Events & Promotions
| Last visit was: 06 Sep 2026, 11:42 |
It is currently 06 Sep 2026, 11:42 |
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 0712:01 AM PDT
-11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT- Sep 07
10:00 AM PDT
-11:00 AM PDT
A complete walkthrough of GMAT Club’s free 12-week GMAT study plan: what you study each week, how progress and goals are tracked, how the error log works, and how your practice unlocks paid tools at no cost. 
Sep 0201:00 PM EDT
-11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 0410:00 AM EDT
-11:59 PM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
Sep 0609:30 AM EDT
-12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.- Sep 09
07:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway - Sep 09
10:00 AM PDT
-11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application... 
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2108:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
Posts: 33
16 Jul 2026, 18:56
Kudos
Bookmarks
From the initial equation, we get either x=1 or y=-3 or xy=-3.
The question is a must be true question and we know x, y are integers.
I.
If x=-1, the equation becomes (y+3)(-y+3)=0
Therefore, 9-y^2=0
or, y^2=9
or, |y|=3
True.
II.
If |y| is not a prime number, y cannot be -3. Therefore, x=1 or x=-3/y
We know x is an integer and y is an integer as well, |y| can only be 1 or 3.
Since |y| cannot be 3, |y|=1.
Therefore, x could be either 1 or 3 or -3
False as x can be 3.
III.
If x is a prime number, x cannot be 1. Therefore, y=-3 or y=-3/x
We know x and y are integers, hence x can only be 3.
Thus, y=-3 or -1
y/x could be -1 or -1/3
False as y/x can be -1.
A
The question is a must be true question and we know x, y are integers.
I.
If x=-1, the equation becomes (y+3)(-y+3)=0
Therefore, 9-y^2=0
or, y^2=9
or, |y|=3
True.
II.
If |y| is not a prime number, y cannot be -3. Therefore, x=1 or x=-3/y
We know x is an integer and y is an integer as well, |y| can only be 1 or 3.
Since |y| cannot be 3, |y|=1.
Therefore, x could be either 1 or 3 or -3
False as x can be 3.
III.
If x is a prime number, x cannot be 1. Therefore, y=-3 or y=-3/x
We know x and y are integers, hence x can only be 3.
Thus, y=-3 or -1
y/x could be -1 or -1/3
False as y/x can be -1.
A
16 Jul 2026, 20:05
Kudos
Bookmarks
(x-1)(y+3)(xy+3)=0, Given x,y:integers
Evaluating I,
x=-1,
.:-2(y+3)(-y+3)=0
.: (y+3)(y-3)=0
.:y^2 -9=0
.|y|=3
This MBT.
Evaluating II.
|y| not a prime number.
Now notice the equation to have x as an integer number, value of y has to be +/-3 or +/-1.
Now y can not be +/-3 because we are considering the case when |y| is not a prime number.
.: when y=-1,
(x-1)*2*(-x+3)=0
.: (x-1)(x-3)=0
.: x can be 3, a prime number.
.:this is not MBT.
Evaluating III,
x is prime number.
say x=3
.: 2*(y+3)(3y+3)=0
.: (y+3)(y+1)=0
.: y can be -3
.:y/x is -1
.:NOT MBT.
THE ANSWER IS A.
Evaluating I,
x=-1,
.:-2(y+3)(-y+3)=0
.: (y+3)(y-3)=0
.:y^2 -9=0
.|y|=3
This MBT.
Evaluating II.
|y| not a prime number.
Now notice the equation to have x as an integer number, value of y has to be +/-3 or +/-1.
Now y can not be +/-3 because we are considering the case when |y| is not a prime number.
.: when y=-1,
(x-1)*2*(-x+3)=0
.: (x-1)(x-3)=0
.: x can be 3, a prime number.
.:this is not MBT.
Evaluating III,
x is prime number.
say x=3
.: 2*(y+3)(3y+3)=0
.: (y+3)(y+1)=0
.: y can be -3
.:y/x is -1
.:NOT MBT.
THE ANSWER IS A.
Bunuel
16 Jul 2026, 20:42
Kudos
Bookmarks
for (x-1)(y+3) ( xy+3) to be 0, atleast one of the terms has to be 0. For must be true questions try and always prove not true.
for I. if x = -1, the first term =-2 so either (y+3) or (xy+3) has to be zero. so for either of them to be zero y has to be 3 or -3. so |y|= 3, a prime no.
for II. |y| is not a prime no. so (y+3) isnt 0. so either (x-1) or (xy+3) has to be 0. so suppose if y =-1 and x=3, xy=-3 and the third term would be 0. therefore x can be a prime no.
for III. suppose x=3 and y = -3, second term is 0. also y/x=-1, an integer. must not be true.
for I. if x = -1, the first term =-2 so either (y+3) or (xy+3) has to be zero. so for either of them to be zero y has to be 3 or -3. so |y|= 3, a prime no.
for II. |y| is not a prime no. so (y+3) isnt 0. so either (x-1) or (xy+3) has to be 0. so suppose if y =-1 and x=3, xy=-3 and the third term would be 0. therefore x can be a prime no.
for III. suppose x=3 and y = -3, second term is 0. also y/x=-1, an integer. must not be true.
Quote:
alamin8
Joined: 24 May 2025
Last visit: 05 Sep 2026
Posts: 183
Given Kudos: 32
Location: Bangladesh
Concentration: Accounting, Technology
Schools: HEC Sept '27 INSEAD Aug '27 UNC '27
GPA: 3.89
16 Jul 2026, 21:12
Kudos
Bookmarks
I. x = -1,
-2* (3+y) *(3-y) = 0
9 - y^2 = 0
y = 3 or -3
Therefore, |y| = 3, a prime number.
II. Let's |y| = 4 (not a prime number)
(x-1)*7*(4x+3) = 0
or, 4x^2 - 4x + 3x - 3 = 0
or, 4x(x - 1) + 3(x-1) = 0
or, (x-1) (4x + 3) = 0
or, x = 1, x = -(3/4)
Let's |y| = -1 not a prime number
(x-1)*2*(3-x) = 0
3x - 3 - x^2 + x = 0
3 (x -1) - x (x - 1) = 0
or, x = 1 or x = 3, which is a prime number
Therefore, statement II is not correct
III. x = 3
2*(y+3)*(3y+3) = 0
or, y = - 3 or y = -1
therefore, y/x = -3/3 = -1 is an integer or y = -1/3 is not an integer. There are two different answer. Not sufficient.
Therefore, A (only I) is the answer.
-2* (3+y) *(3-y) = 0
9 - y^2 = 0
y = 3 or -3
Therefore, |y| = 3, a prime number.
II. Let's |y| = 4 (not a prime number)
(x-1)*7*(4x+3) = 0
or, 4x^2 - 4x + 3x - 3 = 0
or, 4x(x - 1) + 3(x-1) = 0
or, (x-1) (4x + 3) = 0
or, x = 1, x = -(3/4)
Let's |y| = -1 not a prime number
(x-1)*2*(3-x) = 0
3x - 3 - x^2 + x = 0
3 (x -1) - x (x - 1) = 0
or, x = 1 or x = 3, which is a prime number
Therefore, statement II is not correct
III. x = 3
2*(y+3)*(3y+3) = 0
or, y = - 3 or y = -1
therefore, y/x = -3/3 = -1 is an integer or y = -1/3 is not an integer. There are two different answer. Not sufficient.
Therefore, A (only I) is the answer.
