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ab2c3 > 0 a2b2c3 < 0

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ab2c3 > 0 a2b2c3 < 0 [#permalink] New post 18 Sep 2018, 04:33
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Question Stats:

70% (00:28) correct 29% (01:07) wrong based on 17 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?
[Reveal] Spoiler: OA
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Re: ab2c3 > 0 a2b2c3 < 0 [#permalink] New post 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] New post 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] New post 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] New post 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] New post 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] New post 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
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