Inequality and absolute value questions from my collection : Data Sufficiency (DS)

Last visit was: 06 May 2026, 18:48

It is currently 06 May 2026, 18:48

Toolkit

Data Insights Questions

Data Sufficiency (DS)

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

Request Expert Reply

Confirm Cancel

May 06
Paradox & Discrepancy Questions: GMAT Critical Reasoning #3 | GMAT Ninja
08:30 AM PDT
-
09:30 AM PDT
In Episode 3 of our GMAT Ninja Critical Reasoning series, we tackle Discrepancy, Paradox, and Explain an Oddity questions. You know the feeling: the passage gives you two facts that seem completely contradictory....
May 08
$10,000 MBA Scholarships Winners Drawing - Join & Watch the Lucky Winners
08:00 AM PDT
-
08:30 AM PDT
Join the special YouTube live-stream for selecting the winners of GMAT Club MBA Scholarships sponsored by Juno live. Watch who gets these coveted MBA scholarships offered by GMAT Club and Juno.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

[1244]

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

[1244]

16 Nov 2009, 11:33

241

Kudos

1003

Bookmarks

Guys I didn't forget your request, just was collecting good questions to post.

So here are some inequality and absolute value questions from my collection. Not every problem below is hard, but there are a few, which are quite tricky. Please provide your explanations along with the answers.

1. If $6*x*y = x^2*y + 9*y$, what is the value of xy?
(1) $y – x = 3$
(2) $x^3< 0$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653690

2. If y is an integer and $y = |x| + x$, is $y = 0$?
(1) $x < 0$
(2) $y < 1$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653695

3. Is $x^2 + y^2 > 4a$?
(1) $(x + y)^2 = 9a$
(2) $(x – y)^2 = a$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653697

4. Are x and y both positive?
(1) $2x-2y=1$
(2) $\frac{x}{y}>1$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653709
Graphic approach: https://gmatclub.com/forum/inequality-a ... l#p1269802

5. What is the value of y?
(1) $3|x^2 -4| = y - 2$
(2) $|3 - y| = 11$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653731

6. If x and y are integer, is y > 0?
(1) $x +1 > 0$
(2) $xy > 0$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653740

7. $|x+2|=|y+2|$ what is the value of x+y?
(1) $xy<0$
(2) $x>2$, $y<2$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653783 AND https://gmatclub.com/forum/inequality-a ... l#p1111747

8. $a*b \neq 0$. Is $\frac{|a|}{|b|}=\frac{a}{b}$?
(1) $|a*b|=a*b$
(2) $\frac{|a|}{|b|}=|\frac{a}{b}|$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653789

9. Is n<0?
(1) $-n=|-n|$
(2) $n^2=16$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653792

10. If n is not equal to 0, is |n| < 4 ?
(1) $n^2 > 16$
(2) $\frac{1}{|n|} > n$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653796

11. Is $|x+y|>|x-y|$?
(1) $|x| > |y|$
(2) $|x-y| < |x|$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653853

12. Is r=s?
(1) $-s \leq r \leq s$
(2) $|r| \geq s$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653870

13. Is $|x-1| < 1$?
(1) $(x-1)^2 \leq 1$
(2) $x^2 - 1 > 0$

Solution: https://gmatclub.com/forum/inequality-a ... ml#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.

PLEASE READ THE WHOLE DISCUSSION BEFORE POSTING A QUESTION.

Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

Word Problems
Work Problems
Distance Problems
Mixture Problems
Percent and Interest Problems
Overlapping Sets Problems
Statistics Problems
Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore these question.

petocities

Joined: 26 Jul 2015

Last visit: 24 Jun 2016

Posts: 17

Own Kudos:

Given Kudos: 16

Location: Chile

Schools: Sloan '18 Haas '18 HBS '18 (S) Kellogg '18 (S) Wharton '18 CBS '18 Stern '18

GMAT 1: 620 Q38 V37 Verified score

GMAT 2: 690 Q49 V35 Verified score

WE:Analyst (Consulting)

Schools: Sloan '18 Haas '18 HBS '18 (S) Kellogg '18 (S) Wharton '18 CBS '18 Stern '18

GMAT 2: 690 Q49 V35 Verified score

Posts: 17

Kudos: 21

08 Oct 2015, 13:34

Kudos

Bookmarks

Bunuel

jayaddula

Bunuel

7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

Hi Bunuel,

I am getting E and just cannot understand D. Please see my solution below -
I used number picking.

