Question 235976
Let {{{a}}} = grams of 45% mixture to be used
Let {{{b}}} = grams of 75% mixture to be used
given:
(1) {{{a + b = 60}}} grams
(2) {{{.45a + .75b}}} = pure nickel in final mixture
------------------------------
In words:
( grams of pure nickel in final mixture )/(total grams) = 65%
{{{(.45a + .75b) / 60 = .65}}}
{{{.45a + .75b = 39}}}
(3){{{45a + 75b = 3900}}}
Multiply both sides of (1) by {{{45}}} and subtract
(1) from (3)
(3){{{45a + 75b = 3900}}}
(1){{{-45a - 45b = -2700}}}
{{{30b = 1200}}}
{{{b = 40}}}
And, since
{{{a + b = 60}}}
{{{a + 40 = 60}}}
{{{a = 20}}}
20 grams of the 45% mixture and 40 grams of the 75% mixture are needed
check:
{{{(.45a + .75b) / 60 = .65}}}
{{{(.45*20 + .75*40) / 60 = .65}}}
{{{( 9 + 30) / 60 = .65}}}
{{{39 = .65*60}}}
{{{39 = 39}}}
OK