Last visit was: 01 May 2026, 11:18 It is currently 01 May 2026, 11:18
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
fskilnik
Joined: 12 Oct 2010
Last visit: 03 Jan 2025
Posts: 883
Own Kudos:
1,891
 [6]
Given Kudos: 57
Status:GMATH founder
Expert
Expert reply
Posts: 883
Kudos: 1,891
 [6]
1
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
User avatar
GMATRockstar
Joined: 21 Apr 2014
Last visit: 12 Nov 2025
Posts: 90
Own Kudos:
814
 [1]
Given Kudos: 3
Expert
Expert reply
Posts: 90
Kudos: 814
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
fskilnik
Joined: 12 Oct 2010
Last visit: 03 Jan 2025
Posts: 883
Own Kudos:
1,891
 [1]
Given Kudos: 57
Status:GMATH founder
Expert
Expert reply
Posts: 883
Kudos: 1,891
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
CrackverbalGMAT
User avatar
Major Poster
Joined: 03 Oct 2013
Last visit: 01 May 2026
Posts: 4,846
Own Kudos:
Given Kudos: 226
Affiliations: CrackVerbal
Location: India
Expert
Expert reply
Posts: 4,846
Kudos: 9,188
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We can also get to the answer without any additional constructions.

Referring to the figure drawn below.

Let \(\angle{CDA}\) = x

Therefore \(\angle{GCD}\) = 20 + x (Exterior angle is equal to sum of opposite interior angle)

\(\angle{BCG}\) = \(\angle{GCD}\) = 20 + x (Given AC is the angular bisector of \(\angle{BCD}\))

\(\angle{EBC}\) = \(70^o\) (Sum of angles on a straight line is equal to \(180^o\))

\(\angle{FAC}\) = \(160^o\) (Sum of angles on a straight line is equal to \(180^o\))

Figure EBCAF is a 5 sided polygon, who's sum of angles is = \(540^o\) (Sum of angles of a polygon is (n - 2)*180)

Therefore 90 + 90 + 70 + 160 + \(\angle{FCB}\) = \(540^o\)

\(\angle{C}\) = 540 - 410 = \(130^o\)

Now, \(\angle{FCB}\) + \(\angle{BCG}\) = \(180^o\) (Sum of angles on a straight line is equal to \(180^o\))

130 + 20 + x = 180

x = 180 - 150 = 30

Option B

Arun Kumar
Attachments

Angles Sum.jpg
Angles Sum.jpg [ 538.29 KiB | Viewed 3636 times ]

User avatar
nova31
Joined: 11 Jun 2019
Last visit: 04 Jul 2024
Posts: 13
Own Kudos:
Given Kudos: 74
Location: India
Concentration: Technology, Strategy
GMAT 1: 730 Q49 V40
GPA: 3.4
WE:Engineering (Computer Software)
GMAT 1: 730 Q49 V40
Posts: 13
Kudos: 14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
A more intuitive approach:

1. Extend line BC to AD. Let's call this intersection point of BC on AD as "E".
2. Angle AEC = 110 ( corresponding angles) => angle ACE = 50 (180-20-110)
3. Angle BCX(other end of angle bisector) = Angle ACE ( vertically opp.) = 50
4. Since CX is bisector of Angle BCD, angle BCD = 100
5. The larger angle around C must be 360 - smaller angle BCD = 360 - 100 =260
6. Since CA bisects the larger BCD angle as well, angle ACD = 260/2 = 130
7. Angle ADC = 180 - angle CAD - angle ACD = 180 - 20 - 130 = 30. [ Option B]
Moderators:
Math Expert
109997 posts
Tuck School Moderator
852 posts