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solmaris [256]
3 years ago
5

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?

Mathematics
1 answer:
Lina20 [59]3 years ago
8 0
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"?

\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}

\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}
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