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# In the circular region with center O, shown above, the two unshaded se

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Joined: 02 Sep 2009
Posts: 47072
In the circular region with center O, shown above, the two unshaded se [#permalink]

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23 Nov 2017, 23:28
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Question Stats:

92% (00:31) correct 8% (01:28) wrong based on 24 sessions

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In the circular region with center O, shown above, the two unshaded sections comprise 3/7 and 1/3 of the area of the circular region. The shaded section comprises what fractional part of the area of the circular region?

(A) 3/5
(B) 6/7
(C) 2/21
(D) 5/21
(E) 16/21

Attachment:

2017-11-23_2026.png [ 6.6 KiB | Viewed 622 times ]

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In the circular region with center O, shown above, the two unshaded se [#permalink]

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24 Nov 2017, 06:40
Bunuel wrote:

In the circular region with center O, shown above, the two unshaded sections comprise 3/7 and 1/3 of the area of the circular region. The shaded section comprises what fractional part of the area of the circular region?

(A) 3/5
(B) 6/7
(C) 2/21
(D) 5/21
(E) 16/21

Attachment:
2017-11-23_2026.png

If the circle will have the entire fraction of the area 1,

the shaded portion will have area $$1 - \frac{3}{7} - \frac{1}{3} = \frac{21 - 9 - 7}{21} = \frac{5}{21}$$

Another method

Assume area of the circle to be 21.

The portion containing $$\frac{3}{7}$$ of the circle will have area $$9$$
and the part containing $$\frac{1}{3}$$ of the circle to be $$7$$.

The shaded portion of the circle will have an area of $$21 - 9 - 7 = 5$$

Hence shaded portion of the circle contains $$\frac{5}{21}$$ of the circle(Option D)
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Re: In the circular region with center O, shown above, the two unshaded se [#permalink]

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24 Nov 2017, 06:42
In the circular region with center O, shown above, the two unshaded sections comprise 3/7 and 1/3 of the area of the circular region. The shaded section comprises what fractional part of the area of the circular region?

(A) 3/5
(B) 6/7
(C) 2/21
(D) 5/21
(E) 16/21

Lets x to be the whole area of the circle and y be the shaded section =>

$$x = \frac{3}{7} * x + \frac {1}{3} * x + y$$
$$x = \frac{16x}{21} + y$$
$$y = \frac{21x - 16x}{21}$$
$$y = \frac{5x}{21}$$

The circle area is accaunted to shaded region as 5 : 21
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Re: In the circular region with center O, shown above, the two unshaded se [#permalink]

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24 Nov 2017, 09:04
First solution is better and lucid...

pushpitkc wrote:
Bunuel wrote:

In the circular region with center O, shown above, the two unshaded sections comprise 3/7 and 1/3 of the area of the circular region. The shaded section comprises what fractional part of the area of the circular region?

(A) 3/5
(B) 6/7
(C) 2/21
(D) 5/21
(E) 16/21

Attachment:
2017-11-23_2026.png

If the circle will have the entire fraction of the area 1,

the shaded portion will have area $$1 - \frac{3}{7} - \frac{1}{3} = \frac{21 - 9 - 7}{21} = \frac{5}{21}$$

Another method

Assume area of the circle to be 21.

The portion containing $$\frac{3}{7}$$ of the circle will have area $$9$$
and the part containing $$\frac{1}{3}$$ of the circle to be $$7$$.

The shaded portion of the circle will have an area of $$21 - 9 - 7 = 5$$

Hence shaded portion of the circle contains $$\frac{5}{21}$$ of the circle(Option D)

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Re: In the circular region with center O, shown above, the two unshaded se   [#permalink] 24 Nov 2017, 09:04
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