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

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Math Expert
Joined: 02 Sep 2009
Posts: 46305
Is ab < 0? (1) ab^2 < b^2 (2) b^2 < b [#permalink]

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13 Sep 2017, 21:44
00:00

Difficulty:

45% (medium)

Question Stats:

62% (01:15) correct 38% (00:47) wrong based on 78 sessions

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Is $$ab < 0$$?

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

(2) $$b^2 < b$$

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Joined: 25 Feb 2013
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Is ab < 0? (1) ab^2 < b^2 (2) b^2 < b [#permalink]

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Updated on: 14 Sep 2017, 22:54
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

Originally posted by niks18 on 13 Sep 2017, 22:03.
Last edited by niks18 on 14 Sep 2017, 22:54, edited 2 times in total.
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Joined: 19 Sep 2016
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Re: Is ab < 0? (1) ab^2 < b^2 (2) b^2 < b [#permalink]

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

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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.

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

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