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

Algebra ->  Customizable Word Problem Solvers  -> Evaluation -> 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      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

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) About Me  (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

.
.
w=288

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 %286M-16%29%2B%284M-12%29%2B%283M%2B4%29=12M.
.
.