Last visit was: 14 Apr 2025, 00:55

It is currently 14 Apr 2025, 00:55

Decision Tracker

New comers' posts

GRE Prep Club Daily Prep

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 GRE 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.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

PREVIOUS TOPIC

–1 < a < 0 < |a| < b < 1

Tags :

Find Similar Topics About Add a Tag

Show Tags

Hide Tags

Carcass

Verbal Expert

Joined: 18 Apr 2015

Posts: 31546

Own Kudos [?]: 37760 [30]

Given Kudos: 26238

Send PM

–1 < a < 0 < |a| < b < 1 [#permalink]

10 Aug 2018, 11:29

9

Expert Reply

21

Bookmarks

00:00

A

B

C

D

Question Stats:

47% (01:53) correct

52% (01:40) wrong

based on 717 sessions

Hide Show timer Statistics

$–1 < a < 0 < |a| < b < 1$

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.

Kudos for R.A.E

Official Answer

A

Most Helpful Expert Reply

sandy

Retired Moderator

Joined: 07 Jun 2014

Posts: 4815

Own Kudos [?]: 11365 [10]

Given Kudos: 0

GRE 1: Q167 V156

WE:Business Development (Energy and Utilities)

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

24 Sep 2018, 08:16

10

Expert Reply

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.

Most Helpful Community Reply

GreenlightTestPrep

Retired Moderator

Joined: 10 Apr 2015

Posts: 6218

Own Kudos [?]: 12354 [12]

Given Kudos: 136

Send PM

Top Member of the Month

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

01 Jun 2021, 10:41

9

3

Bookmarks

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,
Brent

General Discussion

saifee

Intern

Joined: 10 Sep 2018

Posts: 20

Own Kudos [?]: 63 [3]

Given Kudos: 0

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

10 Sep 2018, 09:51

3

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.

Kp3642

Intern

Joined: 16 Sep 2018

Posts: 1

Own Kudos [?]: 0 [0]

Given Kudos: 0

Send PM

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) ??

Posted from my mobile device

huda

Director

Joined: 22 Jun 2019

Posts: 521

Own Kudos [?]: 744 [0]

Given Kudos: 161

Send PM

Top Member of the Month

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

praveenshahi13

Intern

Joined: 12 Feb 2019

Posts: 1

Own Kudos [?]: 2 [2]

Given Kudos: 0

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

23 Oct 2019, 14:40

1

1

Bookmarks

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?

Carcass

Verbal Expert

Joined: 18 Apr 2015

Posts: 31546

Own Kudos [?]: 37760 [0]

Given Kudos: 26238

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

23 Oct 2019, 15:00

Expert Reply

Not really

mohmu123

Intern

Joined: 18 Oct 2019

Posts: 3

Own Kudos [?]: 3 [1]

Given Kudos: 0

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

Updated on: 31 Oct 2019, 02:59

1

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.

huda

Director

Joined: 22 Jun 2019

Posts: 521

Own Kudos [?]: 744 [0]

Given Kudos: 161

Send PM

Top Member of the Month

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?

mohmu123

Intern

Joined: 18 Oct 2019

Posts: 3

Own Kudos [?]: 3 [0]

Given Kudos: 0

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

31 Oct 2019, 03:00

I meant a and b

huda

Director

Joined: 22 Jun 2019

Posts: 521

Own Kudos [?]: 744 [0]

Given Kudos: 161

Send PM

Top Member of the Month

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

huda

Director

Joined: 22 Jun 2019

Posts: 521

Own Kudos [?]: 744 [0]

Given Kudos: 161

Send PM

Top Member of the Month

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 ?

Narek

Intern

Joined: 17 Oct 2019

Posts: 6

Own Kudos [?]: 9 [0]

Given Kudos: 8

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

24 Apr 2020, 04:44

1

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?

Carcass

Verbal Expert

Joined: 18 Apr 2015

Posts: 31546

Own Kudos [?]: 37760 [0]

Given Kudos: 26238

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

24 Apr 2020, 05:59

Expert Reply

The fastest and simple approach is that above. reduce to a minimum term the two Qs

Then see which is greater.

regards

anjaliad20

Intern

Joined: 20 Apr 2020

Posts: 16

Own Kudos [?]: 4 [0]

Given Kudos: 0

Send PM

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?

Posted from my mobile device

GreenlightTestPrep

Retired Moderator

Joined: 10 Apr 2015

Posts: 6218

Own Kudos [?]: 12354 [0]

Given Kudos: 136

Send PM

Top Member of the Month

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

26 Apr 2020, 16:35

1

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

Raj30

Intern

Joined: 29 Mar 2019

Posts: 10

Own Kudos [?]: 8 [0]

Given Kudos: 0

Send PM

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?

sujoykrdatta

GRE Instructor

Joined: 19 Jan 2020

Status:Entrepreneur | GMAT, GRE, CAT, SAT, ACT coach & mentor | Founder @CUBIX | Edu-consulting | Content creator

Posts: 117

Own Kudos [?]: 272 [0]

Given Kudos: 0

GPA: 3.72

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

04 May 2020, 11:51

2

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

Nikhil4GRE

Manager

Joined: 29 Apr 2020

Posts: 67

Own Kudos [?]: 9 [0]

Given Kudos: 0

Send PM

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

06 May 2020, 08:34

what about the a in the bar

gmatclubot

Re: –1 < a < 0 < |a| < b < 1 [#permalink]

06 May 2020, 08:34

Moderators:

GRE Instructor

108 posts

GRE Forum Moderator

37 posts

Moderator

1131 posts

GRE Instructor

234 posts

Hi Generic [Bot],

Here are updates for you:

ANNOUNCEMENTS

GRE Practice Quiz - Solve 20 Challenging Questions in Timed Conditions

Subscribe to our YouTube Channel to Get Quality GRE Content

INEQUALITIES / Absolute Value / Modulus UPDATED

GRE Quant & Verbal ADVANCED Daily Challenge 2025 NE

ETS Explains GRE - Test Format, Sections, Scoring Algorithm, Free Prep Resources

Subscribe to our YouTube Channel to Get Quality GRE Content

Average GRE Scores - Top 50 MBA Programs (2025)

GRE PREMIUM Quant Question Banks - Topic-Wise 2700 Questions

GRE Percentiles 2024-2025

GRE Math Handout

GMAT Club and Prodigy Finance scholarships

LATEST POSTS

More Latest posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

Copyright © 2025 Prep Club for GRE

GRE® is a registered trademark of the Education Testing Services® (ETS ®). Prep Club for GRE website has not been reviewed or endorsed by ETS

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact