Last visit was: 24 Apr 2026, 22:07 It is currently 24 Apr 2026, 22:07
Home
Messages
Tests
Schools
Events
Close
GMAT 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 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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
nades09
User avatar
Joined: 18 Feb 2010
Last visit: 24 Feb 2014
Posts: 21
Own Kudos:
618
 [11]
Given Kudos: 5
Posts: 21
Kudos: 618
 [11]
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? Updated on: 21 Aug 2017, 08:24
Originally posted by nades09 on 26 Nov 2010, 20:07.
Last edited by Bunuel on 21 Aug 2017, 08:24, edited 1 time in total.
Renamed the topic and edited the question.
1
Kudos
Add Kudos
10
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

44% (02:37) correct 56% (02:47) wrong based on 213 sessions

History

Date
Time
Result
Not Attempted Yet
a is an odd integer and a ≠ -b. Is \(\frac{a^2}{|b+a|}> a-b ?\)


(1) \(ab=0\)

(2) \(a^3>a\)
ShowHide Answer
Official Answer
C
User avatar
skovinsky
User avatar
Kaplan GMAT Instructor
Joined: 21 Jun 2010
Last visit: 17 Dec 2019
Posts: 129
Own Kudos:
635
 [1]
Location: Toronto
Posts: 129
Kudos: 635
 [1]
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 26 Nov 2010, 21:56
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nades09
a is an odd integer and a <>-b

Is \(\frac{a^2}{|b+a|}> a-b ?\)

(1) ab=0
(2)\(a^3>a\)

PLease explain the shortest way to solve this

Thanks
NAD
Show more

I have no clue what "a <> -b" means, but that doesn't seem to be overly important to solve.

Let's dive into the statements:

1) if a is an odd integer, then (1) must mean that b=0. So, the question becomes:

Is a^2/|a|> a?

Let's pick a positive and negative a and see what happens.

If a = 1, we get:

Is 1/1 > 1? NO

If a = -1, we get:

Is 1/1 > -1? YES

Can get a yes and a no: insufficient.

2) if a is an odd integer and a^3 > a, then a must be positive. However, we know nothing about b. If we pick a big negative value of b then a-b will be negative and the left side will be positive, giving us a NO answer. If we pick a small positive value for b then both sides will be positive and we can play with the value of a to produce a YES answer: insufficient.

Together: if a is positive and b=0, the two sides will be equal, give us a "no" answer: sufficient, choose (C)!
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,442
Own Kudos:
79,405
 [3]
Given Kudos: 485
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,442
Kudos: 79,405
 [3]
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 27 Nov 2010, 12:35
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
nades09
a is an odd integer and a <>-b

Is \(\frac{a^2}{|b+a|}> a-b ?\)

(1) ab=0
(2)\(a^3>a\)

PLease explain the shortest way to solve this

Thanks
NAD
Show more

Alternate approach:
I am assuming a <>-b means 'a is not equal to -b' since a + b is in the denominator so it should not be 0.
First thing that strikes me about the question stem is that \(\frac{a^2}{|b+a|}\) is always positive or 0 (if a is 0) while a - b can be +ve, 0 or -ve. I do not know if this observation will help me here but it does give me some level of confidence.

Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.
If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.
Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Stmnt 2: \(a^3>a\). This only happens when a > 1 or when -1 < a < 0. Since a is odd and positive integer, a > 1. This alone is again not sufficient since the answer YES or NO depends on the value of b.

Using both together, b = 0 and a is odd positive integer. Question stem becomes is |a| > a? Answer is definite NO. (They are both equal.) Hence sufficient.
Answer (C).
User avatar
nades09
User avatar
Joined: 18 Feb 2010
Last visit: 24 Feb 2014
Posts: 21
Own Kudos:
618
Given Kudos: 5
Posts: 21
Kudos: 618
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 27 Nov 2010, 12:56
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks Karishma and Stuart. Yes <> means "not equal to". It had a long solution and wanted to know from experts like you if this resembles a GMAT Question. I got it from GMATfix question bank.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,442
Own Kudos:
79,405
Given Kudos: 485
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,442
Kudos: 79,405
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 27 Nov 2010, 13:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nades09
Thanks Karishma and Stuart. Yes <> means "not equal to". It had a long solution and wanted to know from experts like you if this resembles a GMAT Question. I got it from GMATfix question bank.
Show more

