Question 363416
Let {{{a}}} = liters of 40% solution needed
Let {{{b}}} = liters of 10% solution needed
given:
{{{.4a}}} = liters of alcohol in 40% solution
{{{.1b}}} = liters of alcohol in 10% solution
{{{.25*2 = .5}}} = liters of alcohol in final solution
--------------------
(1) {{{(.4a + .1b)/2 = .25}}}
(2) {{{a + b = 2}}}
from (1)
(1) {{{.4a + .1b = .5}}}
(1) {{{4a + b = 5}}}
Subtract (2) from (1)
 {{{4a + b = 5}}}
{{{-a - b = -2}}}
{{{3a = 3}}}
{{{a = 1}}}
and
{{{a + b = 2}}}
{{{1 + b =2}}}
{{{b = 1}}}
1 liter of 40% solution and 1 liter of 10% solution are needed
check:
(1) {{{(.4a + .1b)/2 = .25}}}
(1) {{{(.4*1 + .1*1)/2 = .25}}}
{{{.5/2 = .25}}}
{{{.5 = .5}}}
OK