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26 Aug 2021, 22:08
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Area of the shaded region is 3 times the area of the smaller circular region.
Let the smaller circle have r as radius and larger have R as radius.
pi(R)^2 - pi(r)^2 = 3 x pi r^2
=>4 pi r^2 = pi(R)^2
=> r/R = 1/2
=> R = 2r
Circumference of the larger circle C1= 2 * pi *R
= 2 * pi * 2r
= 4*pi*r
Circumference of the smaller circle C2 = 2 * pi * r
C1/C2 = 4*pi*r / 2*pi*r
= 2:1
(option c)
D.S
GMAT SME
Let the smaller circle have r as radius and larger have R as radius.
pi(R)^2 - pi(r)^2 = 3 x pi r^2
=>4 pi r^2 = pi(R)^2
=> r/R = 1/2
=> R = 2r
Circumference of the larger circle C1= 2 * pi *R
= 2 * pi * 2r
= 4*pi*r
Circumference of the smaller circle C2 = 2 * pi * r
C1/C2 = 4*pi*r / 2*pi*r
= 2:1
(option c)
D.S
GMAT SME
avigutman
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Expert reply
04 Oct 2021, 04:37
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
12 Oct 2021, 01:52
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The area of the larger circle = The area of the smaller circle + The area of the shaded region
Given,
The area of the smaller circle = x
The area of the shaded region = 3x
So,
The area of the larger circle = x+3x = 4x
Let,
The radius of the smaller circle = 1
The radius of the larger circle = a
The area of the smaller circle= π (1)^2 = π
The area of the larger circle = 4π
so,
πa^2 = 4π
a^2 = 4
a= 2
Now,
Circumference of the smaller circle = 2π.1 = 2π
Circumference of the larger circle = 2π.2 = 4π
So, (4π/2π)= 2 times.
The Answer is C.
Is this the right approach Bunuel ?
Given,
The area of the smaller circle = x
The area of the shaded region = 3x
So,
The area of the larger circle = x+3x = 4x
Let,
The radius of the smaller circle = 1
The radius of the larger circle = a
The area of the smaller circle= π (1)^2 = π
The area of the larger circle = 4π
so,
πa^2 = 4π
a^2 = 4
a= 2
Now,
Circumference of the smaller circle = 2π.1 = 2π
Circumference of the larger circle = 2π.2 = 4π
So, (4π/2π)= 2 times.
The Answer is C.
Is this the right approach Bunuel ?
