Question 269223
Let {{{a}}} = liters of 25% solution needed
Let {{{b}}} = liters of 40% solution needed
Let {{{c}}} = liters of 50% solution needed
given:
(1) {{{a + b + c = 200}}} liters
(2) {{{a = 2b}}} liters
{{{(.25a + .4b + .5c)/200 = .32}}}
{{{(.25a + .4b + .5c)/200 = 64}}}
(3) {{{25a + 40b + 50c = 6400}}}
Multiply both sides of (1) by {{{25}}}
and subtract from (3)
(3) {{{25a + 40b + 50c = 6400}}}
(1) {{{-25a - 25b - 25c = 5000}}}
{{{15b + 25c = 1400}}}
{{{3b + 5c = 280}}}
{{{5c = 280 - 3b}}}
{{{c = 56 - (3/5)*b}}}
Going back to (1)
(1) {{{a + b + c = 200}}}
{{{2b + b + 56 - (3/5)*b = 200}}}
{{{3b - (3/5)*b = 144}}}
{{{15b - 3b = 720}}}
{{{12b = 720}}}
{{{b = 60}}}
And, since
{{{a = 2b}}}
{{{a = 2*60}}}
{{{a = 120}}}
(1) {{{a + b + c = 200}}}
{{{120 + 60 + c = 200}}}
{{{c = 20}}}
120 liters of 25% solution are needed
60 liters of 40% solution are needed
20 liters of 50% solution are needed
check:
{{{(.25a + .4b + .5c)/200 = .32}}}
{{{(.25*120 + .4*60 + .5*20)/200 = .32}}}
{{{(30 + 24 + 10)/200 = .32}}}
{{{64/200 = .32}}}
{{{64 = 64 }}} OK