In an alloy of copper and tin, there is 85% of copper. How much of this alloy do you need in order to get 1 13/32 lb of tin?

we know the amount of copper is 85%, that means the rest is tin in the Alloy, so the tin is 15% of the Alloy then.

now, say "x" lbs is the total amount of the Alloy, and we know that 1 and 13/32 which is tin, is the 15% of "x", what the dickens is "x"?

[tex]\bf \begin{array}{ccll} amount&\%\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ x&100\\\\ 1\frac{13}{32}&15 \end{array}\implies \cfrac{x}{1\frac{13}{32}}=\cfrac{100}{15}\implies \cfrac{x}{\frac{1\cdot 32+13}{32}}=\cfrac{20}{3}[/tex]

[tex]\bf \cfrac{\quad x\quad }{\frac{45}{32}}=\cfrac{20}{3}\implies \cfrac{\quad \frac{x}{1}\quad }{\frac{45}{32}}=\cfrac{20}{3}\implies \cfrac{x}{1}\cdot \cfrac{32}{45}=\cfrac{20}{3}\implies \cfrac{32x}{45}=\cfrac{20}{3} \\\\\\ 96x=900\implies x=\cfrac{900}{96}\implies x=\cfrac{75}{8}\implies x=9\frac{3}{8}[/tex]