SOLUTION: How many gallons of a 70% acid solution must be mixed with 30 gallons of a 16% solution to obtain a solution that is 60% acid?

How many gallons of a 70% acid solution must be mixed 

with 30 gallons of a 16% solution to obtain a solution that is 

60% acid?

---------------------------------------------------------

We are mixing a stronger 70% acid solution with a weaker 16% 

acid solution to obtain a medium strength 60% acid solution. 

Let x = the number of gallons of the stronger

We mix the following:

 x gallons of stronger

30 gallons of weaker

and therefore we get

x+30 gallons of medium strength solution

Now we figure the number of gallons of pure alcohol contained 

in all three of these:

The x gallons of stronger solutions contains 0.70x gallons of pure 

alcohol (and the rest is water, which we ignore.)

The 30 gallons of weaker solution contains 0.16(30) or 4.8 gallons 

of pure alcohol (and the rest is water, which we ignore).

The x+30 gallons of the final medium-strength solution contains 0.60(x+30) gallons of pure alsohol (and the rest is water, which we ignore).   

Now 

   The number of gallons of pure alcohol in the stronger +

       The number of gallons of pure alcohol in the weaker solution =

           The number of gallons of pure alcohol in the medium-strength

           solution.  

or  0.70x + 4.8 = 0.60(x+30)

Can you solve that? Answer: x = 132 gallons. 

---------------------------------------------

Check:

The 132 gallons of 70% contains 92.4 gallons of pure alcohol and

39.6 gallons of water.

The 30 gallons of 16% contains 4.8 gallons of pure alcohol and 

25.2 gallons of water.

When they are mixed:

The 162 gallons of 60% contains 97.2 gallons of pure alcohol and 

64.8 gallons of water.

Adding up the liquid: 132 gallons + 30 gallons = 162 gallons

Adding up the pure acid: 92.4 gallons + 4.8 gallons = 97.2 gallons 

Adding up the water: 39.6 gallons + 25.2 gallons = 64.8 gallons

It all adds up!

Edwin