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Is ab < 0? (1) ab^2 < b^2 (2) b^2 < b

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Is ab < 0? (1) ab^2 < b^2 (2) b^2 < b [#permalink]

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New post 13 Sep 2017, 21:44
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A
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C
D
E

Difficulty:

  45% (medium)

Question Stats:

65% (01:09) correct 35% (00:47) wrong based on 57 sessions

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Is ab < 0? (1) ab^2 < b^2 (2) b^2 < b [#permalink]

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New post 13 Sep 2017, 22:03
Bunuel wrote:
Is \(ab < 0\)?


(1) \(ab^2 < b^2\)

(2) \(b^2 < b\)


Statement 1: \(ab^2-b^2<0\) or \(b^2(a-1)<0\). as \(b^2\) is always positive, this implies that \(a-1<0\)
or \(a<1\)--------------(1). but we don't know whether \(a\) is positive or negative and we don't have any information about \(b\). Hence Insufficient

Statement 2: \(b^2<b\). this implies that \(b\) is positive and it is less than \(1\). so \(0<b<1\)
But we have no information about \(a\). Hence Insufficient

Combining 1 & 2 we don't know the sign of \(a\). if \(0<a<1\), then \(ab>0\) and if \(a<0\), then \(ab<0\). Hence insufficient

Option E

Last edited by niks18 on 14 Sep 2017, 22:54, edited 2 times in total.

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

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New post 14 Sep 2017, 19:06
Statement 1 - ab^2<b^2. On simplifying ab^2-b^2<0, further simplifying b^2(a-1)<0, which makes a<1 & b^ can never be negative as it is a square of a number. Now since a<1 means for a=-1, ab will be <0 but if a can be 0 also in which case ab=0 and hence not sufficient.
Statement 2 - b^2<b. Simplifying, b^2-b<0, further simplifying b(b-1)<0. This states that b<1 and as b<0 also falls in the range but this only states about the polarity of b and since the main equation has dependence on value of a also which is not covered here, hence this statement is also insufficient.
Statement 1+2 - states the polarity of b as negative but the dual possibility of a (i.e a can be 0 as well as negative) from Statement 1.hence in either case either ab=0 or ab>0 since (-ve)*(-ve)=+ve, so ab is not <0. hence C.
What is the original OA.

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New post 14 Sep 2017, 19:17
Also from statement 2 b can never be negative as the equation is of a parabola which only touches the x axis at the origin. So either ab=0 or ab<0 so the statements are insufficient. So answer is E.
Just made a correction to my answer. Sorry for the change in answer.

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

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New post 14 Sep 2017, 20:50
E.
From statement 1, a can be either less than zero or between 0 and 1. Not sufficient.
From statement 2, b is between 0 and 1. Not sufficient because no information about a.
Combine 1+2 is not sufficient to answer.

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

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New post 18 Sep 2017, 13:21
Bunuel wrote:
Is \(ab < 0\)?


(1) \(ab^2 < b^2\)

(2) \(b^2 < b\)


Lets try to solve for C.
1) & 2)
a*b^2 - b^2 <0
Hence a<1

b^2-b<0
Hence b(b-1)<0 (only true when b<1)

So now we have a<1 and b<1
Both conditions together are not sufficient.
Hence if answer is not C, then it has to be E.

Ans. : E

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Re: Is ab < 0? (1) ab^2 < b^2 (2) b^2 < b   [#permalink] 18 Sep 2017, 13:21
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