ETS: Circle : Quant Question Archive [LOCKED]
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# ETS: Circle

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Intern
Joined: 06 Aug 2004
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15 Sep 2004, 09:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

could n't figure this out...
pls help
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Senior Manager
Joined: 05 Feb 2004
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15 Sep 2004, 10:25
I did a silly mistake previously..hence edited!!
B..
basically, 2 squares are formed......two sides of larger square are 80*2 = 160 cm in lengtjh. Similarly, 2 sides of smaller one are 60 *2 = 120 cm in length. Total is 280 cm. Now add 3/4th of the 2 circumferemnces as well:
Big circlr: 3/4*2* Pi*80 = 120Pi
Small circle: 3/4*2*pi*60 = 90Pi

Total = 120pi+90pi+280 = B!!
Manager
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16 Sep 2004, 02:51
B it is ,

you don't need to complete calculation if you see there is some cord which is connecting the two discs and and two sectors are left free without cord .

calculate disc 1 = 270/360 * 2* pi*80
cl . disc 2 - 270/360 *2 * pi* 60

= 90pi + 120 pi = 210 pi + the extra length
if look at the choice there is only once choice which is 210 oi + extra B
so the rest you need not calculate .

let me know if their is any flaw in my reasoning .
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16 Sep 2004, 03:31
My approach was
1. the staight tangent lines connecting the circle = 140+140 = 280.
2. It has to be 280+some Pi stuff (because portion of the circumference of the two circles also need to be added - which would invariably be a pi something).
3. We are now left with B and E options.
4. E is too big - why? 2(pi)(r) of both the circles = 280pi. Surely we need smaller than that (because the cord does not wound the entire circle and only portion of it) and we have only one option in our shortlist and that is B.

We can calculate B for sure if we have time as already outlined by CBRf3 and Rahul.
16 Sep 2004, 03:31
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