Question 1050211
A quantity of steel rods are used to produce 10,000 pins. Allow 1/16" for waste in cutting each pin. 
Description of pins: 
1/2" diameter. 
2" in length. 

Description of rods: 
1/2" diameter. 
12' in length. 

Determine: Number and weight of rods. 
Steel weighs .283 lbs./cu. ft. 
My attempt: 
The rods and pins are cylindrical. 
3.1416 * .25 * .25 * 2.0625 = .404 cu. in. (volume of 1 pin).
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Don't include the volume of the waste in the volume of the pins.
Vol = 3.1416 * .25 * .25 * 2 = 0.3927 cu. in. (volume of 1 pin).
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3.1416 * .25 * .25 * 144 = 28.27 cu. in. (volume of 1 rod).

28.27 / .404 = 69.97 = 70 pins / rod. ****** The waste is included here.
But, you have to round down.
69 pins*0.404 = 27.876 cubic inches
There's not enough left to make another pin.
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10,000 / 70 = 142.85 = 143 rods. 
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10000/69 =~ 144.93 rods which mean 145 rods will be needed.


28.27 * 143 = 4042.61 = 4043 cu. in. (total vol. of rods). 
4043 * .283 = 1144.169 = 1144 lbs. (total wt. of rods).