Question 601799
Let {{{ a }}} = ml of 50% solution needed
Let {{{ b }}} = ml of 20% solution needed
Let {{{ c }}} = ml of 5% solution needed
given:
(1) {{{ a + b + c = 600 }}} ml
(2) {{{ ( .5a + .2b + .05c ) / 600 = .1 }}}
This is only 2 equations and 3 unknowns,
but I can guess at a solution
--------------------------------
(2) {{{ .5a + .2b + .05c = 60 }}}
(2) {{{ 50a + 20b + 5c = 6000 }}}
Multiply both sides of (1) by {{{ 5 }}} and
subtract (1) from (2)
(2) {{{ 50a + 20b + 5c = 6000 }}}
(1) {{{ -5a - 5b - 5c = 3000 }}}
{{{ 45a + 15b = 3000 }}}
{{{ 3a + b = 200 }}}
I'll guess that {{{ a = 50 }}} and {{{ b = 50 }}}
then
(2) {{{ 50a + 20b + 5c = 6000 }}}
(2) {{{ 50*50 + 20*50 + 5c = 6000 }}}
(2) {{{ 2500 + 1000 + 5c = 6000 }}}
(2) {{{ 3500 + 5c = 6000 }}}
(2) {{{ 5c = 2500 }}}
(2) {{{ c = 500 }}}
That works, so I can use
50 ml of 50% solution needed
50 ml of 20% solution needed
500 ml of 5% solution needed
There can be other solutions, but this works