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)?	Choose the correct choice from the following choices.
A) 13
B) 25
C) 30
D) 35
E) 40
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 -&gt; h/3=20 =&gt; h=60   2nd case: h=10 -&gt; r^2 = sqrt(600)   sqrt(600) = 10*sqrt(6)   sqrt(6)&gt;sqrt(4)   sqrt(6)&lt;sqrt(9)   so radius must be between 20 and 30. this approximation is enough.   60-30= 
<gadget id="calculator">60-30</gadget>
<output>30</output>
30   60-20= 
<gadget id="calculator">60-20</gadget>
<output>40</output>
40   so we are looking for smth between 30 and 40.   only 35 fits in here.
<result>D</result>
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)?