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magical cook: Master | Next Rank: 500 Posts; Posts: 484; Joined: Sun Jul 30, 2006 7:01 pmFollowed by: 1 member; Thanked: 2 times

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

Hi,

I dont have OA but hope someone could help

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Source: Beat The GMAT — Problem Solving | Sat Dec 29, 2007 7:53 am

StarDust845: Master | Next Rank: 500 Posts; Posts: 158; Joined: Mon Dec 03, 2007 8:32 am; Thanked: 7 times

by StarDust845 » Sat Dec 29, 2007 9:17 am

Answer is 2.

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.

Calista.

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camitava: Legendary Member; Posts: 645; Joined: Wed Sep 05, 2007 4:37 amFollowed by: 5 members; Location: India; Thanked: 34 times

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

Regards,

Amitava

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samirpandeyit62: Master | Next Rank: 500 Posts; Posts: 460; Joined: Sun Mar 25, 2007 7:42 am; Thanked: 27 times

by samirpandeyit62 » Sun Dec 30, 2007 10:57 pm

agree with amitava D

Regards
Samir

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beingAndNothing: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Fri Sep 14, 2007 10:35 am; Thanked: 1 times

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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hopetobeat: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Thu Dec 20, 2007 6:06 am; Thanked: 1 times

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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camitava: Legendary Member; Posts: 645; Joined: Wed Sep 05, 2007 4:37 amFollowed by: 5 members; Location: India; Thanked: 34 times

by camitava » Sun Dec 30, 2007 11:15 pm

Sorry guys! It's my mistake - I admit ...

Correct me If I am wrong

Regards,

Amitava

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diya123: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Tue Dec 11, 2007 12:59 pm

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

i think answer is 3.

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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Auzbee: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Mon Jul 09, 2007 6:54 pm; Location: US of A

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.
Correct answer is 2.

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II: Master | Next Rank: 500 Posts; Posts: 400; Joined: Mon Dec 10, 2007 1:35 pm; Location: London, UK; Thanked: 19 times; GMAT Score:680

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.
Correct answer is 2.

how do you get from
"((#R^2)-(#r^2)/(#r^2))=3/1"
to
"=> #R^2=4#r^2"

???

thanks.

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Auzbee: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Mon Jul 09, 2007 6:54 pm; Location: US of A

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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hemanth28: Master | Next Rank: 500 Posts; Posts: 157; Joined: Tue Dec 04, 2007 9:05 am; Thanked: 9 times; GMAT Score:680

by hemanth28 » Sat Jan 05, 2008 4:29 am

I guess the answer should be C.

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II: Master | Next Rank: 500 Posts; Posts: 400; Joined: Mon Dec 10, 2007 1:35 pm; Location: London, UK; Thanked: 19 times; GMAT Score:680

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 = ∏R²
Small circle area = ∏r²

Shaded area = Big circle area - Small circle area
Shaded area = ∏R² - ∏r²

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

Shaded area = 3 x ∏r²
∏R² - ∏r² = 3(∏r²)
∏R² = 3(∏r²)+(∏r²)
∏R² = 4(∏r²)

Lets get rid of the squares ... so we have to apply √ to both sides of the equation:

√∏R² = √4(∏r²)
∏R = 2∏r
∏R/∏r = 2∏r/∏r
∏R/∏r = 2
Remove ∏ 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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StarDust845: Master | Next Rank: 500 Posts; Posts: 158; Joined: Mon Dec 03, 2007 8:32 am; Thanked: 7 times

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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gmatguy16: Master | Next Rank: 500 Posts; Posts: 124; Joined: Thu Aug 23, 2007 5:11 am; Thanked: 2 times

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

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