Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Magoosh is excited to offer you a free GRE practice test with video answers and explanations. If you’re thinking about taking the GRE or want to see how effective your GRE test prep has been, pinpoint your strengths and weaknesses with this quiz!

Real GRE® prep. Really free. Twice monthly, join a Manhattan Prep instructor for a free one-hour GRE prep session. Next sessions are June 11 or June 28th

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

2

This post received KUDOS

Expert's post

00:00

Question Stats:

37% (01:31) correct
62% (01:15) wrong based on 247 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

This post received KUDOS

Expert's post

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
_________________

Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Dayemails

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

2

This post received KUDOS

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