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Manonamission wrote:
Is b < 0 ?

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

Great Sum to explore the anomaly of [-1,1]
#1
(-2)^3 < (-2)
(.5)^3 < (.5) Insufficient

#2
2^2 > 2
(-2)^2 > 2. Insufficient.

#1+2

.5^2 can't be more than .5, But
-2^2 is always more than -2, so b is -ve. C
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