A. xy<0,
x=+ and y=- For this condition choosing different values of x and y (x=2,y=-6: x=3, y=-7)satisfies the given condition in modulus. Hence x=y can be different value
or x=- and y=+ - This condition doesn't satisfy the modulus condiotion

B- x>2 and y<2 - As per the above stmt 1 - condition 1, there can be various values for x and y, hence x+y is different.

Hence E. I know I am going wrong some where, please help.

thanks
jay

In your example, both pairs give the same value for x+y: 2-6=-4 and 3-7=-4.

We can solve this question in another way:

7. |x+2|=|y+2| what is the value of x+y?

Square both sides: $x^2+4x+4=y^2+4y+4$ --> $x^2-y^2+4x-4y=0$ --> $(x+y)(x-y)+4(x-y)=0$ --> $(x-y)(x+y+4)=0$ --> either $x=y$ or $x+y=-4$.

(1) xy<0 --> the first case is not possible, since if $x=y$, then $xy=x^2\geq{0}$, not $<0$ as given in this statement, hence we have the second case: $x+y=-4$. Sufficient.

(2) x>2 and y<2. This statement implies that $x\neq{y}$, therefore $x+y=-4$. Sufficient.

Answer: D.

Hope it's clear.

Hi @Buenel, i'm having a really hard time understanding this question. First, I don't understand why x=y should imply a unique answer for x+y. Same for the second stantement, I don't fully understand why having the equation x+y=-4 ensures a unique answer. Maybe I am missing some steps. Would greatly appreciate your help, or any1 else's help (maybe different approaches will help me understand better).

Thanks in advance!

GMATinsight

Major Poster

Joined: 08 Jul 2010

Last visit: 06 May 2026

Posts: 7,008

Own Kudos:

16,965

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 Verified score

WE:Education (Education)

Products:

Expert

Expert reply

Schools: IIM (A) ISB '24

GMAT 1: 750 Q51 V41 Verified score

Posts: 7,008

Kudos: 16,965

09 Oct 2015, 21:22

Kudos

Bookmarks

petocities

7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

Hi @Buenel, i'm having a really hard time understanding this question. First, I don't understand why x=y should imply a unique answer for x+y. Same for the second stantement, I don't fully understand why having the equation x+y=-4 ensures a unique answer. Maybe I am missing some steps. Would greatly appreciate your help, or any1 else's help (maybe different approaches will help me understand better).

Thanks in advance!

Hi petocities,

I think this question requires more of observation about given information

|x+2|=|y+2| will be true for two possible cases of x and y

Case 1: when x = y
Case 2: For Values like (x=1 and y=-5) or (x=2 and y=-6) or (x=3 and y=-7) ...etc.

Case 1 gives inconsistent answers because for each different value of x and y, x+ywill be different
but
Case 2 always gives a consistent value of x+y=-4 (Check all set of values mentioned above in Case 2)

Statement 1 suggests that x and y are not equal (for x and y equal, their product must be Non-Negative) i.e. Case 2 prevails which always gives us x+y=-4
i.e. SUFFICIENT

Statement 2 also rules out the scenario in which x and y may be equal i.e. Case 2 prevails again leading to a consistent value of x+y=-4
i.e. SUFFICIENT

Answer: Option D

BrainLab

Current Student

Joined: 10 Mar 2013

Last visit: 26 Jan 2025

Posts: 342

Own Kudos:

3,223

Given Kudos: 200

Location: Germany

Concentration: Finance, Entrepreneurship

Schools: WHU MBA"20 (A$)

GMAT 1: 580 Q46 V24 Verified score

GPA: 3.7

WE:Marketing (Telecommunications)

Schools: WHU MBA"20 (A$)

GMAT 1: 580 Q46 V24 Verified score

Posts: 342

Kudos: 3,223

18 Oct 2015, 13:18

Kudos

Bookmarks

Bunuel

10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

Hi Bunuel, I've solved this one correctly, but have one question. A is ok - no questions.
Can we manipulate Statement 2 and say |n|*n<1 as |n| is always positive we must be able to do this - but |n|*n can be also positive as it's not stated that n must be an integer, let's say 1/2*1/2<1 and it can be also any negative value as stated above.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

18 Oct 2015, 13:25

Kudos

Bookmarks

BrainLab

Bunuel

$\frac{1}{| n |}>n$ --> multiply by $|n|$ (we can safely do that since |n|>0): $n*|n| < 1$.

