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14 Jul 2017, 12:14
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In the circular region with center 0, shown above, the two unshaded sections comprise \(\frac{3}{7}\) and \(\frac{1}{3}\) of the area of the circular region. The shaded section comprises what fractional part of the area of the circular region?
(A) \(\frac{3}{5}\)
(B) \(\frac{6}{7}\)
(C)\(\frac{2}{21}\)
(D) \(\frac{5}{21}\)
(E) \(\frac{16}{21}\)
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14 Jul 2017, 12:25
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Assume, the area of the circular region to be 1.
Since the two unshaded portions comprise \(\frac{3}{7}\) and \(\frac{1}{3}\) of the total area,
the shaded portion must contain 1 - (\(\frac{3}{7} + \frac{1}{3})\) of the area.
1 - (\(\frac{3}{7} + \frac{1}{3})\) = 1 - (\(\frac{9+7}{21}\)) = 1 - \(\frac{16}{21}\) =\(\frac{5}{21}\)(Option D)
Since the two unshaded portions comprise \(\frac{3}{7}\) and \(\frac{1}{3}\) of the total area,
the shaded portion must contain 1 - (\(\frac{3}{7} + \frac{1}{3})\) of the area.
1 - (\(\frac{3}{7} + \frac{1}{3})\) = 1 - (\(\frac{9+7}{21}\)) = 1 - \(\frac{16}{21}\) =\(\frac{5}{21}\)(Option D)
24 Nov 2017, 06:40
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Bunuel
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)
