Last visit was: 25 Nov 2024, 05:14 It is currently 25 Nov 2024, 05:14
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

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
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

–1 < a < 0 < |a| < b < 1

 1   2   
Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30021
Own Kudos [?]: 36389 [27]
Given Kudos: 25928
Send PM
–1 < a < 0 < |a| < b < 1 [#permalink] New post   10 Aug 2018, 11:29
9
Expert Reply
18
Bookmarks
00:00

Question Stats:

47% (01:53) correct 52% (01:40) wrong based on 697 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
ShowHide Answer
Official Answer
A
Most Helpful Expert Reply
User avatar
sandy
Retired Moderator
Joined: 07 Jun 2014
Posts: 4813
Own Kudos [?]: 11197 [10]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   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
User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12197 [12]
Given Kudos: 136
Send PM

Top Member of the Month
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   01 Jun 2021, 10:41
9
3
Bookmarks
Image

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
avatar
saifee
Intern
Intern
Joined: 10 Sep 2018
Posts: 20
Own Kudos [?]: 63 [3]
Given Kudos: 0
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   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.
avatar
Kp3642
Intern
Intern
Joined: 16 Sep 2018
Posts: 1
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
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) ??

Posted from my mobile device Image
User avatar
huda
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 711 [0]
Given Kudos: 161
Send PM

Top Member of the Month
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
avatar
praveenshahi13
Intern
Intern
Joined: 12 Feb 2019
Posts: 1
Own Kudos [?]: 2 [2]
Given Kudos: 0
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   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?
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30021
Own Kudos [?]: 36389 [0]
Given Kudos: 25928
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   23 Oct 2019, 15:00
Expert Reply
Not really :)
avatar
mohmu123
Intern
Intern
Joined: 18 Oct 2019
Posts: 3
Own Kudos [?]: 3 [1]
Given Kudos: 0
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   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.
User avatar
huda
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 711 [0]
Given Kudos: 161
Send PM

Top Member of the Month
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?
avatar
mohmu123
Intern
Intern
Joined: 18 Oct 2019
Posts: 3
Own Kudos [?]: 3 [0]
Given Kudos: 0
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   31 Oct 2019, 03:00
I meant a and b
User avatar
huda
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 711 [0]
Given Kudos: 161
Send PM

Top Member of the Month
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
User avatar
huda
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 711 [0]
Given Kudos: 161
Send PM

Top Member of the Month
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 ?
avatar
Narek
Intern
Intern
Joined: 17 Oct 2019
Posts: 6
Own Kudos [?]: 9 [0]
Given Kudos: 8
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   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?
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30021
Own Kudos [?]: 36389 [0]
Given Kudos: 25928
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   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
avatar
anjaliad20
Intern
Intern
Joined: 20 Apr 2020
Posts: 16
Own Kudos [?]: 4 [0]
Given Kudos: 0
Send PM
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?

Posted from my mobile device Image
User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12197 [0]
Given Kudos: 136
Send PM

Top Member of the Month
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   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 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
avatar
Raj30
Intern
Intern
Joined: 29 Mar 2019
Posts: 10
Own Kudos [?]: 8 [0]
Given Kudos: 0
Send PM
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?
User avatar
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 [?]: 265 [0]
Given Kudos: 0
GPA: 3.72
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   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
avatar
Nikhil4GRE
Manager
Manager
Joined: 29 Apr 2020
Posts: 67
Own Kudos [?]: 9 [0]
Given Kudos: 0
Send PM
Re: –1 < a < 0 < |a| < b < 1 [#permalink] New post   06 May 2020, 08:34
what about the a in the bar
Prep Club for GRE Bot
gmatclubot
Re: –1 < a < 0 < |a| < b < 1 [#permalink]
06 May 2020, 08:34
 1   2   
Moderators:
adewale223
GRE Instructor
84 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1111 posts
VibhuAnurag
GRE Instructor
234 posts

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

Main navigation

Partners

GRE Resources

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