If $n>0$, then we'll have $n^2<1$ --> $-1<n<1$. Since we consider the range when $n>0$, then for this range we'll have $0<n<1$.
If $n<0$, then we'll have $-n^2<1$ --> $n^2>-1$. Which is true for any n from the range we consider. So, $n*|n| < 1$ holds true for any negative value of n.

Thus $\frac{1}{| n |}>n$ holds true if $n<0$ and $0<n<1$.

Johnbreeden85

Joined: 28 Sep 2015

Last visit: 06 Apr 2018

Posts: 29

Own Kudos:

Given Kudos: 16

Location: United States (TX)

Concentration: Strategy, Other

Schools: Jones '19 (M$) McCombs '19 (WL)

GMAT 1: 670 Q45 V36 Verified score

GMAT 2: 710 Q49 V38 Verified score

GPA: 2.97

WE:Military Officer (Military & Defense)

Schools: Jones '19 (M$) McCombs '19 (WL)

GMAT 2: 710 Q49 V38 Verified score

Posts: 29

Kudos: 11

04 Nov 2015, 19:46

Kudos

Bookmarks

Bunuel

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
$2x-2y=1$ --> $x=y+\frac{1}{2}$
$\frac{x}{y}>1$ --> $\frac{x-y}{y}>0$ --> substitute x --> $\frac{1}{y}>0$ --> $y$ is positive, and as $x=y+\frac{1}{2}$, $x$ is positive too. Sufficient.

Answer: C.

How did you figure out that $\frac{x}{y}>1$ --> $\frac{x-y}{y}>0$ ? Any help is appreciated. Thank you.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

[1]

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

[1]

05 Nov 2015, 01:28

Kudos

Bookmarks

Johnbreeden85

Bunuel

How did you figure out that $\frac{x}{y}>1$ --> $\frac{x-y}{y}>0$ ? Any help is appreciated. Thank you.

This is explained couple of times on the previous pages:
$\frac{x}{y}>1$

$\frac{x}{y}-1>0$

$\frac{x}{y}-\frac{y}{y}>0$

$\frac{x-y}{y}>0$.

Hope it's clear.

nishantdoshi

Joined: 23 Sep 2015

Last visit: 30 May 2017

Posts: 27

Own Kudos:

Given Kudos: 99

Posts: 27

Kudos: 3

23 Feb 2016, 00:25

Kudos

Bookmarks

Bunuel

9. Is n<0?
(1) -n=|-n|
(2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.

hey bunuel
can you please clear my doubt?
in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: math-absolute-value-modulus-86462.html

correct me if i'm wrong!

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

23 Feb 2016, 00:43

Kudos

Bookmarks

nishantdoshi

Bunuel

An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits:
-n=|-n| --> -0 = |-0| --> 0 = 0.

nishantdoshi

Joined: 23 Sep 2015

Last visit: 30 May 2017

Posts: 27

Own Kudos:

Given Kudos: 99

Posts: 27

Kudos: 3

23 Feb 2016, 00:53

Kudos

Bookmarks

Bunuel

nishantdoshi

Bunuel

An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits:
-n=|-n| --> -0 = |-0| --> 0 = 0.

thanks for the reply
my understanding about this topic is that...

if x>=0 then |x|=x
and if x<0 then |x|=-x

am i wrong?
please reply!!!

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

[1]

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

[1]

23 Feb 2016, 00:55

Kudos

Bookmarks

nishantdoshi

Bunuel

nishantdoshi

An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits:
-n=|-n| --> -0 = |-0| --> 0 = 0.

thanks for the reply
my understanding about this topic is that...

if x>=0 then |x|=x
and if x<0 then |x|=-x

am i wrong?
please reply!!!

Yes, you are wrong.

We can say that when x<=0, then |x| is also equal to -x:

|0| = -0.

ameyaprabhu

Joined: 28 Apr 2016

Last visit: 09 Aug 2017

Posts: 66

Own Kudos:

Given Kudos: 79

Posts: 66

Kudos: 34

14 May 2016, 00:34

Kudos

Bookmarks

In your step you have taken it as (x-3)^2 but isn't (x^2-6x+9) = (x-3)^2 or (3-x)^2?

Also, what does 'reduce the expression by y' mean?

Bunuel

SOLUTIONS:

1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

First let's simplify given expression $6*x*y = x^2*y + 9*y$:

$y*(x^2-6x+9)=0$ --> $y*(x-3)^2=0$. Note here that we CAN NOT reduce this expression by $y$, as some of you did. Remember we are asked to determine the value of $xy$, and when reducing by $y$ you are assuming that $y$ doesn't equal to $0$. We don't know that.

