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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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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 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]  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]  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]  23 Oct 2019, 14:40
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]  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 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 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]  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]  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]  24 Apr 2020, 05:59
Expert's post
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]  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]  26 Apr 2020, 16:35
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Ans choice A is right beacouse sum of A is greater then B?

Posted from my mobile device

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]  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]  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]  06 May 2020, 08:34
what about the a in the bar
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Re: –1 < a < 0 < |a| < b < 1 [#permalink]  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?
Re: –1 < a < 0 < |a| < b < 1   [#permalink] 12 May 2020, 11:56
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