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?
1 answer:
Answer is:
7.8 lb of 21% aluminum and 33.2 ib of <span>
42% aluminum.</span>
ω₁<span> = 21% ÷ 100% = 0.21.
ω</span>₂<span> = 42% ÷ 100% = 0.42.
ω</span>₃<span> = 38% ÷ 100% = 0.38.
</span>m₁ = ?.
m₂<span> = ?.
</span>m₃ = m₁ + m₂<span>.
</span>m₃ = 41 pounds.
m₁ = 41 lb - m₂<span>.
ω</span>₁ · 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<span> = 7.8 lb.</span>
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