Next: we can conclude that either $x=3$ or/and $y=0$. Which means that $xy$ equals to 0, when y=0 and x any value (including 3), OR $xy=3*y$ when y is not equal to zero, and x=3.

(1) $y-x=3$. If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case $xy=18$. But if y=0 then x=-3 and $xy=0$. Two possible scenarios. Not sufficient.

OR:

$y-x=3$ --> $x=y-3$ --> $y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0$ --> either $y=0$ or $y=6$ --> if $y=0$, then $x=-3$ and $xy=0$ $or$ if $y=6$, then $x=3$ and $xy=18$. Two different answers. Not sufficient.

(2) $x^3<0$. x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.

This one was quite tricky and was solved incorrectly by all of you.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

14 May 2016, 01:06

Kudos

Bookmarks

ameyaprabhu

In your step you have taken it as (x-3)^2 but isn't (x^2-6x+9) = (x-3)^2 or (3-x)^2?

Also, what does 'reduce the expression by y' mean?

Bunuel

Both (x-3)^2 and (3-x)^2 are the same.

rpradhan25

Joined: 02 Nov 2015

Last visit: 28 Feb 2020

Posts: 1

Given Kudos: 1

Posts: 1

Kudos: 0

02 Sep 2016, 23:52

Kudos

Bookmarks

Bunuel

13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

Hi Bunuel,

Can you please explain why have you not considered (for option1) the other case. I mean, x(x-2)<=0 can lead to two possiblities, one that you have mentioned, and the other one could be just the opposite x<=0 and x>=2. I need to understand this, please reply.

Thanks.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

04 Sep 2016, 04:50

Kudos

Bookmarks

rpradhan25

Bunuel

x(x-2)<=0 is true for 0<=x<=2 and not true for any other range.

Check the links below:
Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach
Inequality tips
Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

Hope it helps.

naren01

Joined: 17 Sep 2016

Last visit: 01 Nov 2017

Posts: 3

Given Kudos: 2

Posts: 3

Kudos: 0

27 Dec 2016, 05:11

Kudos

Bookmarks

[quote="Bunuel"]SOLUTIONS:

1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

First let's simplify given expression $6*x*y = x^2*y + 9*y$:

$y*(x^2-6x+9)=0$ --> $y*(x-3)^2=0$. Note here that we CAN NOT reduce this expression by $y$, as some of you did. Remember we are asked to determine the value of $xy$, and when reducing by $y$ you are assuming that $y$ doesn't equal to $0$. We don't know that.

Next: we can conclude that either $x=3$ or/and $y=0$. Which means that $xy$ equals to 0, when y=0 and x any value (including 3), OR $xy=3*y$ when y is not equal to zero, and x=3.

(1) $y-x=3$. If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case $xy=18$. But if y=0 then x=-3 and $xy=0$. Two possible scenarios. Not sufficient.

OR:

$y-x=3$ --> $x=y-3$ --> $y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0$ --> either $y=0$ or $y=6$ --> if $y=0$, then $x=-3$ and $xy=0$ $or$ if $y=6$, then $x=3$ and $xy=18$. Two different answers. Not sufficient.

(2) $x^3<0$. x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.
This one was quite tricky and was solved incorrectly by all of you.

Hi Bunuel,

I am a big fan of your posts and all your easy explanations, just want to point out a possible correction in your explanation of question 1 here.
From the statement (ii) x^3<0, it is clear that x<0. But how did you arrive at the value of x=3 without combining statement (i) which gives two values of x=3,-3?
That is the reason I think the answer is (C) when we get x=-3 using both statement (i) and statement (ii) and hence the value of xy.

Let me know if I am correct.

Thanks!

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

27 Dec 2016, 06:52

Kudos

Bookmarks

naren01

Bunuel

SOLUTIONS:

1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

First let's simplify given expression $6*x*y = x^2*y + 9*y$:

$y*(x^2-6x+9)=0$ --> $y*(x-3)^2=0$. Note here that we CAN NOT reduce this expression by $y$, as some of you did. Remember we are asked to determine the value of $xy$, and when reducing by $y$ you are assuming that $y$ doesn't equal to $0$. We don't know that.

