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....

\(b^3 < b\) 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.

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

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.

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.

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

Target question: Is b < 0 ?

Statement 1: b³ < b
There are several values of b that satisfy statement 1. Here are two:
Case a: b = -2 (which means b³ = -8, and -8 < -2). In this case, the answer to the target question is YES, b IS less than 0
Case b: b = 1/2 (which means b³ = 1/8, and 1/8 < 1/2). In this case, the answer to the target question is NO, b is NOT less than 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: b² > b
There are several values of b that satisfy statement 2. Here are two:
Case a: b = -1 (which means b² = 1, and 1 > -1). In this case, the answer to the target question is YES, b IS less than 0
Case b: b = 2 (which means b² = 4, and 4 > 2). In this case, the answer to the target question is NO, b is NOT less than 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that b³ < b
Statement 2 tells us that b² > b
Combine the inequalities to get: b³ < b < b²
This tells us that b³ < b²
Subtract b² from both sides to get: b³ - b² < 0
Factor to get: b(b² - b) < 0
If b(b² - b) < 0, then there are TWO POSSIBLE CASES:
Case a: b < 0 and (b² - b) > 0
OR
Case b: b > 0 and (b² - b) < 0
If we take statement 2 (b² > b) and subtract b from both sides, we get: b² - b > 0
This RULES OUT case b (above), which means case a must be true.
If case a is true, then it must be the case that b < 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

RELATED VIDEO

can someone explain this using wavy line method?
@karsihmaveritas egmat

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