Question 220821
Let {{{a}}} = liters of 20% juice needed
Let {{{b}}} = liters of 10% juice needed
given:
(1) {{{a + b = 10}}} liters of 12% mixture
OJ in 20% juice is {{{.2a}}} liters
OJ in 10% juice is {{{.1b}}} liters
OJ in 12% juice is {{{.12*10 = 1.2}}} liters
In words:
(liters OJ in 20% mix + liters OJ in 10% mix)/(total liters of 12% mix) = 12%
{{{(.2a + .1b)/10 = .12}}}
{{{.2a + .1b = 1.2}}}
(2) {{{2a + b = 12}}}
Subtract (1) from (2)
(2) {{{2a + b = 12}}}
(1) {{{-a - b = -10}}}
{{{a = 2}}} liters
and, since
{{{a + b = 10}}}
{{{2 + b = 10}}}
{{{b = 8}}} liters
2 liters of the 20% juice and 8 liters of the 10%
juice need to be mixed
check answer:
{{{(.2a + .1b)/10 = .12}}}
{{{(.2*2 + .1*8)/10 = .12}}}
{{{(.4 + .8)/10 = .12}}}
{{{1.2/10 = .12}}}
{{{1.2 = 1.2}}}
OK