Last visit was: 16 Jul 2025, 00:34 It is currently 16 Jul 2025, 00:34
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 16 July 2025
Posts: 102,584
Own Kudos:
Given Kudos: 98,191
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,584
Kudos: 741,756
 [21]
1
Kudos
Add Kudos
19
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 16 July 2025
Posts: 102,584
Own Kudos:
741,756
 [7]
Given Kudos: 98,191
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,584
Kudos: 741,756
 [7]
1
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
General Discussion
avatar
lucassandrade
Joined: 16 Jun 2015
Last visit: 13 May 2016
Posts: 5
Own Kudos:
Given Kudos: 1
Posts: 5
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 16 July 2025
Posts: 102,584
Own Kudos:
741,756
 [1]
Given Kudos: 98,191
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,584
Kudos: 741,756
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
lucassandrade
Bunuel
Official Solution:


(1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \leq 0\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).

So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.

(2) \(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \leq 0\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.


Answer: D

Bunuel,

Why in (1) you made X-2 = 0? I didn't understand why LHS should be the least value.

BR,

Have you read the highlighted part?

In \(|x - 2| \lt 2 - y\), the left hand side of the inequality (|x - 2|) is an absolute value. An absolute value of a number cannot be negative, so the left hand side of the inequality (|x - 2|) is not negative, which means that the right hand side of the inequality, which is greater than the left hand side, also cannot be negative. This gives \(0 \lt 2 - y\).
User avatar
danjbon
Joined: 14 Oct 2015
Last visit: 25 Oct 2021
Posts: 11
Own Kudos:
GMAT 1: 640 Q45 V33
GMAT 1: 640 Q45 V33
Posts: 11
Kudos: 38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I think this is a high-quality question and I agree with explanation. This question is repeated.
User avatar
GMATDemiGod
Joined: 23 Sep 2015
Last visit: 05 Feb 2017
Posts: 64
Own Kudos:
52
 [4]
Given Kudos: 213
Concentration: General Management, Finance
GMAT 1: 680 Q46 V38
GMAT 2: 690 Q47 V38
GPA: 3.5
GMAT 2: 690 Q47 V38
Posts: 64
Kudos: 52
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Hey guys

Here is what helped me solve the problem - I had trouble at first and trouble understanding the solution. So hopefully this helps someone.

So we know X>0 and Y>0

Lets focus on 1

|x-2|<2-y

with x>0 lets use x=1, x=2. x=3 as cases

x=1 ; 1 < 2 - y (y >0 and integer) Not possible. --- BAD CASE
x=2 ; 0 < 2 - y (y>) and interger, there fore y=2) -- GOOD
x=3 ; 1 < 2- y -- BAD
x=4 ; 2 < 2 - y -- BAD

Therefore for this case x must 2 and 2 is prime ..... SUFFICIENT

Prompt 2

x + y - 3 = |1-y|

again, y>0 and integer, so choose y=1, y=2. y=3 y=4

y=1 ; x+1-3=0 --- x=2 --- Good Case
y=2 ; x+2-3=1 ----- x=2 --- Good case
y=3 ; x+3-3=2 --- X=2 --- GooD Case
y=4 ; x+4-3=3 --- x=2 -- Good Case


Therefore x=2, x is prime and this is Sufficient

Therefore overall D is answer
User avatar
grsm
Joined: 20 Jul 2012
Last visit: 28 Jul 2019
Posts: 12
Own Kudos:
56
 [2]
Given Kudos: 67
Products:
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
dear Bunuel,
I got your solution, it really good ! how ever can you please see where i am going wrong in my approach for choice A.

A. |x−2|<2−y
i intent to open the mod and test the values of x and Y. Therefore 2 cases come up while opening the mod

Case 1 x-2>0 i.e x>2
The equation becomes
x-2 < 2-y
x < 4-y
Now according the above inequality and the condition X>2 there is no value of x or y which satisfies the 2<x<4-y

What am i missing here ?

Case 2 x-2<0 i.e x<2
if we analyze this condition carefully the only value that x can take here is 1 coz, 0<x<2 . However the equations becomes,
-x+2 <2-y
x>y
Now this cannot be correct as y does not have any valid +integer value. 0<y<x<2


This is not a optimal approach, but i am keen to know gap do i have in my concepts ?

thanks







Bunuel
Official Solution:


(1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \leq 0\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).

So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.

(2) \(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \leq 0\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.


Answer: D
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 16 July 2025
Posts: 102,584
Own Kudos:
Given Kudos: 98,191
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,584
Kudos: 741,756
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sandysilva
grsm
dear Bunuel,
I got your solution, it really good ! how ever can you please see where i am going wrong in my approach for choice A.

A. |x−2|<2−y
i intent to open the mod and test the values of x and Y. Therefore 2 cases come up while opening the mod

Case 1 x-2>0 i.e x>2
The equation becomes
x-2 < 2-y
x < 4-y
Now according the above inequality and the condition X>2 there is no value of x or y which satisfies the 2<x<4-y

What am i missing here ?

Case 2 x-2<0 i.e x<2
if we analyze this condition carefully the only value that x can take here is 1 coz, 0<x<2 . However the equations becomes,
-x+2 <2-y
x>y
Now this cannot be correct as y does not have any valid +integer value. 0<y<x<2


This is not a optimal approach, but i am keen to know gap do i have in my concepts ?

thanks







Bunuel
Official Solution:


(1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \leq 0\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).

So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.

(2) \(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \leq 0\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.


Answer: D

I have the same query.Can any experts please address what GRSM said. I used the same approach.

Thanks

The problem with that solution is that it does not consider the case when x = 2. When you open the modules for |x - 2| you should consider the case x <= 2 and x > 2 OR x < 2 and x >= 2, so you should include = sign for 2 either in the first case or in second. If you do, then you'll find that x = 2 and y = 1 are possible.

Hope it's clear.
avatar
ankitamundhra28
Joined: 14 Mar 2018
Last visit: 23 Dec 2020
Posts: 20
Own Kudos:
Given Kudos: 249
Location: India
Concentration: Finance, Marketing
GMAT 1: 760 Q50 V42
GPA: 3.48
GMAT 1: 760 Q50 V42
Posts: 20
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Superb Question! Tests concepts very well.
If one is familiar with the concept, it can be solved under 2 minutes.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 16 July 2025
Posts: 102,584
Own Kudos:
Given Kudos: 98,191
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,584
Kudos: 741,756
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Duplicate of M02-25. Unpublished.................
Moderators:
Math Expert
102583 posts
Founder
41103 posts