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# –1 < a < 0 < |a| < b < 1

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–1 < a < 0 < |a| < b < 1 [#permalink]  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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[Reveal] Spoiler: OA

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Re: –1 < a < 0 < |a| < b < 1 [#permalink]  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?
bests xx
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Re: –1 < a < 0 < |a| < b < 1 [#permalink]  14 Aug 2018, 15:42
Expert's post
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]  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]  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]  23 Sep 2018, 21:17
I cant solve this, can someone help?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink]  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]  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]  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]  23 Oct 2019, 15:00
Expert's post
Not really
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Re: –1 < a < 0 < |a| < b < 1 [#permalink]  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]  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]  31 Oct 2019, 03:00
I meant a and b
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Re: –1 < a < 0 < |a| < b < 1 [#permalink]  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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