Last visit was: 11 Jul 2025, 14:29 It is currently 11 Jul 2025, 14:29
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
avatar
JJ2014
Joined: 06 Sep 2012
Last visit: 26 Aug 2013
Posts: 35
Own Kudos:
199
 [36]
Given Kudos: 42
Concentration: Social Entrepreneurship
Posts: 35
Kudos: 199
 [36]
4
Kudos
Add Kudos
32
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
mikemcgarry
User avatar
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Last visit: 06 Aug 2018
Posts: 4,480
Own Kudos:
30,105
 [12]
Given Kudos: 130
Expert
Expert reply
Posts: 4,480
Kudos: 30,105
 [12]
7
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
General Discussion
User avatar
shanmugamgsn
Joined: 04 Oct 2011
Last visit: 31 Dec 2014
Posts: 141
Own Kudos:
Given Kudos: 44
Location: India
Concentration: Entrepreneurship, International Business
GMAT 1: 440 Q33 V13
GPA: 3
GMAT 1: 440 Q33 V13
Posts: 141
Kudos: 154
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
mikemcgarry
User avatar
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Last visit: 06 Aug 2018
Posts: 4,480
Own Kudos:
30,105
 [1]
Given Kudos: 130
Expert
Expert reply
Posts: 4,480
Kudos: 30,105
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
shanmugamgsn
I do have a doubt !
Corresponding angle means the angle made on its side?

No. "Corresponding angles" is a technical term from Euclidean Geometry. It's the name of a particular pair of angles formed when a transversal crosses a pair of parallel lines.
Attachment:
parallel line diagram.JPG
parallel line diagram.JPG [ 22.96 KiB | Viewed 25786 times ]
The following pairs are corresponding angles
1 & 5
2 & 6
3 & 7
4 & 8
Corresponding angles are congruent if and only if the lines are parallel.

The following pairs are alternate interior angles
3 & 6
4 & 5
Alternate interior angles are congruent if and only if the lines are parallel

The following pairs are alternate exterior angles
1 & 8
2 & 7
Alternate exterior angles are congruent if and only if the lines are parallel

The following pairs are same side interior angles
3 & 5
4 & 6
Same side interior angles are supplementary if and only if the lines are parallel

The following pairs are same side exterior angles
1 & 7
2 & 8
Same side exterior angles are supplementary if and only if the lines are parallel

Those are all the names relating pairing an angle at one vertex with an angle at the other vertex, when a transversal intersects a pair of parallel lines.

Does all this make sense?

Mike :-)
User avatar
priyamne
Joined: 24 Apr 2012
Last visit: 15 Feb 2014
Posts: 36
Own Kudos:
Given Kudos: 1
Posts: 36
Kudos: 54
Kudos
Add Kudos
Bookmarks
Bookmark this Post
JJ2014
Attachment:
Screen shot 2012-12-09 at 1.40.06 PM.png
In the figure above, QRS is a straight line and QR = PR. Is it true that lines TR and PQ parallel?

(1) Length PQ = Length PR
(2) Line TR bisects angle PRS

Ans: For lines TR and PQ to be parallel angle PQR= angle TRS. From statement 1 we get angle PQR=x=60 but nothing about angle TRS.
From statement 2 we get PRQ=180-2X , therefore PRS=180-(180-2x)=2x and TR bisects it so angle TRS=x which is equal to PQR. Therefore the answer is (B).
User avatar
jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,316
Own Kudos:
2,719
 [2]
Given Kudos: 355
Concentration: Finance
Posts: 1,316
Kudos: 2,719
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let's solve. So we need to know if TR is parallel to PQ.

Now then, let's hit the first statement. We are told that PQ=QR. Now we know that PQR is an equilateral triangle but still no info on TR. Therefore, insufficient.

Statement 2, we have that TR bisects PRS. Now let's see. So we know that QR=PR from the question stem. Hence angle QRP is 180-2x, 'x' being the angles P and Q respecively. Therefore angle PRS will be 2x since QRS is a straight line with total measure of 180 degrees. Now if TRS bisects then angle TRS is x only. Which means that the angles Q and R are equal and thus PQ // TR.

B stands

Cheers!
J :)
User avatar
BrainLab
User avatar
Current Student
Joined: 10 Mar 2013
Last visit: 26 Jan 2025
Posts: 350
Own Kudos:
3,009
 [3]
Given Kudos: 200
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.7
WE:Marketing (Telecommunications)
GMAT 1: 580 Q46 V24
Posts: 350
Kudos: 3,009
 [3]
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
JJ2014
Attachment:
The attachment Screen shot 2012-12-09 at 1.40.06 PM.png is no longer available
In the figure above, QRS is a straight line and QR = PR. Is it true that lines TR and PQ parallel?

(1) Length PQ = Length PR
(2) Line TR bisects angle PRS

(1) this tells us that all 3 sides are equal = Equilateral Triangle. Not sufficient, as there is no info about TR
(2) Now look at the attachment, I'll use the property of the "Exterior Angles of a Triangle" Angle Y is the sum of the angles on the both sides of TR

Exterior angle of a triangle is equal to the sum of the opposite interior angles

Angle Y=2X and if TR bisects angle Y it means that Angle TRS = \(\frac{Y}{2}=\frac{2X}{2}=X\) and when this two angles of to different triangles are equal then QP and TR are parallel.
Answer B
Attachments

Screen shot 2012-12-09 at 1.40.06 PM.png
Screen shot 2012-12-09 at 1.40.06 PM.png [ 5.36 KiB | Viewed 22862 times ]

User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,375
Own Kudos:
Posts: 37,375
Kudos: 1,010
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderator:
Math Expert
102634 posts