Question

How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy

Answer 1

`300\; \rm lb`.

Explanation:

Let
`x` represent the mass (in pounds) of that
`15\%` copper alloy required, such that the final mixture would contain
`25.5\%` copper by mass.

Consider: if
`x` pounds of that
`15\%` copper alloy is mixed with
`700` pounds that
`30\%` copper alloy, what would be the mass of copper in the mixture?

• Mass of copper in
  `x` pounds of that
  `15\%` copper alloy:
  `(0.15\, x)\; \rm lb`.
• Mass of copper in
  `700` pounds of that
  `30\%` copper alloy:
  `700 * 0.30 = 210\; \rm lb`.

Therefore, the mixture would contain
`(210 + 0.15\, x) \; \rm lb` of copper.

The mass of that mixture would be
`(700 + x)\; \rm lb`. The mass fraction of copper in that mixture would be:

`\displaystyle ((210 + 0.15\, x)\; \rm lb)/((700 + x)\; \rm lb) * 100\%`.

This ratio is supposed to be equal to
`25.5\%`. These two pieces of equations combine to give an equation about
`x`:

`\displaystyle ((210 + 0.15\, x)\; \rm lb)/((700 + x)\; \rm lb) * 100\% = 25.5\%`.

`\displaystyle (210 + 0.15\, x)/(700 + x) = 0.255`.

Simplify and solve for
`x`:

`210 + 0.15\, x= 0.255\, (700 + x)`.

`(0.255 - 0.15)\, x= 210 - 0.255 * 700`.

`\displaystyle x = (210 - 0.255 * 700)/(0.255 - 0.15) = 300`.

Therefore,
`300\; \rm lb` of that
`15\%` alloy would be required.