SOLUTION: 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

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).

3.1416 * .25 * .25 * 144 = 28.27 cu. in. (volume of 1 rod).
28.27 / .404 = 69.97 = 70 pins / rod.
10,000 / 70 = 142.85 = 143 rods.

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

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
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).
******************
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).
==================================
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.
-------------
10,000 / 70 = 142.85 = 143 rods.
*******
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).

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