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20 Mar 2018, 22:13
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
63% (01:08) correct
38%
(01:24)
wrong
based on 136
sessions
History
Date
Time
Result
Not Attempted Yet
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%
A. 0%
B. -25%
C. -33.33%
D. 25%
E. 50%
20 Mar 2018, 22:27
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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%
Sent from my Lenovo A7020a48 using GMAT Club Forum mobile app
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%
Sent from my Lenovo A7020a48 using GMAT Club Forum mobile app
21 Mar 2018, 10:33
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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
% 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
