<PRE><B>Question 203665</B>
needs 20 quarts of 50% antifreeze solution in her radiator.
plans to obtain this by mixing some pure antifreeze with an appropriate
 amount of a 40% antifreeze solution. How many quarts of antifreeze?
:
Let x = amt of pure antifreeze
The resulting amt is to be 20 qts, therefore
(20-x) = amt of 40% solution required
:
A typical mixture problem
:
x + .40(20-x) = .50(20)
;
x + 8 - .4x = 10
:
x - .4x = 10 - 8
:
.6x = 2
x = {{{2/.6}}}
x = 3{{{1/3}}} qts of pure anti-freeze
and
20 - 3{{{1/3}}} = 16{{{2/3}}} qts of 40% mixture </PRE>