Question 295596
.
How many ounces of pure alcohol must be added to 40 ounces of a 70% alcohol solution 
to produce an 85% alcohol solution?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



        The solution in the post by @mananth is incorrect.

        I came to bring a correct solution.



let quantity of pure alcohol to be added be x ounces
70% alcohol is 40 ounces
the total will be x+40 ounces
x + 0.7*40 = (x+40)*0.85
x + 28 = 0.85x + 34
0.15x = 34 - 28
0.15x = 6
x = 6/0.15 = 40.


<U>ANSWER</U>.  &nbsp;&nbsp;40 &nbsp;ounces of pure alcohol must be added.


<U>CHECK</U> &nbsp;&nbsp;for the concentration: &nbsp;&nbsp;&nbsp;&nbsp;{{{(40 + 0.7*40)/(40+40)}}} = 0.85  &nbsp;&nbsp;! &nbsp;Precisely correct &nbsp;!


Solved correctly.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Notice that &nbsp;85% &nbsp;concentration is precisely midpoint between &nbsp;70% &nbsp;and &nbsp;100%, 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;so the answer is obvious and can be guessed &nbsp;MENTALLY.