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### circle

by magical cook » Sat Dec 29, 2007 7:53 am
Hi,

I dont have OA but hope someone could help
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by StarDust845 » Sat Dec 29, 2007 9:17 am

Just plug the equations in assuming the bigger circle is of radius R and the smaller circle is of radius r. you will get the answer.

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by camitava » Sun Dec 30, 2007 10:47 pm
Sorry StarDust845! I cann't be agree with u. According to me,
Let R = radius of large circle
r = radius of small circle
Now 3 * pi * r^2 = pi * (R - r)^2
So (R - r) = sqrt(3) * r

Now circumference of big circle : circum of small circle = 2 * pi * (R - r) / 2 * pi * r = (R - r) / r = sqrt(3).
So IMO D
Correct me If I am wrong

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by samirpandeyit62 » Sun Dec 30, 2007 10:57 pm
agree with amitava D
Regards
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by beingAndNothing » Sun Dec 30, 2007 11:06 pm
Area of Shaded Region = pi* (R^2 - r^2)
Area of Smaller circle = pi*r^2

Hence (R^2 - r^2)/r^2 = 3

R/r = sqrt(2)

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by hopetobeat » Sun Dec 30, 2007 11:08 pm
camitava wrote:Sorry StarDust845! I cann't be agree with u. According to me,
Let R = radius of large circle
r = radius of small circle
Now 3 * pi * r^2 = pi * (R - r)^2
So (R - r) = sqrt(3) * r

Now circumference of big circle : circum of small circle = 2 * pi * (R - r) / 2 * pi * r = (R - r) / r = sqrt(3).
So IMO D
I am afraid the equiation should be 3*pi*r^2=pi * (R^2-r^2), answer should be 2.

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by camitava » Sun Dec 30, 2007 11:15 pm
Sorry guys! It's my mistake - I admit ...
Correct me If I am wrong

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Amitava

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by diya123 » Mon Dec 31, 2007 10:23 am

3 pi r^2 = pi(R-r)^2
R-r =sqrt 3 r
R = 1.7r + r
R/r =1.7 +1
= 2.7 or approx 3

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by Auzbee » Mon Dec 31, 2007 6:56 pm
Let R be the radius of bigger circle and r be the radius of the smaller one.
Then
*Note Pi is denoted by #
((#R^2)-(#r^2)/(#r^2))=3/1
=> #R^2=4#r^2
=> R=2r

Ratio of Circumference of the two circles is the ratio of the radius
Hence R/r=2 from above.

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by II » Fri Jan 04, 2008 4:14 pm
Auzbee wrote:Let R be the radius of bigger circle and r be the radius of the smaller one.
Then
*Note Pi is denoted by #
((#R^2)-(#r^2)/(#r^2))=3/1
=> #R^2=4#r^2
=> R=2r

Ratio of Circumference of the two circles is the ratio of the radius
Hence R/r=2 from above.
how do you get from
"((#R^2)-(#r^2)/(#r^2))=3/1"
to
"=> #R^2=4#r^2"

???

thanks.

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by Auzbee » Sat Jan 05, 2008 12:23 am
how do you get from
"((#R^2)-(#r^2)/(#r^2))=3/1"
to
"=> #R^2=4#r^2"
Solve it like this:
=> (a-b)/b=3/1
=> a-b=3b
=> a=4b

where a=#R^2 and b=#r^2.

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by hemanth28 » Sat Jan 05, 2008 4:29 am
I guess the answer should be C.

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by II » Sat Jan 05, 2008 7:24 am
I am with Auzbee on this. I also think that the answer is 2. Does anyone actually have the official answer:

How I came to the answer, with some help from Auzbee !

Let R = big circle radius
Let r = small circle radius

Big circle area = &#8719;R²
Small circle area = &#8719;r²

Shaded area = Big circle area - Small circle area
Shaded area = &#8719;R² - &#8719;r²

The question tells us that the shaded area is 3 time the small circle area.

Shaded area = 3 x &#8719;r²
&#8719;R² - &#8719;r² = 3(&#8719;r²)
&#8719;R² = 3(&#8719;r²)+(&#8719;r²)
&#8719;R² = 4(&#8719;r²)

Lets get rid of the squares ... so we have to apply &#8730; to both sides of the equation:

&#8730;&#8719;R² = &#8730;4(&#8719;r²)
&#8719;R = 2&#8719;r
&#8719;R/&#8719;r = 2&#8719;r/&#8719;r
&#8719;R/&#8719;r = 2
Remove &#8719; from both numerator and denominator from left hand side:
R/r = 2

So the radius of the big circle (R) is twice as big as the small circle radius (r).

Therefore the circumference will also be twice as big, so the answer is C.

Thanks.
II

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by StarDust845 » Sat Jan 05, 2008 8:52 am
I should read all the posts before posting mine. Camitava corrected his solution earlier.. so deleting my comments and posting only the solution.

<deleting>

Given: PI (R^2 - r^2) = 3PIr^2 which implies (R/r)^2 = 4 and hence R/r = 2

That is what we want. Hence the ratio is 2 that is C in the given choices.

Calista.
Last edited by StarDust845 on Sun Jan 06, 2008 12:10 pm, edited 2 times in total.

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by gmatguy16 » Sun Jan 06, 2008 10:42 am
i agree with 2,
amit dont you think that area of shaded circle is
pi*R^2- pi*r^2 and not pi*(R-r)^2