<PRE><B>Question 135619</B>
A chemist is making 200 L of a solution that is 62% acid. He is mixing an 80% acid solution with a 30% acid solution. How much of the 80% acid solution will he use?
--------------------
80% solution DATA:
Amt = x liters; amount of acid is 0.8x liters
----------------------------------
30% solution DATA:
Amt = 12-x liters; amt of acid is 0.3(12-x) = 3.6-0.3x liters
----------------------------------
Misture DATA:
Amt = 12 liters; amt of acid is 0.62*12 = 7.44 liters
---------------------------------------
EQUATION:
acid + acid = acid
0.8x + 2.6-0.3x = 7.44
0.5x = 4.84
x = 9.68 liters of 80% solution in the mixture
=================================================
Cheers,
Stan H.
</PRE>