<PRE><B>Question 103174</B>
how many gallons of a 5% acid solution should be mixed with 30 gallons of a 10% acid solution to obtain a mixture that is 8% acid?
---------------
10% liquid DATA:
Amt = 30 gals ; amt of active ingredient = 0.10*30 = 3 gallons
-----------------
5% liquid DATA:
Amt = x gals ; amt of active = 0.05x gallons
-----------------
Mixture DATA:
Amt = 30+x gallons ; amt of active = 0.08(30+x) = 2.4 + 0.08x gallons
---------------
EQUATION:
active + active = active
3 + 0.05x = 2.4 + 0.08x
0.6 = 0.03x
x = 0.6/0.03 = 20 gallons ( amt of 5% acid solution that must be added)
===============
Cheers,
Stan H.
</PRE>