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30% of the surface area of a right circular cylinder is shad

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30% of the surface area of a right circular cylinder is shad [#permalink] New post 22 Apr 2018, 09:20
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30% of the surface area of a right circular cylinder is shaded. If the diameter of the base of the cylinder is 10 and the height is 4, what is the surface area of the unshaded region?

A. \(27\pi\)

B. \(40\pi\)

C. \(63\pi\)

D. \(70\pi\)

E. \(90\pi\)
[Reveal] Spoiler: OA

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Re: 30% of the surface area of a right circular cylinder is shad [#permalink] New post 24 Apr 2018, 01:29
knowing the formulas for volumes and surface area of solid figure is useful
surface area of a cylinder = 2pi*r(r+h)
since diameter \(= 10\); \(r = 5\)

Find the total surface are of the cylinder \(= 2*pi*5(5+4) = 90pi\)
\(30\)% of \(90pi\) \(= 27pi\) which is the shaded area hence
unshaded area \(= 90pi-27pi = 63pi\)
option c
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Last edited by amorphous on 24 Apr 2018, 01:42, edited 2 times in total.
Edited by Carcass
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Re: 30% of the surface area of a right circular cylinder is shad [#permalink] New post 24 Apr 2018, 01:43
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if you use \(\pi\) copying and paste it from the web, then when you wrap it up with \(tag m\) the result are strange symbols.

Instead, just write simply \pi and wrap with \(tag m\)

the result is this \(\pi\)

:wink:
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Re: 30% of the surface area of a right circular cylinder is shad   [#permalink] 24 Apr 2018, 01:43
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30% of the surface area of a right circular cylinder is shad

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