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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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\(–1 < a < 0 < |a| < b < 1\)

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 Array[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 Array[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] New post 16 Sep 2019, 18:09
sandy Array[WROTE]:
fixzion Array[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.


Good one
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 23 Oct 2019, 14:40
sandy Array[WROTE]:
fixzion Array[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.




Without putting in values to bring in the calculation, I think another way could be:

a^3.b a.b^3

so the comparison is between
a^2 & b^2

as we know |a| < b, and a,b both are fractions(positive or negative, doesn't really matter), 'b' is the larger fraction in magnitude hence would produce a smaller squared fraction

Is anything wrong with this approach?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 23 Oct 2019, 15:00
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Not really :)
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 31 Oct 2019, 02:42
Can we plug in values for a and b with out simplifications?
we will have negative square root for a?

Last edited by mohmu123 on 31 Oct 2019, 02:59, edited 1 time in total.
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 31 Oct 2019, 02:48
mohmu123 Array[WROTE]:
Can we plug in values for a and be with out simplifications?
we will have negative square root for a?


b, or be, which one u refer?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 31 Oct 2019, 03:00
I meant a and b
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post 31 Oct 2019, 03:19
mohmu123 Array[WROTE]:
Can we plug in values for a and b with out simplifications?
we will have negative square root for a?


Yes u can. For Negative value, just 1st put the only values without negative connotation then compare with ur result u got by putting values with the negative connotation which one u put aside before checking the values
Re: –1 < a < 0 < |a| < b < 1   [#permalink] 31 Oct 2019, 03:19
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