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M19-21

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M19-21  [#permalink]

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New post 16 Sep 2014, 01:06
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

51% (01:36) correct 49% (01:50) wrong based on 222 sessions

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Re M19-21  [#permalink]

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New post 16 Sep 2014, 01:06
1
Official Solution:

If the area of a square equals the area of a circle, which of the following is closest to the ratio of the diagonal of the square to the diameter of the circle?

A. 0.95
B. 1.26
C. 1.40
D. 1.57
E. 2.51

If \(x\) is the side of the square, then the area of the square is \(x^2\) and the diagonal of the square is \(x\sqrt{2}\). If \(d\) is the diameter of the circle then the area of the circle is \(\pi(\frac{d}{2})^2\). Because \(x^2 = \pi(\frac{d}{2})^2\), \(d = \frac{2x}{\sqrt{\pi}}\). The required ratio \(=\frac{x\sqrt{2}}{\frac{2x}{\sqrt{\pi}}} = \frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}\) or approximately \(\sqrt{1.57}\). This is slightly smaller than \(\sqrt{1.69} = 1.3\). The best answer is therefore B.

Answer: B
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Re: M19-21  [#permalink]

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New post 16 Sep 2015, 04:58
Most likely a stupid question but why does

x sqrt(2) / [(2x/sqrt(pi)] = sqrt(pi)/sqrt(2)?

Doesnt it simplify to sqrt(2) * sqrt(pi) / 2?

Bunuel wrote:
Official Solution:

If the area of a square equals the area of a circle, which of the following is closest to the ratio of the diagonal of the square to the diameter of the circle?

A. 0.95
B. 1.26
C. 1.40
D. 1.57
E. 2.51

If \(x\) is the side of the square, then the area of the square is \(x^2\) and the diagonal of the square is \(x\sqrt{2}\). If \(d\) is the diameter of the circle then the area of the circle is \(\pi(\frac{d}{2})^2\). Because \(x^2 = \pi(\frac{d}{2})^2\), \(d = \frac{2x}{\sqrt{\pi}}\). The required ratio \(=\frac{x\sqrt{2}}{\frac{2x}{\sqrt{\pi}}} = \frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}\) or approximately \(\sqrt{1.57}\). This is slightly smaller than \(\sqrt{1.69} = 1.3\). The best answer is therefore B.

Answer: B
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Re: M19-21  [#permalink]

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New post 16 Sep 2015, 06:38
dor1209 wrote:
Most likely a stupid question but why does

x sqrt(2) / [(2x/sqrt(pi)] = sqrt(pi)/sqrt(2)?

Doesnt it simplify to sqrt(2) * sqrt(pi) / 2?

Bunuel wrote:
Official Solution:

If the area of a square equals the area of a circle, which of the following is closest to the ratio of the diagonal of the square to the diameter of the circle?

A. 0.95
B. 1.26
C. 1.40
D. 1.57
E. 2.51

If \(x\) is the side of the square, then the area of the square is \(x^2\) and the diagonal of the square is \(x\sqrt{2}\). If \(d\) is the diameter of the circle then the area of the circle is \(\pi(\frac{d}{2})^2\). Because \(x^2 = \pi(\frac{d}{2})^2\), \(d = \frac{2x}{\sqrt{\pi}}\). The required ratio \(=\frac{x\sqrt{2}}{\frac{2x}{\sqrt{\pi}}} = \frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}\) or approximately \(\sqrt{1.57}\). This is slightly smaller than \(\sqrt{1.69} = 1.3\). The best answer is therefore B.

Answer: B


Check below:
\(\frac{x\sqrt{2}}{(\frac{2x}{\sqrt{\pi}})} =x\sqrt{2}* \frac{\sqrt{\pi}}{2x}=\sqrt{2}* \frac{\sqrt{\pi}}{2}=\sqrt{2}* \frac{\sqrt{\pi}}{\sqrt{2}*\sqrt{2}}=\frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}\).

Hope it helps.
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M19-21  [#permalink]

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New post 15 Apr 2017, 23:16
1
I used smart numbers to calculate the ratio and find it even faster than the algebrical approach.

\(A(square)=1\) --> \(a=1\)
\(d(square)=1*\sqrt{2}≈1.4\)

\(A(circle)=1\)
\(r=\sqrt{\frac{1}{pi}}≈1.1\)

\(Ratio=\frac{1.4}{1.1}≈1.3\)
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Re: M19-21  [#permalink]

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New post 23 Aug 2017, 08:12
1
This is really a good Question..

Answer to this one is B.
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Re: M19-21  [#permalink]

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New post 26 Aug 2017, 03:53
22/7 * r^2 = a^2
a/r =sqrt(22/7)
sqrt(2) * a/2r = sqrt(22/14)
=sqrt(1.50..)
approximately equal to 1.26
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Re: M19-21   [#permalink] 26 Aug 2017, 03:53
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