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 May 2016, 01:39
Kudos
Bookmarks
etienneg
You cannot solve the questions by assuming that the diagram is drawn to scale. Two angles which look equal may not actually be equal. You can just assume the basics - straight lines that look straight are straight; points are in the order in which they are drawn. But don't take calls based on relative length of lines, relative measure of angles according to the diagram drawn etc.
31 Mar 2018, 18:33
Kudos
Bookmarks
I think I understand how to solve the problem now, but I am having trouble understanding one assumption that people are making in many of the solutions: that O is at the origin. When I tried to solve this problem in the practice test I got to the point where I knew the radius was 2 and that line QO had a slope that was the negative reciprocal of line PO (since they were perpendicular bisectors), but I couldn't solve for slope m of line QO because I was stuck with 1=-sqrt(3)*m+b.
Can we assume that O is at the origin even though it isn't stated in the question, just because it looks like it is? Why can we assume that and not that the y-axis is equidistant from points P and Q (the diagram shows the y-axis passing directly in the middle of the 90 degree angle shown)?
Can we assume that O is at the origin even though it isn't stated in the question, just because it looks like it is? Why can we assume that and not that the y-axis is equidistant from points P and Q (the diagram shows the y-axis passing directly in the middle of the 90 degree angle shown)?
01 Apr 2018, 02:31
Kudos
Bookmarks
ktjorge
O is the origin because it's on the intersection of x and y axis. Also, if we did not know that, we would not be able to solve the question and since it's a PS questions, we should be able to.
OFFICIAL GUIDE:
Problem Solving
Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.
Data Sufficiency:
Figures:
• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2).
• Lines shown as straight are straight, and lines that appear jagged are also straight.
• The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero.
• All figures lie in a plane unless otherwise indicated.
