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A metallurgist has one alloy containing 21% aluminum and another containing 42% aluminum. how many pounds of each alloy must he use to make 41 pounds of a third alloy containing 38% aluminum?

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Dejanras
Answer is: 7.8 lb of 21% aluminum and 33.2 ib of 42% aluminum.

ω₁ = 21% ÷ 100% = 0.21.
ω₂ = 42% ÷ 100% = 0.42.
ω₃ = 38% ÷ 100% = 0.38.
m₁ = ?.

m₂ = ?.
m₃ = m₁ + m₂.
m₃ = 41 pounds.

m₁ = 41 lb - m₂.
ω₁ · m₁ + ω₂ ·m₂ = ω₃ · m₃.

0.21 · (41 lb - m₂) + 0.42 · m₂ = 0.38 · 41 lb.

8.61 lb - 0.21m₂ + 0.42m₂ = 15.58 lb.

0.21m₂ = 6.97 lb.

m₂ = 6.97 lb ÷ 0.21.

m₂ = 33.2 lb.

m₁ = 41 lb - 33.2 lb.

m₁ = 7.8 lb.



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