Question

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?

Answer 1

9
`(3)/(8)`

Explanation:

I checked in the RSM system

So its write

Sorry if my word using sucks

Answer 2

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(13)/(32)&15 \end{array}\implies \cfrac{x}{1(13)/(32)}=\cfrac{100}{15}\implies \cfrac{x}{(1\cdot 32+13)/(32)}=\cfrac{20}{3}`


`\bf \cfrac{\quad x\quad }{(45)/(32)}=\cfrac{20}{3}\implies \cfrac{\quad (x)/(1)\quad }{(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(3)/(8)`