<PRE><B>Question 114743</B>
How many liters of 80% alcohol must be added to 10 liters of a 50% solution to obtain a 70% solution?
:
A typical mixture equation. If you learn this method you can do all of them:
:
Let x = amt of 80% stuff
then
(x+10) = resulting amt
:
.8x + .5(10) = .7(x+10)
:
.8x + 5 = .7x + 7
:
.8x - .7x = 7 - 5
:
.1x = 2
:
x = 2/.1
:
x = 20 liters of the 80% stuff 
:
:
Resulting amt would be 30 liters, check our solutions
.8(20) + .5(10) = .7(30)
16 + 5 = 21</PRE>