An alloy of tin is 15% tin and weighs 20 pounds. A second alloy is 10% tin. How many pounds of the second alloy must be added to the first to get a 12% mixture?
Answer:
You must add 30 lb of the 10 % alloy.
Explanation:
Let the mass of the 10 % alloy be \(x \mathrm{lb .}\)
\(300+10 x=12(20+x)=240+12 x\)
\(2 x=300-240=60\)
\(x=\frac{60}{2}=30\)
You must add 30 lb of the second alloy.
Check:
\(20 \mathrm{lb} \times 15 \% \mathrm{Sn}+30 \mathrm{lb} \times 10 \% \mathrm{Sn}=50 \mathrm{lb} \times 12 \% \mathrm{Sn}\)
\(3 \mathrm{lb} \mathrm{Sn}+3 \mathrm{lb} \mathrm{Sn}=6 \mathrm{lb} \mathrm{Sn}\)
\(6 \mathrm{lb}=6 \mathrm{lb}\)
It checks!