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? 30 lb 40 lb 60 lb

so, they alloy is hmmm some piece of metal.

it has tin, among other metals, regardless of what other metals it may have, we know that for the first one, 15% is tin, the rest something else, and we also know that the alloy itself, weights 20 lbs.

now, how much is just the weight of the tin metal in it alone?  well, is 15% of 20 lbs, or (15/100) * 20, or 3 lbs, so of all those 20 lbs, only 3 lbs or 15% is tin.

for the second alloy, let's say we'll add "x" lbs, we know the second alloy is 10% tin, how much of that weight is tin?  well 10% of x, or (10/100) * x, or 0.10x lbs.

now, we're looking for a mixture, say the resulting lbs of mix will be "y" lbs, and we know the mixture will be 12% of tin, how much is 12% of y?  (12/100) * y, or 0.12y.

now, regardless of what "x" and "y" is, we know that 20 + x = y, and that 3 + 0.10x = 0.12y, thus

[tex]\bf \begin{array}{lccclll} &\stackrel{lbs}{amount}&\stackrel{tin~\%}{quantity}&\stackrel{tin~lbs}{quantity}\\ &------&------&------\\ \textit{15\% alloy}&20&0.15&3\\ \textit{10\% alloy}&x&0.10&0.10x\\ ------&------&------&------\\ mixture&y&0.12&0.12y \end{array} \\\\\\ \begin{cases} 20+x=\boxed{y}\\ 3+0.10x=0.12y\\ ----------\\ 3+0.10x=0.12\left( \boxed{20+x} \right) \end{cases} \\\\\\ 3+0.10x=2.4+0.12x\implies 3-2.4=0.12x-0.10x \\\\\\ 0.6=0.02x\implies \cfrac{0.6}{0.02}=x\implies 30=x[/tex]