Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
17 Oct 2025, 06:40
Kudos
Bookmarks
Goal: find OA = radius r.
Statement 1. Area = 250.
πr2 = 250 → r = √(250/π). That is one unique value (≈ 8.924 ft). So statement (1) is sufficient.
Note: A few posts in the thread mistakenly asserted r = 10 from area = 250 — that’s wrong because π·102 = 100π ≈ 314, not 250. The correct r is √(250/π) ≈ 8.924 ft.
Statement 2. OC = 20.
O, A, C form a right triangle at A because CA is tangent and OA is radius (OA ⟂ CA). So OA2 + AC2 = OC2 = 400. That single equation does not determine OA uniquely because AC is unknown; many (OA, AC) pairs satisfy OA2 + AC2 = 400. So (2) is not sufficient.
Common incorrect arguments addressed:
“It must be a 12–16–20 triangle.” No — knowing the hypotenuse is 20 doesn’t force integer legs or that particular Pythagorean triple.
“OABC is a square so OA = AC and OC = 20 ⇒ OA = 20/√2.” That mixes up things Clean my speaker : tangent segments from C satisfy CA = CB, but CA is not equal to OA (radius). OA = OB (both radii), and CA = CB (tangent segments from same external point), but that does not force OA = CA. So no square is implied.
Conclusion: (1) sufficient, (2) not sufficient → answer A.
Statement 1. Area = 250.
πr2 = 250 → r = √(250/π). That is one unique value (≈ 8.924 ft). So statement (1) is sufficient.
Note: A few posts in the thread mistakenly asserted r = 10 from area = 250 — that’s wrong because π·102 = 100π ≈ 314, not 250. The correct r is √(250/π) ≈ 8.924 ft.
Statement 2. OC = 20.
O, A, C form a right triangle at A because CA is tangent and OA is radius (OA ⟂ CA). So OA2 + AC2 = OC2 = 400. That single equation does not determine OA uniquely because AC is unknown; many (OA, AC) pairs satisfy OA2 + AC2 = 400. So (2) is not sufficient.
Common incorrect arguments addressed:
“It must be a 12–16–20 triangle.” No — knowing the hypotenuse is 20 doesn’t force integer legs or that particular Pythagorean triple.
“OABC is a square so OA = AC and OC = 20 ⇒ OA = 20/√2.” That mixes up things Clean my speaker : tangent segments from C satisfy CA = CB, but CA is not equal to OA (radius). OA = OB (both radii), and CA = CB (tangent segments from same external point), but that does not force OA = CA. So no square is implied.
Conclusion: (1) sufficient, (2) not sufficient → answer A.
