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–1 < a < 0 < |a| < b < 1

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safana
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   12 May 2020, 11:56
After simplifying both sides quantity A is a^2 while quantity B is b^2.
if i take the value of a to be -1/2 and b to be 3/4, after calculation im getting quantity B to be greater. Can someone tell me where im going wrong?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   12 May 2020, 12:28
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Please Sir.

read the explanation above made by the Tutor sujoykrdatta.

It is neat and clear
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   14 May 2020, 03:13
why the answer will not be "D" if we put a=-1/2; b=+1/2?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   14 May 2020, 14:18
if a is -1/2 then b cannot be +1/2, as b has to be a number between modulus a and 1. modulus a will be +1/2 if a is -1/2.
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   21 May 2021, 22:02
sandy wrote:
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.

Aint it wrong cosidering bigger value of a all the time?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   04 Jun 2021, 22:47
sandy wrote:
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.


sir, if we divide by ab on both side , we get, i am referring to third simplified form , QA: b*a^3 and QB: ab^3,
QA QB
a^2 b^2
then we have to take square root on both side ?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   22 Aug 2021, 05:53
Can someone please help me understand this,

Simplifying everything, I got Quantity A : a^2
Quantity B : b^2

I picked a = -0.5 and b = 0.9

In this case B is greater as 0.81 > 0.25
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   22 Aug 2021, 09:27
1
Quantity A: \(a^3b\), where \(a^3>0\)
Quantity B: \(ab^3\)

cancelling out \(ab\) results in
Quantity A: \(a^2\)
Quantity B: \(b^2\)

Since \(a>-1\), and \(0<b<1\), answer is A



Carcass wrote:
\(–1 < a < 0 < |a| < b < 1\)

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   02 Dec 2021, 06:15
1
Use the placeholder method here
Simplify it to obtain a^3b ? ab^3
Since a<0, the placeholder sign will flip. Now use a flip operator on the placeholder sign, say it is now ?' after the flip
a^2 ?' b^2, since |a|<b, a^2 < b^2
Hence, ?' is <, so ? must be >, hence A>B and that is the answer.
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1 < a < 0 < |a| < b < 1 [#permalink] New post   11 Jun 2022, 14:01
In the left equation, it has a square root of A, which is negative.
I think it will require a use of the imaginary numbers...
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Re: 1 < a < 0 < |a| < b < 1 [#permalink] New post   11 Jun 2022, 17:09
Paul121 wrote:
Is it mathematically legal question (for the GRE level)?
In the left equation, it has a square root of A, which is negative.
I think it will require a use of the imaginary numbers...


Great point!
I didn't consider that while composing my solution.
Given that, I don't think this could ever be an official GRE question.
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Re: 1 < a < 0 < |a| < b < 1 [#permalink] New post   11 Jun 2022, 22:28
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Paul121 wrote:
Is it mathematically legal question (for the GRE level)?
In the left equation, it has a square root of A, which is negative.
I think it will require a use of the imaginary numbers...



The question if the question tests the right concepts for the exam or less. IF the answer is not...just skip the question sir
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Re: 1 < a < 0 < |a| < b < 1 [#permalink] New post   15 Jun 2022, 02:45
Hello,

I am new in this forum. I believe this question has a fundamental issue: By definition, negative numbers don't have square roots in the real numbers' domain. So, I am tempted to choose D as answer since [square_root][a] is undefined (a is a negative number).

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Ariel
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Re: 1 < a < 0 < |a| < b < 1 [#permalink] New post   15 Jun 2022, 07:12
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Yes sir

brent pointed out also above
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Re: 1 < a < 0 < |a| < b < 1 [#permalink] New post   23 Jul 2023, 17:43
I think the problem is invalid, because any square root function is defined only when the input is a non-negative number. In the left-hand side, there is a square root of a, but a is less than zero, so it will be invalid.
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Re: 1 < a < 0 < |a| < b < 1 [#permalink] New post   12 Sep 2023, 10:25
GreenlightTestPrep wrote:

Concept #1: k² ≥ 0 for all values of k

QUANTITY A
Quantity A = (some value)²
Applying concept #1, we can see that Quantity A is greater than or equal to zero


This concept does not apply to the square roots of negative numbers, which is the case here.
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Re: 1 < a < 0 < |a| < b < 1 [#permalink]
12 Sep 2023, 10:25
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