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

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Manager
Joined: 23 Sep 2016
Posts: 220
the surface area of sphere is 4pie r^2. If a sphere is halved into two  [#permalink]

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20 Mar 2018, 22:13
1
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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%
Intern
Joined: 26 Sep 2017
Posts: 23
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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20 Mar 2018, 22:27
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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Manager
Joined: 30 Mar 2017
Posts: 136
GMAT 1: 200 Q1 V1
Re: the surface area of sphere is 4pie r^2. If a sphere is halved into two  [#permalink]

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21 Mar 2018, 10:33
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

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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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21 Mar 2018, 11:46
1
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 [ 17.62 KiB | Viewed 504 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? $$\frac{New-Old}{Old} * 100$$

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

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

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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, 11:46
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