Question 921829
Let {{{ a }}} = ml of 30% alcohol
Let {{{ b }}} = ml of 45% alcohol
Let {{{ c }}} = ml of 60% alcohol
Let {{{ d }}} = ml of 95% alcohol
{{{ .5*20 = 10 }}} ml of alcohol in final mixture
----------------------------------------
{{{ a + b + c + d = 20 }}}
{{{ .3a + .45b + .6c + .95d = 10 }}}
-------------------------------
Actually you don't need any 95% or 30%
so, I can say:
(1) {{{ b + c = 20 }}}
(2) {{{ .45b + .6c = 10 }}}
(2) {{{ 45b + 60c = 1000 }}}
(2) {{{ 9b + 12c = 200 }}}
----------------------
Multiply both sides of (1) by {{{ 9 }}} and
subtract (1) from (2)
(2) {{{ 9b + 12c = 200 }}}
(1) {{{ -9b - 9c = -180 }}}
{{{ 3c = 20 }}}
{{{ c = 6.667 }}}
and
(1) {{{ b + c = 20 }}}
(2) {{{ b = 20 - 6.667 }}}
(2) {{{ b = 13.333 }}}
-------------------
0 ml of 30% alcohol
13.333 ml of 45% alcohol
6.667 ml of 60% alcohol
0 ml of 95% alcohol
---------------------
Hope I got it