Question 1058969
Let {{{ x }}} = gallons of pure acid to be added
{{{ .2*1 = .2 }}} gallons of acid in original solution
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{{{ ( x + .2 ) / ( x + 1 ) = .3 }}}
{{{ x + .2 = .3*( x + 1 ) }}}
{{{ x + .2 = .3x + .3 }}}
{{{ .7x = .1 }}}
{{{ x = 1/7 }}}
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1/7 gallon of pure acid needs to be added
( about .1429 gallons )
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check:
{{{ ( x + .2 ) / ( x + 1 ) = .3 }}}
{{{ ( 1/7 + 1/5 ) / ( 1/7 + 1 ) = 3/10 }}}
{{{ ( 5/35 + 7/35 ) / ( 8/7 ) = 3/10 }}}
{{{ ( 12/35 ) / ( 40/35 ) = 3/10 }}}
{{{ 12/40 = 3/10 }}}
{{{ 3/10 = 3/10 }}}
OK