<PRE><B>Question 529827</B>
A tank contains 20 gallons of antifreeze solution.
 When it is full, it contains 15% antifreeze. 
How many gallons must be replaced by an 80% antifreeze solution to get 20 gallons of a 70% solution?
:
Let x = amt of 15% solution removed, and the amt of 80% solution to be added
:
.15(20-x) + .80x = .70(20)
3 - .15x + .80x = 14
-.15x + .80x = 14 - 3
.65x = 11
x = {{{11/.65}}}
x = 16.9 gal removed and 16.9 gal of 80% antifreeze to be added
:
:
Check this: 20-16.9 = 3.1 gal of 15% solution remains
.15(3.1) + .8(16.9) = 14
.465 + 13.52 = 13.985 ~ 14</PRE>