Question 126875
In words:
(pounds of aluminum in 12% alloy) + (pounds of aluminum in 25% alloy)
divided by (total pounds of mixture) = 17%
Let {{{a}}} = pounds of 12% alloy needed
Let {{{b}}} = pounds of 25% alloy needed
{{{(.12a + .25b) / (a + b) = .17}}}
{{{b = 400}}}
{{{(.12a + .25*400) / (a + 400) = .17}}}
{{{(.12a + 100) / (a + 400) = .17}}}
{{{.12a + 100 = .17(a + 400)}}}
{{{.12a + 100 = .17a + 68}}}
{{{.05a = 32}}}
{{{a = 640}}}
640 pounds of 12% alloy must be used
check answer
{{{(.12a + .25b) / (a + b) = .17}}}
{{{(.12*640 + .25*400) / 1040 = .17}}}
{{{(76.8 + 100) / 1040 = .17}}}
{{{176.8 = 1040*.17}}}
{{{176.8 = 176.8}}}
OK