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

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Carcass
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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. 27π

B. 40π

C. 63π

D. 70π

E. 90π
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amorphous
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Re: 30% of the surface area of a right circular cylinder is shad [#permalink] New post   Updated on: 24 Apr 2018, 01:42
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

Originally posted by amorphous on 24 Apr 2018, 01:29.
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 π copying and paste it from the web, then when you wrap it up with tagm the result are strange symbols.

Instead, just write simply \pi and wrap with tagm

the result is this π

:wink:
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30% of the surface area of a right circular cylinder is shad [#permalink] New post   21 Mar 2019, 07:16
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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. 27π

B. 40π

C. 63π

D. 70π

E. 90π
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Re: 30% of the surface area of a right circular cylinder is shad [#permalink] New post   21 Mar 2019, 07:25
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Carcass wrote:
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. 27π

B. 40π

C. 63π

D. 70π

E. 90π


Total surface area of cylinder = 2πr² + 2πrh

r = 5 (since the DIAMETER = 10)
h = 4

So, surface area = 2π(5²) + 2π(5)(4)
= 50π + 40π
= 90π

If 30% is shaded, then 70% is UNSHADED.
70% of 90π = 63π

Answer: C

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Brent
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Re: 30% of the surface area of a right circular cylinder is shad [#permalink] New post   07 Jul 2024, 11:51
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: 30% of the surface area of a right circular cylinder is shad [#permalink]
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