BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

parallelogram

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times
by goyalsau » Thu Dec 02, 2010 8:47 am
In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Sat Oct 16, 2010 7:27 am
Thanked: 2 times

by pellucide » Thu Dec 02, 2010 8:57 am
goyalsau wrote:In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
My answer (C)

These three points make a triangle. Each side of the triangle can be the diagonal of the parallelogram. Since there are three sides of this triangle, there could be three distinct parallelograms.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members
Location: Milpitas, CA
Thanked: 447 times

by Rahul@gurome » Thu Dec 02, 2010 9:09 am
goyalsau wrote:In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

Image

See the above image.
For three non collinear points in a plane always only three parallelograms can be drawn with those points as vertex.

The correct answer is C.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 332
Joined: Tue Feb 09, 2010 3:50 pmFollowed by: 7 members
Thanked: 41 times
GMAT Score:720

by rishab1988 » Thu Dec 02, 2010 9:22 am
Great question guys.

rahul it would be great if you could shed some light on whether combinatorics can be used to solve this question,because what you showed was quite intuitive and might not have been thought of others.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Thu Dec 02, 2010 9:26 am
Rahul@gurome wrote:
goyalsau wrote:In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

Image

See the above image.
For three non collinear points in a plane always only three parallelograms can be drawn with those points as vertex.

The correct answer is C.
Thanks Rahul for the figure but , I am not able to understand it completely Please explain with some detail so i can understand it.
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 332
Joined: Tue Feb 09, 2010 3:50 pmFollowed by: 7 members
Thanked: 41 times
GMAT Score:720

by rishab1988 » Thu Dec 02, 2010 9:37 am
goyalsau wrote:
Rahul@gurome wrote:
goyalsau wrote:In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

Image

See the above image.
For three non collinear points in a plane always only three parallelograms can be drawn with those points as vertex.

The correct answer is C.
Thanks Rahul for the figure but , I am not able to understand it completely Please explain with some detail so i can understand it.
Goyal try drawing the xy coordinate plan and then plot those points.Join those points and form a triangle.Rahul's concept is based on keeping each side once as a diagonal and the other 2 as one of the 2 parallel sides.

But this is very intuitive method.I got to agree.

I was hoping that this question could be solved using combinatorics.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members
Location: Milpitas, CA
Thanked: 447 times

by Rahul@gurome » Thu Dec 02, 2010 9:51 am
goyalsau wrote:Thanks Rahul for the figure but , I am not able to understand it completely Please explain with some detail so i can understand it.
In the figure, points A, B and C are any three given points in a plane which are non-collinear. Points O, P and Q are possible points each of which with A, B and C gives rise to a parallelogram. Thus the possible parallelograms are OBCA, ABCQ and ABPC.
rishab1988 wrote:rahul it would be great if you could shed some light on whether combinatorics can be used to solve this question,because what you showed was quite intuitive and might not have been thought of others.
The question is based on simple knowledge of geometry (to be precise concept of parallel lines) and can be answered applying the concept. If you insist then combinatorics can be applied as follows,
  • The three given points can be joined by three lines. Now three of them simultaneously cannot be sides of a parallelogram, but any two of them can. Number of ways to select two lines out of three = 3C2 = 3
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 332
Joined: Tue Feb 09, 2010 3:50 pmFollowed by: 7 members
Thanked: 41 times
GMAT Score:720

by rishab1988 » Thu Dec 02, 2010 9:59 am
got it.

thanks for the amazing explanation.

I know that it could be solved the first way,but I always like to know as many methods of solving as question as possible.You never know which one might be more useful...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1