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–1 < a < 0 < |a| < b < 1

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–1 < a < 0 < |a| < b < 1 [#permalink] New post 10 Aug 2018, 11:29
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35% (01:52) correct 64% (00:59) wrong based on 76 sessions
\(–1 < a < 0 < |a| < b < 1\)

Quantity A
Quantity B
\((\frac{a^2 \sqrt{b}}{\sqrt{a}})^2\)
\(\frac{ab^5}{(\sqrt{b})^4}\)


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.

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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 11 Aug 2018, 01:07
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This question can be solved just by looking at the sign that comes up in the end or I am wrong?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 14 Aug 2018, 15:42
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Not really.

We do know that a is a negative fraction and b is a positive fraction and a little more positive on the number line than |a|.
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 10 Sep 2018, 09:51
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obviously a<0 & b>0.

simplifying quantities A & B gives a^3*b and ab^3, respectively, both of which are negative.

taking a=-0.25, b=0.5, and plugging into the above simplified expressions, we get the answer A.
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 16 Sep 2018, 13:05
saifee wrote:
obviously a<0 & b>0.

simplifying quantities A & B gives a^3*b and ab^3, respectively, both of which are negative.

taking a=-0.25, b=0.5, and plugging into the above simplified expressions, we get the answer A.



What if a= -(1/8) and b= (1/2) ??

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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 23 Sep 2018, 21:17
I cant solve this, can someone help?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 24 Sep 2018, 08:16
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fixzion wrote:
I cant solve this, can someone help?


Both \(a\) and \(b\) are fractions. Now you have to select values of a and b such that \(b>a\).

\(a= \frac{-1}{4}\) and \(b= \frac{1}{2}\).

Quantity A
Quantity B
\((\frac{a^2 \sqrt{b}}{\sqrt{a}})^2\)
\(\frac{ab^5}{(\sqrt{b})^4}\)


or

Quantity A
Quantity B
\((\frac{a^4 b}{a})\)
\(\frac{ab^5}{b^2}\)



or

Quantity A
Quantity B
\(a^3b\)
\(ab^3\)



\(a^3b=(\frac{-1}{4})^3\frac{1}{2}=\frac{-1}{128}\)

\(ab^3=\frac{-1}{4}(\frac{1}{2})^3=\frac{-1}{32}\).

A is the larger number.
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Re: –1 < a < 0 < |a| < b < 1   [#permalink] 24 Sep 2018, 08:16
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