Yes, this question could resemble an actual GMAT question. Definitely interesting! Though, it is a higher level question, let us say 49 and above.
User avatar
vaibhavtripathi
User avatar
Joined: 04 Oct 2010
Last visit: 09 Oct 2015
Posts: 67
Own Kudos:
223
 [1]
Given Kudos: 9
Status:I rest, I rust.
Concentration: IT Major
Schools:ISB - Co 2013
GPA: 2.5-3.0
WE 1: IT Professional since 2006
Posts: 67
Kudos: 223
 [1]
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 27 Nov 2010, 19:47
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma
nades09
a is an odd integer and a <>-b

Is \(\frac{a^2}{|b+a|}> a-b ?\)

(1) ab=0
(2)\(a^3>a\)

PLease explain the shortest way to solve this

Thanks
NAD

Alternate approach:
I am assuming a <>-b means 'a is not equal to -b' since a + b is in the denominator so it should not be 0.
First thing that strikes me about the question stem is that \(\frac{a^2}{|b+a|}\) is always positive or 0 (if a is 0) while a - b can be +ve, 0 or -ve. I do not know if this observation will help me here but it does give me some level of confidence.

Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.
If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.
Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Thats a wrong assumption. Question tells us a is an odd integer.

Stmnt 2: \(a^3>a\). This only happens when a > 1 or when -1 < a < 0. Since a is odd and positive integer, a > 1. This alone is again not sufficient since the answer YES or NO depends on the value of b.

Using both together, b = 0 and a is odd positive integer. Question stem becomes is |a| > a? Answer is definite NO. (They are both equal.) Hence sufficient.
Answer (C).
Show more


In equation
\(\frac{a^2}{|b+a|} + b-a > 0\),
the \(b+a\) part can be either positive or negetive, so we must consider both to remove the MOD. The two conditions we get, after applying a little algebra, are:
\(\frac{b^2}{b+a} > 0\) --------------------------------E1
and
\(\frac{b^2 - 2a^2}{b+a} > 0\)-------------------------E2
any statement must satisfy both the equations to be considered good.

S1 gives us b=0 (since a is an odd integer), putting that in E1 and E2 above we get:
E1: LHS = 0, RHS = 0. Answers the question.
E2: \(-2a>0 ?\). Does not answer the question.
S1 Not sufficient

S2 gives us a>0, putting that in E1 and E2 above does not answer the question.
S2 Not sufficient

Putting b=0 and a>0 in E1 and E2 above:
E1: LHS = 0, RHS = 0. Answers the question.
E2: \(-2a>0 ?\)(where a is positive int). Answers the question.


Both together are sufficient. C
User avatar
oldstudent
User avatar
Joined: 20 Jun 2010
Last visit: 31 Jul 2015
Posts: 39
Own Kudos:
226
Given Kudos: 27
Status:Last few days....Have pressed the throttle
Concentration: Entrepreneurship,Strategy, Marketing
WE 1: 6 years - Consulting
Posts: 39
Kudos: 226
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 30 Nov 2010, 04:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma
nades09
a is an odd integer and a <>-b

Is \(\frac{a^2}{|b+a|}> a-b ?\)

(1) ab=0
(2)\(a^3>a\)

PLease explain the shortest way to solve this

Thanks
NAD

Alternate approach:
I am assuming a <>-b means 'a is not equal to -b' since a + b is in the denominator so it should not be 0.
First thing that strikes me about the question stem is that \(\frac{a^2}{|b+a|}\) is always positive or 0 (if a is 0) while a - b can be +ve, 0 or -ve. I do not know if this observation will help me here but it does give me some level of confidence.

Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.
If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.
Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Stmnt 2: \(a^3>a\). This only happens when a > 1 or when -1 < a < 0. Since a is odd and positive integer, a > 1. This alone is again not sufficient since the answer YES or NO depends on the value of b.

