Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 19 Apr 2015, 11:08

### 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Challenge: Yin and Ying addendum

Author Message
TAGS:
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 50 [0], given: 0

Hi:

Someone solved Stolyar's original YIN-YAN problem too quickly so I decided to add a little twist to make it more "interesting."

In the attached diagram, the assumptions are the same as in Stolyar's problem (A, B, C are on diameter, center of circle is B, and all arcs are semicircles). Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded area region above the red line to the area of the shaded region below the red line? (Diagram is not drawn to scale and angles drawn are not accurate).

A) 3/4
B) 5/6
C) 1
D) 7/5
E) 9/7
Attachments

ying.jpg [ 11.69 KiB | Viewed 788 times ]

ying.jpg [ 11.69 KiB | Viewed 792 times ]

_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

AkamaiBrah wrote:
Hi:

Someone solved Stolyar's original YIN-YAN problem too quickly so I decided to add a little twist to make it more "interesting."

In the attached diagram, the assumptions are the same as in Stolyar's problem (A, B, C are on diameter, center of circle is B, and all arcs are semicircles). Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded area region above the red line to the area of the shaded region below the red line? (Diagram is not drawn to scale and angles drawn are not accurate).

A) 3/4
B) 5/6
C) 1
D) 7/5
E) 9/7

stolyar, what if the info YXA = 105 is not given.

can we still solve the problem.

I tried using the property that an angle inscribed in a semicircle is 90.

So i get AYC = 90 and AXC = 90 . Also ABC,BYC and XYB are

isosceles triangles.

I couldnt really go further , any ideas?

thanks
praetorian
SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

I do not think so.
If an angle is not given, then point Y can slide anywhere around the circle, making parts of the shaded region change. We need to fix point Y somehow.
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 50 [0], given: 0

stolyar wrote:
I do not think so.
If an angle is not given, then point Y can slide anywhere around the circle, making parts of the shaded region change. We need to fix point Y somehow.

Stolyar was able to solve this very nicely, but I deleted his post so that everyone else would have a chance to solve it.

P.S.: Stolyar -- everyone knows you are a whiz. Since you are a moderator, please give everyone a chance to solve before solving! (you can always message your solution to me privately and I will let you know how you did and give you credit if you get it right).
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Intern
Joined: 04 Jun 2003
Posts: 17
Followers: 0

Kudos [?]: 0 [0], given: 0

If angle axy=105 then sector acy=210. obviously sector ac=180 right. Thus, sector yc=30 (210-180). Sector yc is enclosed by the central angle ybc, making angle ycb=30 degrees. 30/360 = 1/12. so the area of that portion is pi/12.

the area of the semi circle is .5*(.5)^2*pi or pi/8

so the area below the red line is pi/12 + pi/8 = 5pi/24

the area of the shaded region is pi/2 (see original ying yang problem).

thus the area above the red line is pi/2 -pi/12 - pi/8 = 7pi/24

the ratio requested in the problem is (7pi/24)/(5pi/24) = 7/5

Great problems akamai. If the teachers at that math work shop are half as good as you then i'm sure i would have gotten my money's worth, too bad they're filled up.
Keep em coming
Intern
Joined: 21 Jul 2003
Posts: 42
Location: India
Followers: 1

Kudos [?]: 0 [0], given: 0

how is sector acy = 210?
Intern
Joined: 04 Jun 2003
Posts: 17
Followers: 0

Kudos [?]: 0 [0], given: 0

Soumala wrote:
how is sector acy = 210?

the problem states that axy = 105. For any non-central angle in a circle the sector enclosed by that angle is always twice that angle. For a central angle the sector is the same as the angle. For example if xby = 80 then sector xy would also equal 80.
Intern
Joined: 21 Jul 2003
Posts: 42
Location: India
Followers: 1

Kudos [?]: 0 [0], given: 0

Thanks for the explanation Sandoval.Looks like I need to brush up my fundamentals.
Similar topics Replies Last post
Similar
Topics:
The Challenge 3 14 Jan 2014, 07:52
1 Addendum to improve GOALS in application? 2 20 Mar 2011, 02:49
Challenges 2 18 Jun 2007, 17:23
challenges 0 02 Feb 2007, 21:29
Challenges! 2 07 Nov 2005, 20:43
Display posts from previous: Sort by