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# In the figure to the right, x is both the radius of the larger circle

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In the figure to the right, x is both the radius of the larger circle  [#permalink]

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Updated on: 23 May 2017, 14:07
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15% (low)

Question Stats:

86% (01:10) correct 14% (00:47) wrong based on 28 sessions

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In the figure BELOW, x is both the radius of the larger circle and the diameter of the smaller circle. The area of the shaded region is
Attachment:

NM.jpg [ 14.57 KiB | Viewed 600 times ]

(A) 3/4 πx^2
(B) π/3
(C) 4/3 πx^2
(D) 3/5 πx^2
(E) πx^2

Source: Nova GMAT

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Last edited by Bunuel on 23 May 2017, 14:07, edited 1 time in total.
Edited the question.
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Re: In the figure to the right, x is both the radius of the larger circle  [#permalink]

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23 May 2017, 14:08
In the figure BELOW, x is both the radius of the larger circle and the diameter of the smaller circle. The area of the shaded region is
Attachment:
NM.jpg

(A) 3/4 πx^2
(B) π/3
(C) 4/3 πx^2
(D) 3/5 πx^2
(E) πx^2

Source: Nova GMAT

Similar question: https://gmatclub.com/forum/in-the-figur ... 11321.html
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Re: In the figure to the right, x is both the radius of the larger circle  [#permalink]

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24 May 2017, 12:13
In the figure BELOW, x is both the radius of the larger circle and the diameter of the smaller circle. The area of the shaded region is
Attachment:
NM.jpg

(A) 3/4 πx^2
(B) π/3
(C) 4/3 πx^2
(D) 3/5 πx^2
(E) πx^2

Source: Nova GMAT

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Posts: 701
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Re: In the figure to the right, x is both the radius of the larger circle  [#permalink]

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24 May 2017, 16:32
In the figure BELOW, x is both the radius of the larger circle and the diameter of the smaller circle. The area of the shaded region is
Attachment:
NM.jpg

(A) 3/4 πx^2
(B) π/3
(C) 4/3 πx^2
(D) 3/5 πx^2
(E) πx^2

Source: Nova GMAT

x is radius of larger circle.
Area of larger circle = $${\pi} x^2$$

x is diameter of smaller circle = radius = $$\frac{x}{2}$$
Area of smaller circle = $${\pi} (\frac{x}{2})^2$$ = $$\frac{{\pi} x^2}{4}$$

Area of the shaded region is = Area of Larger Circle - Area of Smaller Circle
$${\pi} x^2$$ - $$\frac{{\pi}x^2}{4}$$ = $$\frac{4{\pi}x^2 - {\pi}x^2}{4}$$ = $$\frac{3}{4}{\pi}x^2$$
Re: In the figure to the right, x is both the radius of the larger circle &nbs [#permalink] 24 May 2017, 16:32
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