Question 987975
{{{ .6*3.6 = 2.16 }}} liters of antifreeze
in radiator to begin with
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Let {{{ x }}} = liters of coolant to be
drained and replaced with water
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{{{ .6x }}} = liters of antifreeze in the
coolant that is drained off
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In words:
[ liters of antifreeze I end up with ] / [ capacity of radiator ] = 50%
{{{ ( 2.16 - .6x ) / 2.16 = .5 }}}
{{{ 2.16 - .6x = .5*2.16 }}}
{{{ .6x = .5*2.16 }}}
{{{ .6x = 1.08 }}}
{{{ x = 1.8 }}}
1.8 liters of coolant should be
drained and replaced with water
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check:
{{{ ( 2.16 - .6*1.8 ) / 2.16 = .5 }}}
{{{ ( 2.16 -1.08 ) / 2.16 = .5 }}}
{{{ 1.08/2.16 = .5 }}}
{{{ .5 = .5 }}}
OK