Question 191327
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Yes, 20 cm is the overall length of the pencil.  Since the conical end has a height of 4 cm, the length of the cylinder part of the pencil that remained after the sharpening process is 16 cm.


It doesn't matter what the radius is, and you couldn't figure it out properly anyway since you have neither diameter, circumference, or actual volume  measurements.  Not sure what you used to get the 1.4 cm, but for the reason stated, it has to be wrong.  Besides, a 1.4 cm radius for a pencil would make a mighty fat pencil -- the diameter would be 2.8 cm which is about 1.1 inch, roughly 3 times as wide as those fat pencils they gave us in the 1st grade.


The volume of a right circular cylinder is given by:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{cyl} = A_bh]


Where *[tex \LARGE A_b] is the area of the base and *[tex \LARGE h] is the height.


The volume of a right circular cone is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{cone} = \frac{A_bh}{3} ]


So, the volume of the pencil before it was sharpened was:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{cyl} = 20A_b]


And the volume of the pencil after it was sharpened was the volume of the cylinder part plus the volume of the cone part:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{p} = 16A_b + \frac{4A_b}{3} = \frac{48A_b}{3} + \frac{4A_b}{3} = \frac{52A_b}{3}]


The difference between the volume of the unsharpened pencil and the sharpened one is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  20A_b - \frac{52A_b}{3} = \frac{60A_b}{3} - \frac{52A_b}{3} = \frac{8A_b}{3}]


Then the percentage decrease from the original is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 100\left(\frac{ \frac{8A_b}{3}}{20A_b}\right) = 100\left(\frac{8}{60}\right) \approx 13.33%]


Sean's 13% is close enough.


By the way, did you notice that no matter how fat or skinny the pencil is, the answer is the same?


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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