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by sanju09 » Tue Jul 06, 2010 6:09 am
The right circular cone attached is sliced horizontally forming two pieces, each of which has the same height. What is the ratio of the volume of the smaller piece to the volume of the larger piece?
(A) ½
(B) 1/3
(C) 1/6
(D) 1/7
(E) 1/8
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by kvcpk » Tue Jul 06, 2010 7:20 am
Volume of cone is 1/3 * pi* r^2 * h
let the height of the cone be 2h and the radius be 2r
then volume of the cone is
1/3 * pi * 4r^2 * 2h

now the smaller cone will have half height and half radius.
so volume of smaller cone is 1/3 * pi * r^2 * h
volume of remanining part is 1/3 * pi * r^2 * h (8-1) = 7 * 1/3 * pi * r^2 * h

ratio = 1/3 * pi * r^2 * h/7*1/3 * pi * r^2 * h = 1/7
pick D
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by jeremy8 » Tue Jul 06, 2010 7:26 am
Are cones tested on the GMAT?
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by raunakrajan » Tue Jul 06, 2010 8:27 am
kvcpk wrote:Volume of cone is 1/3 * pi* r^2 * h
let the height of the cone be 2h and the radius be 2r
then volume of the cone is
1/3 * pi * 4r^2 * 2h

now the smaller cone will have half height and half radius.
so volume of smaller cone is 1/3 * pi * r^2 * h
volume of remanining part is 1/3 * pi * r^2 * h (8-1) = 7 * 1/3 * pi * r^2 * h

ratio = 1/3 * pi * r^2 * h/7*1/3 * pi * r^2 * h = 1/7
pick D
hi, how do we know that the radius also is half the original radius?
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by kvcpk » Tue Jul 06, 2010 8:54 am
raunakrajan wrote: hi, how do we know that the radius also is half the original radius?
Ok.. That requires some geometry.. May be there is anotherway.. But I will explain through Geometry..

Assume a traingle that is drawn from the center of the base to the top.

Now, we know that the base of the traingle is 2R ..since radius of the base is 2R
Height of the triangle is 2H... assuming that the height of the cone is 2H..

Now, draw a line from the mid point of 2h parallel to the base. that will intersect the hypotenuse at some point.

Now, the 2 traingles that are formed are similar.. so the ratios of sides should match

hence H/2H = x/2R -> x=R

I am poor at drawing.. Tried to draw a picture in mspaint.. but it turned out very bad..

Let me know if you do not understand..
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by raunakrajan » Tue Jul 06, 2010 10:33 am
kvcpk wrote:
raunakrajan wrote: hi, how do we know that the radius also is half the original radius?
Ok.. That requires some geometry.. May be there is anotherway.. But I will explain through Geometry..

Assume a traingle that is drawn from the center of the base to the top.

Now, we know that the base of the traingle is 2R ..since radius of the base is 2R
Height of the triangle is 2H... assuming that the height of the cone is 2H..

Now, draw a line from the mid point of 2h parallel to the base. that will intersect the hypotenuse at some point.

Now, the 2 traingles that are formed are similar.. so the ratios of sides should match

hence H/2H = x/2R -> x=R

I am poor at drawing.. Tried to draw a picture in mspaint.. but it turned out very bad..

Let me know if you do not understand..
thanks a ton!! got it! boss amazing explanation
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by kvcpk » Tue Jul 06, 2010 11:11 am
raunakrajan wrote: thanks a ton!! got it! boss amazing explanation
Glad that it was helpful Raunakrajan.. :)
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by sanju09 » Wed Jul 07, 2010 12:30 am
jeremy8 wrote:Are cones tested on the GMAT?
not sure, but cones are not banned on GMAT; pretty much sure!
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