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–1 < a < 0 < a < b < 1 [#permalink]
10 Aug 2018, 11:29
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Question Stats:
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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. Kudos for R.A.E
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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
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) ?? Posted from my mobile device



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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
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





