Answer:
[tex]300\; \rm lb[/tex].
Step-by-step explanation:
Let [tex]x[/tex] represent the mass (in pounds) of that [tex]15\%[/tex] copper alloy required, such that the final mixture would contain [tex]25.5\%[/tex] copper by mass.
Consider: if [tex]x[/tex] pounds of that [tex]15\%[/tex] copper alloy is mixed with [tex]700[/tex] pounds that [tex]30\%[/tex] copper alloy, what would be the mass of copper in the mixture?
Therefore, the mixture would contain [tex](210 + 0.15\, x) \; \rm lb[/tex] of copper.
The mass of that mixture would be [tex](700 + x)\; \rm lb[/tex]. The mass fraction of copper in that mixture would be:
[tex]\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%[/tex].
This ratio is supposed to be equal to [tex]25.5\%[/tex]. These two pieces of equations combine to give an equation about [tex]x[/tex]:
[tex]\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%[/tex].
[tex]\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255[/tex].
Simplify and solve for [tex]x[/tex]:
[tex]210 + 0.15\, x= 0.255\, (700 + x)[/tex].
[tex](0.255 - 0.15)\, x= 210 - 0.255 \times 700[/tex].
[tex]\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300[/tex].
Therefore, [tex]300\; \rm lb[/tex] of that [tex]15\%[/tex] alloy would be required.
