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24 Jan 2019, 05:09
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E
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Difficulty:
35%
(medium)
Question Stats:
71% (02:02) correct
29%
(01:38)
wrong
based on 174
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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
M36-133
(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
M36-133
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16 Nov 2021, 09:29
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Bunuel
Official Solution:
What is the volume of a right cylindrical can X?
\(Volume=\pi r^2h\), where \(r\) is the radius of the circular base and \(h\) is the height,
(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
The above means that \(\pi r^2h=2*(\pi r^2*r)\). This gives \(h=2r\). Not sufficient.
(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
The above means that \(4*(\pi r^2h)=\pi h^2*h\). This also gives \(h=2r\). Not sufficient.
(1)+(2) We only know the ratio of height to the radius but we know nothing about actual lengths, so we cannot get the volume. Not sufficient.
Answer: E
General Discussion
KanishkM
Joined: 09 Mar 2018
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24 Jan 2019, 05:39
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Bunuel
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
