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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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sandy
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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] New post   01 Jun 2021, 10:41
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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


QUANTITY B
We're told that a is NEGATIVE
We're also told that b is POSITIVE, which means b⁵ is POSITIVE
So, the numerator = (NEGATIVE)(POSITIVE) = NEGATIVE

In the denominator, we have (some value)⁴, which means the denominator must be POSITIVE

So, Quantity B = NEGATIVE/POSITIVE = NEGATIVE
So, Quantity B is some negative number


Answer: A

Cheers,
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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   16 Sep 2019, 18:09
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.


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




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   Updated on: 31 Oct 2019, 02:59
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Can we plug in values for a and b with out simplifications?
we will have negative square root for a?

Originally posted by mohmu123 on 31 Oct 2019, 02:42.
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 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 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
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   01 Jan 2020, 02:27
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.


why not more simplification? Like cancel a's from both sides and also b's ?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   24 Apr 2020, 04:44
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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.




Why don't we simplify this further by dividing both sides by ab? In that case we will have to compare a^2 with b^2 and there the answer is B. What am I missing here? Can anyone explain?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   24 Apr 2020, 05:59
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The fastest and simple approach is that above. reduce to a minimum term the two Qs

Then see which is greater.

regards
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   26 Apr 2020, 16:10
Ans choice A is right beacouse sum of A is greater then B?

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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   26 Apr 2020, 16:35
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anjaliad20 wrote:
Ans choice A is right beacouse sum of A is greater then B?

Posted from my mobile device Image


Here's what each answer choice means:
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.

Here's a video overview of this question type:


Cheers,
Brent
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   04 May 2020, 05:22
Carcass wrote:
The fastest and simple approach is that above. reduce to a minimum term the two Qs

The see which is greater.

regards


It comes down to a^2 & b^2
And here B is greater than A
What am I missing?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   04 May 2020, 11:51
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Raj30 wrote:
Carcass wrote:
The fastest and simple approach is that above. reduce to a minimum term the two Qs

The see which is greater.

regards


It comes down to a^2 & b^2
And here B is greater than A
What am I missing?




\(((a^2 √b)/√a)^2\) VS \((ab^5)/(√b)^4\)
\((a^4 b)/a\) VS \((ab^5)/b^2\)
\(a^3*b\) VS \(a*b^3\) … (i)

\(–1<a<0<|a|<b<1\) => Thus, a is negative and bis positive
We have: \(|a|<b\)
Squaring: \(a^2<b^2\)
Multiply b: \(a^2 b<b^3\)
Multiply a (negative; hence reverse inequality): \(a^3 b>ab^3\)
Thus: A is greater
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   06 May 2020, 08:34
what about the a in the bar
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Re: –1 < a < 0 < |a| < b < 1 [#permalink]
06 May 2020, 08:34
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