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In the diagram above, point B is the center of Circle #1 and point D i

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In the diagram above, point B is the center of Circle #1 and point D i [#permalink]

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New post 10 Mar 2015, 06:00
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In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?

A. 1.5
B. \(\sqrt{3}\)
C. 3
D. \(\sqrt{6}\)
E. 6

Kudos for a correct solution.

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Re: In the diagram above, point B is the center of Circle #1 and point D i [#permalink]

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New post 10 Mar 2015, 06:22
Hi

let's say R1 is the radius of smaller circle and R2 is of the bigger circle

According to question
A1/A2 = 3/2
[pi(R1)^2 ] / [pi(R2)^2] = 3/2
R1/R2 = root3 / root 2

Now CE= 2[R2] & BC =R1

therefore CE/BC = [2(root3)] / root2 = root6
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Re: In the diagram above, point B is the center of Circle #1 and point D i [#permalink]

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New post 10 Mar 2015, 19:01
1
Assume radius of circle 1 = R & radius of circle 2 = r;
given, Area of Circle 1/Area of Circle 2 = 3/2 = R^2/r^2; therefore R/r = Root 3/Root 2
Required, CE:BC = 2*R: r = 2 * Root 3/Root 2; simplify (multiply numerator and denominator by root 2), and answer = Root 6
Therefore answer choice D Root 6
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Re: In the diagram above, point B is the center of Circle #1 and point D i [#permalink]

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New post 15 Mar 2015, 21:04
Bunuel wrote:
Image
In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?

A. 1.5
B. \(\sqrt{3}\)
C. 3
D. \(\sqrt{6}\)
E. 6

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This is tricky. We are given the ratio of areas, and we want to know a ratio involving the radii. Let R be the radius of the bigger circle, and r be the radius of the smaller circle.
\(ratio \ of \ areas = \frac{3}{2} = \frac{\pi{R^2}}{\pi{r^2}}=\frac{\pi{R^2}}{\pi{r^2}}\).

Take the square root and rationalize the denominator:
\(\frac{R}{r}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}\)

Well, CE = 2R, and BC = r, so CE/BC = 2R/r, which is twice this ratio \(\frac{CE}{BC}=\sqrt{6}\).

Answer = (D)
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the diagram above, point B is the center of Circle #1 and point D i [#permalink]

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New post 20 Oct 2015, 09:52
[quote="Bunuel"]
Attachment:
gpp_img7.png
In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?

A. 1.5
B. \(\sqrt{3}\)
C. 3
D. \(\sqrt{6}\)
E. 6


Solution :

Lets assume radius of circle 1 - r, and circle 2 as - R

Area of circle 1 - pi r*2
Area of circle 2 - pi R*2

Therfore pi R*2/pi r*2 = 3/2
R*2/r*2 = 3/2
R/r = root 3/ root 2

We have to find out the ratio of the diameter of circle 2/ radius of circle 1, which can be written mathematically as 2R/r

2R/r = 2* root 3/ root 2
= root 2 * root 3
= root 6

Answer : D
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Re: In the diagram above, point B is the center of Circle #1 and point D i [#permalink]

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New post 06 Apr 2018, 02:36
Bunuel wrote:
Attachment:
gpp_img7.png
In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?

A. 1.5
B. \(\sqrt{3}\)
C. 3
D. \(\sqrt{6}\)
E. 6

Kudos for a correct solution.


Area(C2) : Area (C1) = 3 :2
Diameter(C2) : Diameter(C1) = \(\sqrt{(3/2)}\)
CE: AB = \(\sqrt{(3/2)}\)
CE: 2BC = \(\sqrt{(3/2)}\)
CE:BC = 2*\(\sqrt{(3/2)}\) = \(\sqrt{4*(3/2)}\) = \(\sqrt{6}\)

Answer D
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Re: In the diagram above, point B is the center of Circle #1 and point D i   [#permalink] 06 Apr 2018, 02:36
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