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In the figure above, an unshaded circle with diameter x is s
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25 May 2020, 10:36
25 May 2020, 10:36
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47% (02:10) correct
52% (03:27) wrong based on 63 sessions
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In the figure above, an unshaded circle with diameter x is surrounded by a shaded region with uniform width of 2. For what values of x is the area of the unshaded circle greater than the area of the shaded region?
Indicate all such values.
❑ 6
❑ 7
❑ 8
❑ 9
❑ 10
❑ 11
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Re: In the figure above, an unshaded circle with diameter x is s
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25 May 2020, 10:37
25 May 2020, 10:37
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Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Re: In the figure above, an unshaded circle with diameter x is s
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28 May 2020, 05:11
28 May 2020, 05:11
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Radius of unshaded circle = \(\frac{x}{2}\)
Area of unshaded circle = \(\pi\) * \((\frac{x}{2})^2\)
Area of shaded + unshaded circle = \(\pi\) * \((\frac{x}{2 }+2 )^2\)
Area of shaded circle = \(\pi\) * \((\frac{x}{2 }+2 )^2\) - \(\pi\) * \((\frac{x}{2})^2\)
if area of unshaded circle is greater than area of shaded circle then
\(\pi\) * \((\frac{x}{2})^2\) > \(\pi\) * \((\frac{x}{2 }+2 )^2\) - \(\pi\) * \((\frac{x}{2})^2\)
Equation simplifies to \((\frac{x}{2})^2\) > \((\frac{x}{2 }+2 )^2\) - \((\frac{x}{2})^2\)
\(\frac{x^2 }{4}\) > \(\frac{x^2 }{4}\) + 4 + 2x - \(\frac{x^2 }{4}\)
\(x^2\) > 16 + 8x
Now substituting all values of x in answer choices we see that only x = 10 and x=11 satisfies above equation for x
Answer is E,F
Area of unshaded circle = \(\pi\) * \((\frac{x}{2})^2\)
Area of shaded + unshaded circle = \(\pi\) * \((\frac{x}{2 }+2 )^2\)
Area of shaded circle = \(\pi\) * \((\frac{x}{2 }+2 )^2\) - \(\pi\) * \((\frac{x}{2})^2\)
if area of unshaded circle is greater than area of shaded circle then
\(\pi\) * \((\frac{x}{2})^2\) > \(\pi\) * \((\frac{x}{2 }+2 )^2\) - \(\pi\) * \((\frac{x}{2})^2\)
Equation simplifies to \((\frac{x}{2})^2\) > \((\frac{x}{2 }+2 )^2\) - \((\frac{x}{2})^2\)
\(\frac{x^2 }{4}\) > \(\frac{x^2 }{4}\) + 4 + 2x - \(\frac{x^2 }{4}\)
\(x^2\) > 16 + 8x
Now substituting all values of x in answer choices we see that only x = 10 and x=11 satisfies above equation for x
Answer is E,F
