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

Answer:

$300\; \rm lb$.

Step-by-step 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: $700 \times 0.30 = 210\; \rm lb$.

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:

$\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%$.

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

$\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%$.

$\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255$.

Simplify and solve for $x$:

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

$(0.255 - 0.15)\, x= 210 - 0.255 \times 700$.

$\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300$.

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