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What is the volume of a right cylindrical can?

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What is the volume of a right cylindrical can?  [#permalink]

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New post 24 Jan 2019, 05:09
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  45% (medium)

Question Stats:

61% (01:47) correct 39% (01:34) wrong based on 58 sessions

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What is the volume of a right cylindrical can X?


(1) If the height of can X were equal to the radius of the base of can X, the volume of the new cylinder would have been half of the volume of can X
(2) If the radius of the base of can X were equal to height of can X, the volume of the new cylinder would have been four times the volume of can X

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Re: What is the volume of a right cylindrical can?  [#permalink]

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New post 24 Jan 2019, 05:39
Bunuel wrote:
What is the volume of a right cylindrical can X?

(1) If the height of can X were equal to the radius of the base of can X, the volume of the new cylinder would have been half of the volume of can X
(2) If the radius of the base of can X were equal to height of can X, the volume of the new cylinder would have been four times the volume of can X


Volume of right cylindrical can X = pi rx^2 hx
Vn new volume
Vi initial volume

rx and hx are radius and height of the cylinder, respectively

(1) If the height of can X were equal to the radius of the base of can X, the volume of the new cylinder would have been half of the volume of can X

Now if hx = rx, then
Vn = 1/2 Vi
2 Vn = Vi
2 pi rx^3 = pi rx^2 *hx
2 pi rx^3 - pi rx^2 *hx = 0
rx^2 ( 2rx - hx) = 0

2rx = hx, cant say anything from here

(2) If the radius of the base of can X were equal to height of can X, the volume of the new cylinder would have been four times the volume of can X

Now if rx = hx
Vn = 4 Vi
pi rx^3 = 4 pi rx^2 *hx
4 pi rx^2 *hx - pi rx^3 = 0
rx^2 ( 4hx - rx) = 0

4hx = rx, cant say anything from here

From combination of both, i can
4hx = rx and 2rx = hx

Will we get a definite value from this, No i doubt it .

E
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Re: What is the volume of a right cylindrical can?  [#permalink]

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New post 24 Jan 2019, 05:44
Bunuel wrote:
What is the volume of a right cylindrical can X?


(1) If the height of can X were equal to the radius of the base of can X, the volume of the new cylinder would have been half of the volume of can X
(2) If the radius of the base of can X were equal to height of can X, the volume of the new cylinder would have been four times the volume of can X


IMO E
#1
h=r
pi * r^2* h= pi * r^2*h/2
r=h/2
vol
pi * h^3/4
no info about h
#2
r=h
pi * r^3 =4 * pi * r^2 * h
r=4h
again no info about r & h
from 1 & 2
we cannot get any info
IMO E
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What is the volume of a right cylindrical can?  [#permalink]

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New post 24 Jan 2019, 07:06
Bunuel wrote:
What is the volume of a right cylindrical can X?


(1) If the height of can X were equal to the radius of the base of can X, the volume of the new cylinder would have been half of the volume of can X
(2) If the radius of the base of can X were equal to height of can X, the volume of the new cylinder would have been four times the volume of can X


1) Can create an equation but no info about length of any thing provided hence not possible.

2) Can create an equation but no info about length of any thing provided hence not possible.

1+2) Still can not find exact volume as no information provided about length of any part of the cylindrical can X.

Hence E
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What is the volume of a right cylindrical can?   [#permalink] 24 Jan 2019, 07:06
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