Question 883785
Let {{{ x }}} = ml of the 5% solution to be added
{{{ .05x }}} = ml of the Minoxidil in 5% solution
{{{ .01*100 = 1 }}} ml of the Minoxidil in 1% solution
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(1) {{{ x + 100 }}} = ml of 5% plus 1% solution
(2) {{{ ( .05x + 1 ) / ( x + 100 ) = .03 }}}
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(2) {{{ .05x + 1 = .03*( x + 100 ) }}}
(2) {{{ .05x + 1 = .03x + 3 }}}
(2) {{{ .02x = 2 }}}
(2) {{{ x = 100 }}}
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She needs 100 ml of the 5% solution
check:
(2) {{{ ( .05x + 1 ) / ( x + 100 ) = .03 }}}
(2) {{{ ( .05*100 + 1 ) / ( 100 + 100 ) = .03 }}}
(2) {{{ ( 5 + 1 ) / ( 100 + 100 ) = 3/100 }}}
(2) {{{ 6/200 = 3/100 }}}
(2) {{{ 3/100 = 3/100 }}}
OK