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Is b < 0 ?
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08 Oct 2016, 06:55
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64% (01:26) correct 36% (01:21) wrong based on 123 sessions
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Is b < 0 ? (1) b^3 < b (2) b^2 > b
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Re: Is b < 0 ?
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08 Oct 2016, 07:03
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 ?
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08 Oct 2016, 07:55
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 ?
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08 Oct 2016, 08:20
Manonamission wrote: Is b < 0 ?
(1) b^3 < b (2) b^2 > b Hi, (1) \(b^3<b.......b^3b<0......b(b^21)<0\) Two cases.. b is positive then b^21<0 or b^2<1.... b is between 0and1.. possible b is negative then b^21>0 or b^2>1...b is less than 1... possible So b can be >0 or <0.. Insufficient.. (2) \(b^2>b.....b^2b>0.......b(b1)>0\) Again two cases.. b is positive, then b1also has to be >0....b1>0....b>1.... possible b is negative, then b1<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 ?
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08 Oct 2016, 08:46
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 \(b(b1)>0\), 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 ?
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10 Oct 2016, 01:34
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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(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 ?
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21 Oct 2016, 02:25
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 ?
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21 Oct 2016, 02:30
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 ApproachInequality tipsWavy Line Method Application  Complex Algebraic InequalitiesDS Inequalities Problems PS Inequalities Problems 700+ Inequalities problemsinequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.htmleverythingislessthanzero108884.htmlgraphicapproachtoproblemswithinequalities68037.htmlHope it helps.
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Re: Is b < 0 ?
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18 Jul 2018, 01:54
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 \(b^3 b^2 < 0\) \(b^2(b1) <0\) 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 ?
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18 Jul 2018, 02:07
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 \(b^3 b^2 < 0\) \(b^2(b1) <0\) 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 ?
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18 Jul 2018, 03:08
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 \(b^3 b^2 < 0\) \(b^2(b1) <0\) 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(b1)>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 \(b(b^21b)+1<0\) we find that only for b< 1 this eqn is satisfied. I was simply doing \(b^3 b^2 <bb\) \(b^3 b^2 <0\) \(b^2(b1) <0\) so this eqn. does not give the true values rather \(b(b^21b)+1<0\) gives the true values of b.
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Re: Is b < 0 ?
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27 Jun 2019, 08:49
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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