SOLUTION: In a mixture of copper, tin, and lead, one half of the whole, minus 16 lb, was copper; one third of the whole, minus 12 lb, was tin, and one fourth of the whole, plus 4 lb, was le

Question 1206112: In a mixture of copper, tin, and lead, one half of the whole, minus 16 lb, was copper; one third of the whole, minus 12 lb, was tin, and one fourth of the whole, plus 4 lb, was lead: how much did the whole mixture weigh?
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
One can transcribe the description exactly as written,
here using w for the whole quantity of mixture.

Copper          (w/2)-16
Tin             (w/3)-12
Lead            (w/4)+4
WHOLE             w

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Another way, the whole mixture is some 12 equal parts, so 12M is this mixture quantity.

Copper       6M-16
Tin          4M-12
Lead         3M+4
WHOLE        12M

Remember M here is a factor.
Equation to solve can be .
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