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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. _________________
(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
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.
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. _________________
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. _________________
Is b < 0 ?
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Updated on: 17 May 2021, 07:36
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Manonamission wrote:
Is b < 0 ?
(1) b³ < b (2) b² > b
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
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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