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

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CEO
Joined: 15 Aug 2003
Posts: 3452

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

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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

Kudos [?]: 926 [0], given: 781

Manager
Joined: 28 Aug 2004
Posts: 205

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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]

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Director
Joined: 31 Aug 2004
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10 Oct 2004, 08:08
Praet,

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

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CEO
Joined: 15 Aug 2003
Posts: 3452

Kudos [?]: 926 [0], given: 781

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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

Kudos [?]: 926 [0], given: 781

Director
Joined: 31 Aug 2004
Posts: 606

Kudos [?]: 153 [0], given: 0

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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...

Kudos [?]: 153 [0], given: 0

Joined: 31 Dec 1969

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Location: Russian Federation
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)
Re: Manhattan GMAT problem of the week [#permalink]

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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

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Intern
Joined: 10 Oct 2004
Posts: 12

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Manhattan GMAT problem of this week [#permalink]

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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.

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CEO
Joined: 15 Aug 2003
Posts: 3452

Kudos [?]: 926 [0], given: 781

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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

Kudos [?]: 926 [0], given: 781

12 Oct 2004, 00:19
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# This is one of Akamai's .. nice one...

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