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A cylindrical tank, with radius and height both of 10 feet, is to be r

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Bunuel
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A cylindrical tank, with radius and height both of 10 feet, is to be r [#permalink] New post   29 Oct 2014, 08:18
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Tough and Tricky questions: Geometry.



A cylindrical tank, with radius and height both of 10 feet, is to be redesigned as a cone, capable of holding twice the volume of the cylindrical tank. There are two proposed scenarios for the new cone: in scenario (1) the radius will remain the same as that of the original cylindrical tank, in scenario (2) the height will remain the same as that of the original cylindrical tank. What is the approximate difference in feet between the new height of the cone in scenario (1) and the new radius of the cone in scenario (2)?

(A) 13
(B) 25
(C) 30
(D) 35
(E) 40
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D
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Re: A cylindrical tank, with radius and height both of 10 feet, is to be r [#permalink] New post   29 Oct 2014, 19:35
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Answer = D = 35

Area of Cylindrical tank \(= \pi 10^2 10 = 1000\pi\)

Required area of cone \(= 2 * 1000\pi = 2000\pi\)

Scenario (1) the radius will remain the same

Area of cone \(= \frac{1}{3} \pi 10^2 h = 2000\pi\)

h = 60

Scenario (2) the height will remain the same

Area of cone\(= \frac{1}{3} \pi r^2 10 = 2000\pi\)

\(r^2 = 600\)

\(r = 24.5 \approx 25\) (\(\sqrt{600}\) lies between \(\sqrt{576} = 24\) & \(\sqrt{625} = 25\))

Answer = 60 - 25 = 35
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Re: A cylindrical tank, with radius and height both of 10 feet, is to be r [#permalink] New post   09 May 2016, 19:20
Bunuel wrote:

Tough and Tricky questions: Geometry.



A cylindrical tank, with radius and height both of 10 feet, is to be redesigned as a cone, capable of holding twice the volume of the cylindrical tank. There are two proposed scenarios for the new cone: in scenario (1) the radius will remain the same as that of the original cylindrical tank, in scenario (2) the height will remain the same as that of the original cylindrical tank. What is the approximate difference in feet between the new height of the cone in scenario (1) and the new radius of the cone in scenario (2)?

(A) 13
(B) 25
(C) 30
(D) 35
(E) 40


to be honest, I have never seen any official problem to be asking the volume of a cone..
volume of a cylinder = pi*r^2 *h
r=10
h=10
volume = 1000pi

volume of the cone: 2000pi

volume of the cone = pi*r^2 * h/3
2000=r^2 * h/3
1st case: r=10 -> h/3=20 => h=60

2nd case: h=10 -> r^2 = sqrt(600)
sqrt(600) = 10*sqrt(6)
sqrt(6)>sqrt(4)
sqrt(6)<sqrt(9)

so radius must be between 20 and 30. this approximation is enough.

60-30=30
60-20=40

so we are looking for smth between 30 and 40.
only D fits in here.
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Re: A cylindrical tank, with radius and height both of 10 feet, is to be r [#permalink] New post   12 Dec 2020, 15:22
I just didn't know the formula for a cone.

1/3*r^2*h*pi = 2*r^2*h*pi

(1)
1/3h = 2
h=6 --> 6(10)=60

(2)
1/3r^2 = 2
r^2 = 6
r=sqrt(6) --> 10(sqrt(6)) ~= 24

60-24 = 36 -->35 is closest
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Re: A cylindrical tank, with radius and height both of 10 feet, is to be r [#permalink] New post   08 Jul 2023, 13:21
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Re: A cylindrical tank, with radius and height both of 10 feet, is to be r [#permalink]
08 Jul 2023, 13:21
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