Cheese balls problem! : Quant Question Archive [LOCKED]
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# Cheese balls problem!

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27 Sep 2007, 09:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

hello,

I found this problem in the beginners' section and I have some difficulty.

We have 3 cheese balls with diameters of 1, 2, and 6. We melt the cheese balls and make one big cheese ball. What's the diameter of the new cheese ball?

So I understand we need to find the volume of each (4/3*pi*r^3), so
v1=4/3*pi*.5^3 = pi/6
v2=4/3*pi*1^3 = 4*pi/3
v3=4/3*pi*3^3 = 36*pi

so total volume is the sum and = 75*pi/2
so to solve for Diameter we do
4/3*pi*(D/2)^3 = 75*pi/2 well, I don't know how to solve that
the next line in the answer is D^3 = 225, but I am missing the in-between (there is an in-between explanation right?). Would someone be able to help me?

Zaza.
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27 Sep 2007, 10:13
Yes, cancel out pi s from each sides of equation, you will have 4/3*(r^3)=37.5

than multiply each sides by 3/4 than you will be left with,

r^3=225/8 as you know r=d/2 so,

r^3=(d/2)^3=(d^3)/2^3=(d^3)/8=225/8 if you multiply both sides by 8 you will see the last expression left is the same as you showed.
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27 Sep 2007, 12:33
I understand now. It's the elimination of pi that, for some strange reason, panicked me!!!

27 Sep 2007, 12:33
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