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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

I FEEL LIKE THIS Q IS POSTED IN THE FORUM EARLIER BUT CUD NO

Expert replies
Source: — Problem Solving |

by pops » Mon Mar 29, 2010 9:21 pm
I am not very convinced with the options you gave as all the options are more than 2 pi (which means whole circle)
But here is my approach:
angle (PRO) = 35
since, CR = CP (Radii of same circle) => angle( CRP) = angle (CPR) = 35
hence, angle (PCR)=110
hence, angle (PCO) = 70
hence, angle (CPO) = 70 (alternate angles of parallel lines)

since, CP=CQ (radii) => angle (CPQ)=angle (CQP) = 70
hence, angle (PCQ) = 180 - 70 - 70 = 40
now, 360 --- > 2pi
hence, 40 ----> 2 pi * 40 / 360 = 2 pi / 9
Image
Join the discussion

by thephoenix » Mon Mar 29, 2010 10:29 pm
pops u have made a slight mistake
360=2 pi r
u have missed it
well thanks for your approach
ans is 2pi


is there any other approach
Join the discussion

by solankijignesh » Mon Mar 29, 2010 10:53 pm
i think we can also solve it using inner angles in circle.

angle PCO = 2 x angle PRO = 70
similarly, angle QCR = 70

major arc PQ = 180+70+70 = 320
Minor arc PQ = (40 / 360) * 18 pi = 2pi

is this approach correct?
Join the discussion

by pops » Mon Mar 29, 2010 11:20 pm
thephoenix wrote:pops u have made a slight mistake
360=2 pi r
u have missed it
well thanks for your approach
ans is 2pi


is there any other approach
perfect... my mistake :(

thanks !!
Join the discussion

by eaakbari » Mon Mar 29, 2010 11:49 pm
Since RC = PC
<CPR = <CRP
Hence <p = 70

Alsp <p=<q

Hence <pcq = 180 - 70 - 70 = 40

Hence arc length = 40/360 * 2 *pi*9

Answer = 2pi
Join the discussion