Next: we can conclude that either $x=3$ or/and $y=0$. Which means that $xy$ equals to 0, when y=0 and x any value (including 3), OR $xy=3*y$ when y is not equal to zero, and x=3.

(1) $y-x=3$. If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case $xy=18$. But if y=0 then x=-3 and $xy=0$. Two possible scenarios. Not sufficient.

OR:

$y-x=3$ --> $x=y-3$ --> $y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0$ --> either $y=0$ or $y=6$ --> if $y=0$, then $x=-3$ and $xy=0$ $or$ if $y=6$, then $x=3$ and $xy=18$. Two different answers. Not sufficient.

(2) $x^3<0$. x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.
This one was quite tricky and was solved incorrectly by all of you.

Hi Bunuel,

I am a big fan of your posts and all your easy explanations, just want to point out a possible correction in your explanation of question 1 here.
From the statement (ii) x^3<0, it is clear that x<0. But how did you arrive at the value of x=3 without combining statement (i) which gives two values of x=3,-3?
That is the reason I think the answer is (C) when we get x=-3 using both statement (i) and statement (ii) and hence the value of xy.

Let me know if I am correct.

Thanks!

No, you are not correct.

From $y*(x-3)^2=0$ it follows that either $x=3$ or/and $y=0$. (2) says that $x^3<0$, thus x is not 3, therefore y must be 0 --> xy = 0.

RR88

Joined: 18 Oct 2016

Last visit: 16 Oct 2019

Posts: 107

Own Kudos:

150

Given Kudos: 91

Location: India

WE:Engineering (Energy)

Posts: 107

Kudos: 150

30 Dec 2016, 07:50

Kudos

Bookmarks

Bunuel

Hi Bunuel, thank you so much for such an amazing post, so so so helpful.

A quick query, regarding statement 2 here:
(2) $\frac{1}{|n|} > n$, Shouldn't it be true for all values of n such that n<1 (n#0) ?

Eg: n =1/2: $\frac{1}{|(1/2)|} > \frac{1}{2}$ : $2 > \frac{1}{2}$

PS: It doesn't alter the final answer though.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

30 Dec 2016, 08:09

Kudos

Bookmarks

RR88

Bunuel

Yes, but the fact that it's true for all negative n's was enough to discard this statement. So, we did not need to find the actual range. Still if interested here it is:

$\frac{1}{| n |}>n$ --> multiply by $|n|$ (we can safely do that since |n|>0): $n*|n| < 1$.

If $n>0$, then we'll have $n^2<1$ --> $-1<n<1$. Since we consider the range when $n>0$, then for this range we'll have $0<n<1$.
If $n<0$, then we'll have $-n^2<1$ --> $n^2>-1$. Which is true for any n from the range we consider. So, $n*|n| < 1$ holds true for any negative value of n.

Thus $\frac{1}{| n |}>n$ holds true if $n<0$ and $0<n<1$.

ManishKM1

Current Student

Joined: 22 Apr 2017

Last visit: 03 Jan 2019

Posts: 81

Own Kudos:

329

Given Kudos: 75

Location: India

Schools: MBS '20 (D) ISB '19 (D) ESADE '20 (A) IIM (A)

GMAT 1: 620 Q47 V29 Verified score

GMAT 2: 630 Q49 V26 Verified score

GMAT 3: 690 Q48 V35 Verified score

GPA: 3.7

Schools: MBS '20 (D) ISB '19 (D) ESADE '20 (A) IIM (A)

GMAT 3: 690 Q48 V35 Verified score

Posts: 81

Kudos: 329

25 Jul 2017, 13:48

Kudos

Bookmarks

Bunuel

8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Answer: A.

Hi Bunuel,
From 2, can't we say, |a|/|b|>=0, in this case, both a & B will be either +ve or -ve. Hence, sufficient? Pls pardon my ignorance.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 06 May 2026

Posts: 110,125

Own Kudos:

813,343

Given Kudos: 106,073

Products:

Expert

Expert reply

Posts: 110,125

Kudos: 813,343

25 Jul 2017, 20:55

Kudos

Bookmarks

ManishKM1

Bunuel

Hi Bunuel,
From 2, can't we say, |a|/|b|>=0, in this case, both a & B will be either +ve or -ve. Hence, sufficient? Pls pardon my ignorance.

Absolute value of a number is non-negative, so both |a| and |b| are >=0. So, naturally |a|/|b|>=0. But we cannot know whether a and b are positive or negative? Consider a = 1 and b = 2 OR a = -1 and b = 2.