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# cylindrical container volume

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Intern
Joined: 17 Apr 2010
Posts: 6
Followers: 0

Kudos [?]: 1 [0], given: 4

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20 Apr 2010, 00:29
00:00

Difficulty:

(N/A)

Question Stats:

100% (05:01) correct 0% (00:00) wrong based on 3 sessions

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A cylindrical container, used for holding oil, had a diameter of 16 m and a height of 3 m. To increase the volume, such that if X m are added to either the radius or the height, the increase in volume is the same. Thus, X will be

A. 16 m
B. 5.33 m
C. 6.77 m
D. 3.56 m
E. 4 m

OA will follow..
Intern
Joined: 19 Apr 2010
Posts: 4
Followers: 0

Kudos [?]: 0 [0], given: 2

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20 Apr 2010, 00:39
gmatstar10 wrote:
A cylindrical container, used for holding oil, had a diameter of 16 m and a height of 3 m. To increase the volume, such that if X m are added to either the radius or the height, the increase in volume is the same. Thus, X will be

A. 16 m
B. 5.33 m
C. 6.77 m
D. 3.56 m
E. 4 m

OA will follow..

Volume = pi *R^2* H
New volume = pi *8^2* (3 + X) =pi* (8 + X)^2 *3
X = 16/3 m = 5.33 m.
what's OA ?
Intern
Joined: 08 Apr 2010
Posts: 28
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21 Apr 2010, 22:01
gmatstar10 wrote:
A cylindrical container, used for holding oil, had a diameter of 16 m and a height of 3 m. To increase the volume, such that if X m are added to either the radius or the height, the increase in volume is the same. Thus, X will be

A. 16 m
B. 5.33 m
C. 6.77 m
D. 3.56 m
E. 4 m

OA will follow..

hi,
diameter = 16 hence radius is 8
as the increase in volume is same,
$$\pi r^2 h_1 = \pi r_1^2 h$$
$$8^2 (3+x) = (8+x)^2 3$$
$$64(3+x) = 3(x^2+16x+64)$$
$$64x = 3x^2 + 48x$$
$$x = \frac {16}{3}$$
_________________

CHEERS
fivezero7

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Joined: 18 Apr 2010
Posts: 5
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Kudos [?]: 6 [0], given: 4

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22 Apr 2010, 02:05
thanks guys, OA
[Reveal] Spoiler:
B
Re: cylindrical container volume   [#permalink] 22 Apr 2010, 02:05
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