Question 121689
Let {{{x}}} = the amount of 20% solution that I need to
drain off.
Then {{{x}}} will also be the amount of pure antifreeze 
that I add afterwards.
My basic equation in words is:
(% antifreeze solution) = (amount of antifreeze) / (antifreeze + water)
My (antifreeze + water) does not change. I started with {{{70}}} l
and I end up with {{{70}}} l.
I'm looking for a 40% solution
.4 = (amount of antifreeze)/70
I start off with {{{70}}} l of 20% solution.
That means that it contains
{{{.2*70 = 14}}}l of antifreeze
I first drained off {{{x}}} amount of 20% solution.
That means I drained off {{{.2x}}} amount of antifreeze
The I add {{{x}}} amount of antifreeze
{{{.4 = (14 - .2x + x) / 70}}}
{{{.4 = (14 + .8x) / 70}}}
{{{28 = 14 + .8x}}}
{{{.8x = 14}}}
{{{x = 17.5}}}l answer
I'd like to check this
What did I start with?
{{{14}}} l antifreeze and {{{56}}} l water for a {{{14 + 56 = 70}}}l solution
I drain off {{{17.5}}}l of this 20% solution
I drain off {{{.2*17.5 = 3.5}}}l antifreeze
and {{{.8*17.5 = 14}}}l water
Now I've got {{{14 - 3.5 = 10.5}}}l antifreeze
and {{{56 - 14 = 42}}}l water
The last step: I want to add {{{17.5}}} l of antifreeze
Now I've got {{{10.5 + 17.5 = 28}}}l of antifreeze
and still {{{42}}}l of water
going back to:
(% antifreeze solution) = (amount of antifreeze) / (antifreeze + water)
(% antifreeze solution) = {{{28 / (42 + 28)}}}
(% antifreeze solution) = {{{28 / 70}}}
(% antifreeze solution) = 40%
OK