Last visit was: 22 Apr 2026, 13:22 It is currently 22 Apr 2026, 13:22
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: 22 Apr 2026
Posts: 109,754
Own Kudos:
Given Kudos: 105,823
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,754
Kudos: 810,658
 [22]
Kudos
Add Kudos
22
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
CareerGeek
Joined: 20 Jul 2017
Last visit: 21 Apr 2026
Posts: 1,286
Own Kudos:
4,431
 [6]
Given Kudos: 162
Location: India
Concentration: Entrepreneurship, Marketing
GMAT 1: 690 Q51 V30
WE:Education (Education)
GMAT 1: 690 Q51 V30
Posts: 1,286
Kudos: 4,431
 [6]
6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
vg18
Joined: 30 Nov 2017
Last visit: 17 Mar 2021
Posts: 60
Own Kudos:
85
 [6]
Given Kudos: 95
GMAT 1: 690 Q49 V35
GMAT 1: 690 Q49 V35
Posts: 60
Kudos: 85
 [6]
5
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 22 Apr 2026
Posts: 8,627
Own Kudos:
5,190
 [1]
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
Products:
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,627
Kudos: 5,190
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Is x>0?

(1) |2x−12|<10
(2) x2−10x≥−21

#1
2x-12<10
2x<22
x<11
and
-2x+12<10
-2x<-2
x>1
sufficient

1<x<11
yes insufficient
#2
x2−10x≥−21
(x-7)(x-3)>=0
x>=7 and x<=3
insufficient
IMO A
User avatar
madgmat2019
Joined: 01 Mar 2019
Last visit: 17 Sep 2021
Posts: 584
Own Kudos:
640
 [1]
Given Kudos: 207
Location: India
Concentration: Strategy, Social Entrepreneurship
GMAT 1: 580 Q48 V21
GPA: 4
Products:
GMAT 1: 580 Q48 V21
Posts: 584
Kudos: 640
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
(1) |2x-12|<10

From this we can get 11>x>1

Sufficient


(2) x^2-10x >= -21

We get (x-7)(x-3)>=0

x>=7 or x<=3
Not sufficient

OA:A

Posted from my mobile device
User avatar
exc4libur
Joined: 24 Nov 2016
Last visit: 22 Mar 2022
Posts: 1,680
Own Kudos:
1,469
 [1]
Given Kudos: 607
Location: United States
Posts: 1,680
Kudos: 1,469
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Quote:
Is \(x>0\)?

(1) \(|2x−12|<10\)
(2) \(x^2−10x≥−21\)

(1) \(|2x−12|<10\) sufic.

\(|2x−12|<10…(2x-12)^2<10^2…4x^2+44-48x<0…x^2-12x+11<0…(x-11)(x-1)<0\)
\((x-11)(x-1)<0…[less.than.sign=inside.range]…1<x<11\)

(2) \(x^2−10x≥−21\) insufic.

\(x^2−10x≥−21…x^2-10x+21≥0…(x-7)(x-3)≥0\)
\((x-7)(x-3)≥0…[greater.than.sign=outside.range]…x≥3…or…x≤7\)
\(x=[-9,-1,0,1,2,3,7,8,100…]\)

Answer (A)
User avatar
eakabuah
User avatar
Retired Moderator
Joined: 18 May 2019
Last visit: 15 Jun 2022
Posts: 774
Own Kudos:
1,144
 [1]
Given Kudos: 101
Posts: 774
Kudos: 1,144
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We are to determine if x>0?

1. |2x-12|<10
when x>0, 2x-12<10
hence x<11

when x<0
-(2x-12)<10
2x-12>-10
x>1
Hence x>1
This results in a range of 1<x<11
Statement 1 alone is therefore sufficient since x>1 in the range above.

2. x^2 -10x ≥-21
x^2 -10x + 21≥0
(x-7)(x-3)≥0
This results in the range x≥7 and x≤3
This is insufficient because when x≥7 then x is always greater than 0. But when x≤3, then x is not always greater than 0, because x can be -1, which is less than 0.

The answer is therefore A.
User avatar
lacktutor
Joined: 25 Jul 2018
Last visit: 23 Oct 2023
Posts: 658
Own Kudos:
1,446
 [1]
Given Kudos: 69
Posts: 658
Kudos: 1,446
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Is x >0?

(Statement1): -10 < 2x—12< 10
—> 2< 2x< 22
1 < x < 11
Sufficient

(Statement2): \(x^{2} —10x+ 21>=0\)
(x—3)(x—7) >=0

x <=3 and x>=7
—> If x=2, then YES
—> If x= —1, then NO
Insufficient

The answer is A

Posted from my mobile device
avatar
suchithra
Joined: 31 Oct 2015
Last visit: 14 Oct 2022
Posts: 89
Own Kudos:
119
 [1]
Given Kudos: 179
Posts: 89
Kudos: 119
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Question : is X>0

(1) Given : |2x−12|<10
i.e |x-6|<5
This implies 1<x<11

Statement 1 is sufficient

(2) x^2−10x≥−21
i.e x^2−10x+21≥0
This imples x<3 or x>7
According to this X could take positive and negative values.

Therefore answer is A.
User avatar
Raxit85
Joined: 22 Feb 2018
Last visit: 02 Aug 2025
Posts: 761
Own Kudos:
1,202
 [2]
Given Kudos: 135
Posts: 761
Kudos: 1,202
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Can anyone please explain regarding x <=3 rather than x >=3??

Posted from my mobile device
User avatar
Vibhatu
Joined: 18 May 2021
Last visit: 19 Jan 2026
Posts: 184
Own Kudos:
Given Kudos: 187
Posts: 184
Kudos: 55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel please post the official solution
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 22 Apr 2026
Posts: 109,754
Own Kudos:
Given Kudos: 105,823
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,754
Kudos: 810,658
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Is \(x > 0\)?

(1) \(|2x - 12| < 10\).

Get rid of the modulus sign:
    \(-10<2x - 12 < 10\);

Add 12 to all three parts:
    \(2<2x < 22\);

Divide by 2:
    \(1<x < 11\).

Sufficient.


(2) \(x^2 - 10x \geq {-21}\)

Here I'd use number plugging:

    If x is large enough, say if x = 1,000, the inequality holds. So, x can be more than 0.
    If x = 0, the inequality holds too. So, in this case x > 0 is not true.

Two different answers. Not sufficient.

Another approach:

Transitions points are x = 3 and x = 7. "\(\geq\)" sign indicates that the solution is to the left of the smaller root and to the right of the larger root. Thus \(x \leq 3\) and \(x \geq 7\). Check Solving Quadratic Inequalities - Graphic Approach: https://gmatclub.com/forum/solving-quad ... 70528.html

So, x may or may not be greater than 0. Not sufficient.

Answer: A.

Hope it's clear.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,968
Own Kudos:
Posts: 38,968
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109754 posts
498 posts
212 posts