Using both together, b = 0 and a is odd positive integer. Question stem becomes is |a| > a? Answer is definite NO. (They are both equal.) Hence sufficient.
Answer (C).
Show more

Hi I have a question on solving \(a^3>a\). I see that you do it directly, but then if I need to solve it ->
\(a^3>a\) => \(a^3-a>0\) =>\(a(a^2-1)>0\) => thats means either a>0 (i) or \(a^2>1\) -> a>+/- 1 => a>1 or a<-1 (ii)
Final value for a from (i) and (ii) => a<-1 and a>1

@Karishma -Please confirm whether this solution to a is correct by inequality method.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,442
Own Kudos:
79,405
Given Kudos: 485
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,442
Kudos: 79,405
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 30 Nov 2010, 06:27
Kudos
Add Kudos
Bookmarks
Bookmark this Post
oldstudent
Hi I have a question on solving \(a^3>a\). I see that you do it directly, but then if I need to solve it ->
\(a^3>a\) => \(a^3-a>0\) =>\(a(a^2-1)>0\) => that means either a>0 (i) or \(a^2>1\) -> a>+/- 1 => a>1 or a<-1 (ii)
Final value for a from (i) and (ii) => a<-1 and a>1

@Karishma -Please confirm whether this solution to a is correct by inequality method.
Show more

The problem lies in your interpretation as shown by the colored part.
Think ab > 0. When will this happen? When either both a and b are positive or both a and b are negative.
Similarly when ab < 0, we say that either a < 0 and b > 0 or a > 0 and b < 0 i.e. one and only one of a and b is negative.
Only when ab = 0, can we say that either a = 0 or b = 0 or both are 0.

Here is how you will solve the inequality:
\(a^3>a\) => \(a^3-a>0\) =>\(a(a^2-1)>0\)=>\((a-1)a(a + 1)>0\)
For \((a-1)a(a + 1)\) to be positive, either all (a - 1), a and (a + 1) should be positive, or exactly two of them should be negative. Then the entire product will be positive.
To deal with this easily, take them on the number line. Plot 1, 0 and -1 on the number line. Now your number line is divided into 4 sections. Start from the rightmost section. That will always be positive. The next section will be negative, next positive again, next negative again and so on...
Attachment:
Ques1.jpg
Ques1.jpg [ 3.59 KiB | Viewed 6258 times ]

The green portions show where the expression will be positive and blue portions show where it will be negative.
So you get x > 1 or -1 < x < 0.
For detailed explanation of why this method works, check out my earlier post:
https://gmatclub.com/forum/inequalities-trick-91482.html#p804990
User avatar
CAMANISHPARMAR
User avatar
Joined: 12 Feb 2015
Last visit: 13 Mar 2022
Posts: 1,016
Own Kudos:
2,552
Given Kudos: 77
Posts: 1,016
Kudos: 2,552
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 30 Nov 2019, 04:56
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasKarishma


Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.
If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.
Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Show more

It is mentioned in the question stem that "a is an odd integer". How can we consider a scenario "either a = 0..." a cannot be 0, right?

Please correct me if I am missing something here?
User avatar
GMATinsight
User avatar
Major Poster
Joined: 08 Jul 2010
Last visit: 24 Apr 2026
Posts: 6,977
Own Kudos:
16,914
 [1]
Given Kudos: 128
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Products:
My Reviews
Expert
Expert reply
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
Posts: 6,977
Kudos: 16,914
 [1]
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 27 Jan 2020, 23:55
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
nades09
a is an odd integer and a ≠ -b. Is \(\frac{a^2}{|b+a|}> a-b ?\)


(1) \(ab=0\)

(2) \(a^3>a\)
Show more

Please find the solution as attached.

