GRE Prep 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.
In the figure above, if AD is parallel to BC, then ∠ADC =
[#permalink]
28 Apr 2020, 10:08
28 Apr 2020, 10:08
Expert Reply
2
Bookmarks
00:00
Question Stats:
52% (02:00) correct
48% (03:33) wrong based on 25 sessions
Hide Show timer Statistics
Attachment:
#greprepclub In the figure above, if AD is parallel to BC, then.jpg [ 21.84 KiB | Viewed 2099 times ]
In the figure above, if AD is parallel to BC, then \(∠ADC = \)
A. 11°
B. 22°
C. 33°
D. 46°
E. 134°
Kudos for the right answer and explanation
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: In the figure above, if AD is parallel to BC, then ∠ADC =
[#permalink]
15 Jan 2022, 12:19
15 Jan 2022, 12:19
1
Carcass wrote:
Attachment:
#greprepclub In the figure above, if AD is parallel to BC, then.jpg
In the figure above, if AD is parallel to BC, then \(∠ADC = \)
A. 11°
B. 22°
C. 33°
D. 46°
E. 134°
Kudos for the right answer and explanation
First, since angles in a triangle must add to 180°, we can see that the missing angle in the red triangle must be 180° - (x + 44)°
Simplify this measurement to get (136 - x)°
Finally, since AD is parallel to BC, we know that the two highlighted angles below must add to 180°.
So, we can write: (136 - x)° + 2x° + 3x° = 180°
Simplify: 136 + 4x = 180
Subtract 136 from both sides: : 4x = 44
Solve: x = 11
Our goal is to find the measurement of ∠ADC
Since ∠ADC = 3x°, we can replace x with 11 to get: ∠ADC = 3x° = 3(11)° = 33°
Answer: C
General Discussion
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Re: In the figure above, if AD is parallel to BC, then ∠ADC =
[#permalink]
07 Jun 2020, 04:16
07 Jun 2020, 04:16
1
AD is parallel to BC
So, Angle ADC + Angle BCD = 180°
Now, Angle BCD = Angle BCE + 2x
[Refer attached Image]
= 136°-x + 2x = 136° + x
Angle ADC + Angle BCD = 180°
=> 3x + 136° + x = 180
=> 4x = 180°-136° = 44°
=> x = 11°
So, Angle ADC = 3x = 3*11° = 33°
So, answer will be C
Hope it helps!
So, Angle ADC + Angle BCD = 180°
Now, Angle BCD = Angle BCE + 2x
[Refer attached Image]
= 136°-x + 2x = 136° + x
Angle ADC + Angle BCD = 180°
=> 3x + 136° + x = 180
=> 4x = 180°-136° = 44°
=> x = 11°
So, Angle ADC = 3x = 3*11° = 33°
So, answer will be C
Hope it helps!
Attachments
image-5.jpg [ 41.13 KiB | Viewed 1941 times ]
