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A piece of metal is 1 ft. long, 6 in. wide, and 4 in. thick, and weighs 3037 oz. avoirdupois.
It is composed of an alloy of gold and copper. Determine percentage of gold.
Gold weighs .70 lbs. per cu. in.
Copper weighs .32 lbs. per cu. in.
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When they ask "Determine percentage of gold", they mean the ratio of the gold contents by WEIGHT to the total WEIGHT of the alloy;
the ratio, expressed as percentage.
My first step is to convert 3037 oz. avoirdupois to pounds:
Wikipedia says https://en.wikipedia.org/wiki/Avoirdupois
The avoirdupois system is a measurement system of weights which uses pounds and ounces as units.
It was first commonly used in the 13th century and was updated in 1959.
Shortly speaking, "1 oz. avoirdupois" = of a pound.
So, 3037 oz. avoirdupois = = 189.8125 pounds.
Then the density of the alloy is = 0.659 lb/in^3.
Now, having the weight and the density expressed in consistent units, I can write the system of equations.
Let 0 <= x <= 1 be the fraction (by the weight) of gold in the alloy (i.e. percentage is 100x).
Then the weight fraction of copper is (1-x), obviously.
Then the "fraction" equation is
+ = ,
where D stands for density, or
0.70*x + 0.32*(1-x) = 0.659.
Simplify and solve for x:
0.70x + 0.32 - 0.32x = 0.659 ====> 0.38x = 0.659 - 0.32 = 0.339 ====> x = = 0.8921.
Thus the gold fraction by the weight is 0.8921, or 89.21%.
Solved.