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–1 < a < 0 < |a| < b < 1 [#permalink]
10 Aug 2018, 11:29

2

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Expert's post

00:00

Question Stats:

39% (01:30) correct
60% (01:18) wrong based on 280 sessions

\(–1 < a < 0 < |a| < b < 1\)

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.

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

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 _________________

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?

Re: –1 < a < 0 < |a| < b < 1 [#permalink]
26 Apr 2020, 16:35

1

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anjaliad20 wrote:

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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Brent Hanneson – Creator of greenlighttestprep.com If you enjoy my solutions, you'll like my GRE prep course. Sign up for GRE Question of the Dayemails

Re: –1 < a < 0 < |a| < b < 1 [#permalink]
04 May 2020, 11:51

2

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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 _________________

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?

greprepclubot

Re: –1 < a < 0 < |a| < b < 1
[#permalink]
12 May 2020, 11:56