<PRE><B>Question 752567</B>
 how many gallons of 20% antifreeze should be mixed with 10 gallons of 83% antifreeze to obtain 55% antifreeze mixture?
---------
Equation:
alcohol + alcohol = alcohol
0.20x + 0.83*10 = 0.55(x+10)
-----
20x + 83*10 = 55x + 55*10
----------
35x = 28*10
x = (2/7)28
x = 8 gallons (amt. of 20% solution needed)
==============================================
Cheers,
Stan H.
================</PRE>