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 →

GMATPrep Geometry

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
Senior | Next Rank: 100 Posts
Posts: 51
Joined: Sun Mar 23, 2008 6:54 pm
Location: San Francisco
Thanked: 2 times

by sukrant26 » Tue Mar 25, 2008 10:34 pm
As QR=RS, <RQS=<RSQ. Similarly as TS=TU, <TSU=<TUS.----------- (1)

Also, <RSQ+<x+<TSU=180 degrees.
Using (1) <RQS +x+<SUT=180 degrees.--------------------(2)

Now sum of angles of a quadrilateral is 360 degrees.
Hence <x+<SQP+<SUP=270 degrees.---------------------------(3)

<RQS+<SQP=180 degrees----------------------------(4)
<SUT+<SUP=180 degrees----------------------------(5)


Using (2), (4) & (5)..... 180 - <SQP+ x + 180-<SUP = 180

180+X=<SQP+<SUP--------------------------(6)

Using (3) & (6)

180+x=270-x

x=45 degrees.......Hence answer is C.

Am not sure if there is any easy way to solve this question



Sukrant........
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Mon Feb 25, 2008 10:36 pm

by jaydeer44 » Wed Mar 26, 2008 6:05 pm
thanks!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Mon Feb 25, 2008 10:36 pm

anyone have any thoughts on this?

by jaydeer44 » Wed Apr 02, 2008 4:47 pm
I think this might be an easier method?

After redrawing the triangle to fit assumptions (1) and (2), I realized that the entire triangle must actually be an isosceles triangle (someone please correct me if I am wrong). Accordingly, angles QRS and UTS each equal 45 degrees. From there, angles QSR and UST are solved to equal 67.5 degrees each, leaving angle x to equal 180 - (67.5 + 67.5) = 45 degrees.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Fri May 23, 2008 11:14 am

by aditikedia » Thu Aug 07, 2008 2:10 am
Think this might be a shorter way...

Let's call angle SUT and TSU "y"
Let's call angle RQS and RSQ "p"

Since angle SUT and TSU are equal, angle STU is 180-2y
Since RQS and RSQ are equal, angle QRS is 180 -2p

But we also know that angle QRS = 180-(RPT+PTR)
180-2p= 180-(90+180-2y)
2y +2p = 270
y+p=135

Therefore x= 180-135=45
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1