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

 It is currently 24 Aug 2016, 14:40

### 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

# This is one of Akamai's .. nice one...

Author Message
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 68

Kudos [?]: 830 [0], given: 781

This is one of Akamai's .. nice one... [#permalink]

### Show Tags

10 Oct 2004, 01:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

This is one of Akamai's .. nice one...

http://www.manhattangmat.com

Praetorian
Manager
Joined: 28 Aug 2004
Posts: 205
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

10 Oct 2004, 06:07
7/5.

Draw a radius from B to BZ (to from diameter YZ).

r = 360/pi (from the length-of-an-arc formula).

Draw a radius from B to A, results in a 'small' semicircle with diameter of 360/pi. Area of this 'small' circle is 90(180)/pi. You have to add it to the area of the sector YBC in the denominator of the ratio, and subtract it from the area of the semicircle in the numerator (together with the area of sector ABZ).

Now, angle YXA = 105, draw a line from X to Z, forming angle YXZ inscribed in a semicircle, = 90 degrees. Then angle AXZ is 105-90=15, and results in arc AZ of 30, with angle ABZ = 30 degrees.

From there, get the area of sector ABZ = 30/360pir^2 = (3*360)/pi

For the ratio, in the numerator, you have Area of semicircle - area of ABZ - area of 'small' semicircle

in the denominator, you have area of ABZ + area of small 'semicircle'

solve it out, and you should get 180*210/180*150 = 210/150=7/5.

After this one, I guess I will retire for today!
[/b]
Director
Joined: 31 Aug 2004
Posts: 610
Followers: 3

Kudos [?]: 105 [0], given: 0

### Show Tags

10 Oct 2004, 08:08
Praet,

Could you please repost it I guess there is a pb ...
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 68

Kudos [?]: 830 [0], given: 781

### Show Tags

10 Oct 2004, 08:26
twixt wrote:
Praet,

Could you please repost it I guess there is a pb ...

repost? ok, this might work...

http://www.manhattangmat.com/index.cfm?cid=11&sethome=1

Praetorian
Director
Joined: 31 Aug 2004
Posts: 610
Followers: 3

Kudos [?]: 105 [0], given: 0

### Show Tags

10 Oct 2004, 13:30
Thanks Praet,

Same method, same pain...

Answer is 7/5 after some nice plays with isoceles

Took me 9 mn...
Joined: 31 Dec 1969
Location: India
Concentration: Entrepreneurship, Technology
GMAT 1: 710 Q49 V0
GMAT 2: 700 Q V
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 7: Q42 V44
GMAT 8: Q42 V44
GMAT 9: 740 Q49 V42
GMAT 10: 740 Q V
GMAT 11: 500 Q47 V33
GPA: 3.3
WE: Sales (Investment Banking)
Followers: 0

Kudos [?]: 169 [0], given: 98436

Re: Manhattan GMAT problem of the week [#permalink]

### Show Tags

10 Oct 2004, 19:00
Knowing this theorem can help a lot:
The opposite angles of quadrilaterals in circles are equal to two right angles.

If you complete the quadrilateral YCAX, then according to above theorem, ang YXC + ang YCA = 180

so,ang yca = 180-105=75

Another theorem to remember:
Angle at the centre is double the angle at the circumference if the angles have the same circumference as base.

Based on the above thm, ang YBA is double of the ang YCA.
So, ang YBA = 150

So ang YBC=180-150=30

Area of arc ybc = 0.5*r^2*pi/6

since 30 deg = pi/6

Area below = 0.5*r^2*pi/6 + pi*r^2/8

Area above = 0.5*r^2*5*pi/6 - pi*r^2/8

so ration above/below=7/5

Praetorian wrote:
This is one of Akamai's .. nice one...

http://www.manhattangmat.com

Praetorian
Intern
Joined: 10 Oct 2004
Posts: 12
Followers: 0

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

Manhattan GMAT problem of this week [#permalink]

### Show Tags

11 Oct 2004, 21:28
EDITED: anish, Manhattan gmat sells the archives of these questions. We may not want to be in copyright trouble :) thats why we usually only post the link, not the problem.
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 68

Kudos [?]: 830 [0], given: 781

### Show Tags

12 Oct 2004, 00:19
yes, the answer is indeed 7/5. go to the website to see the legend akamai's explanation. thanks all
Display posts from previous: Sort by