Question 682458
The capacity of the radiator is 3.3 liters.
The radiator is filled with a solution of 75% antifreeze
which is {{{ .75*3.3 = 2.475 }}} liters of antifreeze
Let {{{ x }}} = liters,of coolant to be drained and
replaced with water
{{{ .75x }}} liters of coolant is in the liters of the 
{{{ x }}} liters drained
It started with {{{ 3.3 }}} liters and ended with {{{ 3.3 }}} liters
--------------------
{{{ ( 2.475 - .75x ) / 3.3 = .5 }}} 
{{{ 2.475 - .75x = .5*3.3 }}}
{{{ 2.475 - .75x = 1.65 }}}
{{{ .75x = 2.475 - 1.65 }}}
{{{ .75x = .825 }}}
{{{ x = 1.1 }}}
1.1 liters of the 75% solution must be drained and
replaced with pure water
check:
{{{ 3.3 - 1.1 = 2.2 }}} liters left after draining
{{{ .75*2.2 = 1.65 }}} liters of antifreeze left
Now add {{{ 1.1 }}} liters of water
The concentration is 
{{{ 1.65 / 3.3 = .5 }}} or 50%
OK