Answer: Option C
Attachments

Screenshot 2020-01-28 at 12.23.17 PM.png
Screenshot 2020-01-28 at 12.23.17 PM.png [ 545.67 KiB | Viewed 3744 times ]

User avatar
abhishekmayank
User avatar
Joined: 26 Apr 2016
Last visit: 28 Jan 2024
Posts: 198
Own Kudos:
61
Given Kudos: 6
GMAT 1: 640 Q44 V33
GMAT 1: 640 Q44 V33
Posts: 198
Kudos: 61
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 24 Apr 2021, 08:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasKarishma
nades09
a is an odd integer and a <>-b

Is \(\frac{a^2}{|b+a|}> a-b ?\)

(1) ab=0
(2)\(a^3>a\)

PLease explain the shortest way to solve this

Thanks
NAD

Alternate approach:
I am assuming a <>-b means 'a is not equal to -b' since a + b is in the denominator so it should not be 0.
First thing that strikes me about the question stem is that \(\frac{a^2}{|b+a|}\) is always positive or 0 (if a is 0) while a - b can be +ve, 0 or -ve. I do not know if this observation will help me here but it does give me some level of confidence.

Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.
If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.
Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Stmnt 2: \(a^3>a\). This only happens when a > 1 or when -1 < a < 0. Since a is odd and positive integer, a > 1. This alone is again not sufficient since the answer YES or NO depends on the value of b.

Using both together, b = 0 and a is odd positive integer. Question stem becomes is |a| > a? Answer is definite NO. (They are both equal.) Hence sufficient.
Answer (C).
Show more

Hi,

I don't find in the stem of the question that "a" +ve integer. It is simply an odd integer, which can be either +ve or -ve.
User avatar
Fdambro294
User avatar
Joined: 10 Jul 2019
Last visit: 20 Aug 2025
Posts: 1,331
Own Kudos:
772
Given Kudos: 1,656
Posts: 1,331
Kudos: 772
a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? 07 Aug 2021, 17:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hey, hopefully I can help a bit...

We are given in the Question Stem that A must be an Odd Integer. This means it can not be a Fraction.

From statement 2: (a)^3 > a

the only values of A that satisfy this inequality are:

-1 < a < 0 ----------> which is not possible given our constraint that A = an Odd INTEGER

or

a > 1 ----------> which is possible


therefore, the Inference we can draw from Statement 2 is the following:

a > 1


combined with statement 1: in which we made the Inference that B must = 0, the Question becomes:

Is: (a)^2 / [a] > a ?



(a)^2 = [a] * [a] -------> the Square of A can be written as the Absolute Value of [A] multiplied by the Absolute Value of [a]


Is: [a] * [a] / [a] > a?

----cancel one of the [a] in the NUM and DEN ------

Is: [a] > a?


statement 2 tells us that A must be greater than > 1

no matter which value greater than 1 that we plug into the question stem, we will always find that:

[a] = a


-C- Together Sufficient


Basically it's that one Constraint that A must be an Integer that rules out the case in statement 2 that A could fall in the Range of:

-1 < a < 0


I hope that helps a little bit? if not, please send any follow up questions and I'll do my best.

All the best!

abhishekmayank
VeritasKarishma
nades09
a is an odd integer and a <>-b

Is \(\frac{a^2}{|b+a|}> a-b ?\)

(1) ab=0
(2)\(a^3>a\)

PLease explain the shortest way to solve this

Thanks
NAD

Alternate approach:
I am assuming a <>-b means 'a is not equal to -b' since a + b is in the denominator so it should not be 0.
First thing that strikes me about the question stem is that \(\frac{a^2}{|b+a|}\) is always positive or 0 (if a is 0) while a - b can be +ve, 0 or -ve. I do not know if this observation will help me here but it does give me some level of confidence.

Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.
If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.
Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Stmnt 2: \(a^3>a\). This only happens when a > 1 or when -1 < a < 0. Since a is odd and positive integer, a > 1. This alone is again not sufficient since the answer YES or NO depends on the value of b.

Using both together, b = 0 and a is odd positive integer. Question stem becomes is |a| > a? Answer is definite NO. (They are both equal.) Hence sufficient.
Answer (C).

Hi,

I don't find in the stem of the question that "a" +ve integer. It is simply an odd integer, which can be either +ve or -ve.
Show more
Moderators:
Bunuel
Math Expert
109818 posts
kingbucky
498 posts
Gmat860sanskar
212 posts