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29 Oct 2014, 08:18
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
66% (03:04) correct
34%
(03:01)
wrong
based on 342
sessions
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Not Attempted Yet
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
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
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
mvictor
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09 May 2016, 19:20
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Bunuel
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.
