Question 408464
In words:
(liters of salt in 40% solution + liters of salt in 60% solution)/(liters of solution) = 55%
Let {{{a}}} = liters of 40% solution needed
Let {{{b}}} = liters of 60% solution needed
---------------------------------
given:
{{{.4a}}} = salt in 40% solution
{{{.6b}}} = salt in 60% solution
------------------------
(1) {{{(.4a + .6b)/8 = .55}}}
(2) {{{a + b = 8}}}
This is 2 equations and 2 unknowns, so it's solvable
----------------------
(1) {{{(.4a + .6b)/8 = .55}}}
(1) {{{.4a + .6b = .55*8}}}
(1) {{{.4a + .6b = 4.4}}}
(1) {{{40a + 60b = 440}}}
and, also,
(2) {{{a + b = 8}}}
(2) {{{40a + 40b = 320}}}
Subtract (2) from (1)
(1) {{{40a + 60b = 440}}}
(2) {{{-40a - 40b = -320}}}
{{{20b = 120}}}
{{{b = 6}}}
and, since
(2) {{{a + b = 8}}}
{{{a = 2}}}
2 liters of 40% solution and 6 liters of 60% solution are needed
check answer:
(1) {{{(.4a + .6b)/8 = .55}}}
(1) {{{(.4*2 + .6*6)/8 = .55}}}
{{{(.8 + 3.6)/8 = .55}}}
{{{4.4/8 = .55}}}
{{{4.4 = 4.4}}}
OK