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If O is the center of the circle above, what fraction of the circular region is shaded?

(A) 1/12 (B) 1/9 (C) 1/6 (D) 1/4 (E) 1/3

150°+150°=300° of the circle is not shaded, hence 360°-300°=60° of the circle is shaded, which is \(\frac{60}{360}=\frac{1}{6}\) of the circular region.

Re: If O is the center of the circle above, what fraction of the [#permalink]

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11 Sep 2012, 09:56

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Bunuel wrote:

Attachment:

Circle.png

If O is the center of the circle above, what fraction of the circular region is shaded?

(A) 1/12 (B) 1/9 (C) 1/6 (D) 1/4 (E) 1/3

Total angle in a circle = 360 degree One of Non-shaded angle = 150 degree Other Non-shaded angle = 150 degree (?) because it is opposite angle Total Non-shaded angle = 150 + 150 = 300 degree So, shaded angle = 360 - 300 = 60 degree

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I would say (C), but it seems such an easy answer. Anyways, here is the explanation: 360-2*150=60 60 degrees is the total of shaded regions, so 60/360 = 1/6. So, please correct me if I went awry.

p.s. where is trick here? those shaded regions opposite to each other do not have equal angles?
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Re: If O is the center of the circle above, what fraction of the [#permalink]

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12 Sep 2012, 04:37

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Bunuel wrote:

Attachment:

Circle.png

If O is the center of the circle above, what fraction of the circular region is shaded?

(A) 1/12 (B) 1/9 (C) 1/6 (D) 1/4 (E) 1/3

Since both Lines pass through centers they are alternate angles. Hence shaded degree measure is 360- (150x2)=60

So fraction of circle shaded = 60/360 = 1/6.

Hence Answer Choice C.
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Re: If O is the center of the circle above, what fraction of the [#permalink]

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Fraction of Shaded region = Angle made at the center by shaded region/ Angle at center of a circle = (360-150*2)/360 = 60/360 = 1/6 Answer C
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If O is the center of the circle above, what fraction of the circular region is shaded?

(A) 1/12 (B) 1/9 (C) 1/6 (D) 1/4 (E) 1/3

150°+150°=300° of the circle is not shaded, hence 360°-300°=60° of the circle is shaded, which is \(\frac{60}{360}=\frac{1}{6}\) of the circular region.

Answer: C.

Kudos points given to everyone with correct solution. Let me know if I missed someone.
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04 Feb 2016, 12:56

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