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the surface area of sphere is 4pie r^2. If a sphere is halved into two

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the surface area of sphere is 4pie r^2. If a sphere is halved into two  [#permalink]

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New post 20 Mar 2018, 21:13
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A
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E

Difficulty:

  25% (medium)

Question Stats:

74% (01:05) correct 26% (01:21) wrong based on 38 sessions

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Q. The surface area of sphere is \(4\pi r^2\). If a sphere is halved into two hemispheres, what is the percentage change in the total surface area of the solid(s) before and after?
A. 0%
B. -25%
C. -33.33%
D. 25%
E. 50%
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Joined: 26 Sep 2017
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GMAT 1: 640 Q48 V30
Re: the surface area of sphere is 4pie r^2. If a sphere is halved into two  [#permalink]

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New post 20 Mar 2018, 21:27
Surface area of sphere = 4.pi.r^2
Surface area of 2 new hemispheres = 3.pi.r^2 + 3.pi.r^2
Change = ((6.pi.r^2 - 4.pi.r^2)/4.pi.r^2)*100
Ans = E. 50%

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Re: the surface area of sphere is 4pie r^2. If a sphere is halved into two  [#permalink]

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New post 21 Mar 2018, 09:33
We're asked to find the change in surface area when a sphere is cut in half. Visualize what's happening -- the 2 hemispheres will have ALL of the surface area of the one non-cut sphere, PLUS the surface areas of the 2 circles where the sphere is cut. This means that the change in surface area will be positive. So A, B, or C can't be the answer.

% change = \(\frac{AmountOfChange}{OriginalAmount} * 100\)

Amount of change: this is the area of the 2 circles where the sphere is cut, i.e. \(2 * πr^2\)
Original amount: \(4πr^2\)

% change = \(\frac{2πr^2}{4πr^2} * 100\) = 50

Answer: E
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Re: the surface area of sphere is 4pie r^2. If a sphere is halved into two  [#permalink]

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New post 21 Mar 2018, 10:46
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rishabhmishra wrote:
Q. The surface area of sphere is \(4\pi r^2\). If a sphere is halved into two hemispheres, what is the percentage change in the total surface area of the solid(s) before and after?
A. 0%
B. -25%
C. -33.33%
D. 25%
E. 50%

Attachment:
surfarea.png
surfarea.png [ 17.62 KiB | Viewed 626 times ]

• Surface area of sphere = \(4\pi r^2\)

• Surface area TOTAL of ONE hemisphere =
(sphere S.A.*\(\frac{1}{2}\)) PLUS (S.A. of circular face created by the cut)
(\(4\pi r^2\)*\(\frac{1}{2}\)) = \(2\pi r^2\) PLUS extra circle face area= \(\pi r^2\)
\((2\pi r^2 + \pi r^2)\)\(= 3\pi r^2\)

S.A. of TWO hemispheres: \((2 * 3\pi r^2)\) = \(6\pi r^2\)
S.A. of original sphere: \(4\pi r^2\)

• Percent Change? \(\frac{New-Old}{Old} * 100\)

\(\frac{6\pi r^2 - 4\pi r^2}{4\pi r^2} * 100\)


\((\frac{2\pi r^2}{4\pi r^2}*100)=(\frac{2}{4}*100)=(.5*100)=50\) %

Answer E
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Re: the surface area of sphere is 4pie r^2. If a sphere is halved into two &nbs [#permalink] 21 Mar 2018, 10:46
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