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ab2c3 > 0 a2b2c3 < 0 [#permalink]
18 Sep 2018, 04:33
Question Stats:
61% (00:35) correct
38% (00:47) wrong based on 47 sessions
\(ab^2c^3 > 0\) \(a^2b^2c^3 < 0\)
Quantity A 
Quantity B 
\(ab^2\) 
\(ac\) 
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given. How to approach this question?




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Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]
18 Sep 2018, 04:46
the first inequality is true if a<0 & c<0 or a>0 & c>0.
second inequality is true if c<0.
that's why it must be that a<0.
so we have a<0, b^2>0, and c<0.
thus the answer is B.



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Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]
18 Sep 2018, 04:54
saifee wrote: the first inequality is true if a<0 & c<0 or a>0 & c>0.
second inequality is true if c<0.
that's why it must be that a<0.
so we have a<0, b^2>0, and c<0.
thus the answer is B. I did not understand this  > " that's why it must be that a<0"



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Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]
18 Sep 2018, 05:00
read the first line. because we've got c<0 from the second inequality, we must conclude a<0, too.



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Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]
18 Sep 2018, 15:53
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Start from the second inequality \(a^2b^2c<0\). Since \(]a^2b^2\) cannot be less than zero, as they are squares. c must be less than 0. Now with that information look at the first inequality. \(ab^2c^3>0\) since c<0 a must be less than 0.
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Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]
19 Sep 2018, 09:05
sandy wrote: Start from the second inequality
\(a^2b^2c<0\). Since \(]a^2b^2\) cannot be less than zero, as they are squares. c must be less than 0. Now with that information look at the first inequality.
\(ab^2c^3>0\) since c<0 a must be less than 0. If we in the beginning itself, divide both the sides by "a", then we will have \(b^2\) and \(c\) left, in that case, wouldn't A become bigger?



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Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]
19 Sep 2018, 17:25
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rajlal wrote: If we in the beginning itself, divide both the sides by "a", then we will have \(b^2\) and \(c\) left, in that case, wouldn't A become bigger?
You cant divide with a if you dont know its sign. Division with a negative number would flip the inequality. For example 1 < 5 divide both sides with 1 you get 1> 5.
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Re: ab2c3 > 0 a2b2c3 < 0
[#permalink]
19 Sep 2018, 17:25





