Question 399944
Let {{{a}}} = ml of 2% alcohol needed
Let {{{b}}} = ml of 10% alcohol needed
It follows, then:
{{{.02a}}} = ml of alcohol in 2% solution
{{{.1b}}} = ml of alcohol in 10% solution
In words:
(ml of alcohol in final solution)/(ml of solution) = 8%
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(1) {{{(.02a + .1b)/2000 = .08}}}
(2) {{{a + b = 2000}}}
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(1) {{{.02a + .1b = 160}}}
(1) {{{2a + 10b = 16000}}}
Multiply both sides of (2) by {{{2}}}
and subtract (2) from (1)
(1) {{{2a + 10b = 16000}}}
(2) {{{-2a - 2b = -4000}}}
{{{8b = 12000}}}
{{{b = 1500}}} ml
And,
{{{a + b = 2000}}}
{{{a = 2000 - 1500}}}
{{{a = 500}}} ml
500 ml of 2% alcohol are needed
1500 ml of 10% alcohol are needed
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check answer:
(1) {{{(.02a + .1b)/2000 = .08}}}
(1) {{{(.02*500 + .1*1500)/2000 = .08}}} 
(1) {{{(10 + 150)/2000 = .08}}}
{{{160/2000 = .08}}}
{{{160 = 160}}}
OK