<PRE><B>Question 101977</B>
A car radiator has a capacity of 16 qts and is filled w/a 25% anti-freeze solution. How much must be drained off and replaced w/pure antifreeze to obtain a 40% anti-freeze solution?
:
Let x = the amt drained and the amt added
:
Drained  + Added = required mixture
.25(16-x) + 1.0x = .40(16)
:
4 - .25x + 1x = .40(16)
:
4 + .75x = 6.4
:
.75x = 6.4 - 4
:
.75x = 2.4
:
x = 2.4/.75
:
x = 3.2 gal of pure antifreeze
:
:
Check solution using x = 3.2:
.25(16-3.2) + 3.2 = .40(16)
.25(12.8) + 3.2 = 6.4
   3.2  +  3.2  = 6.4</PRE>