1 Answer

Huw Davies · Answer 1 · 2021-03-26T05:30:15+0000

Answer:

$300\; \rm lb$ .

Explanation:

Let
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
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
pounds of that
$15\%$ copper alloy:
$(0.15\, x)\; \rm lb$ .
Mass of copper in
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
:

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

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

Simplify and solve for
:

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

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

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

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

Please log in or register to add a comment.