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 Question from roger: I have 6800 gallons of a 17% glycol solution. I need to add how many gallons of 100% glycol to this to raise the concentration to 22%, also 20%?

Hi Roger.

Let x = the number of gallons of 100% glycol you need and let p = the percentage concentration you need at the end (as a decimal: 20% = 0.20).

Then
(6800)(0.17) + x(1.00) = the total amount of glycol in the final solution.
and
6800 + x = the total amount of solution.

The ratio of these two is p.

p = [ 6800(0.17) + x(1.00) ] / (6800 + x)

To solve for x, we have:

p = (1156 + x) / (6800 + x)
6800p + xp = 1156 + x
xp - x = 1156 - 6800p
x(p-1) = 1156 - 6800p

x = (6800p - 1156) / (1-p)

So if you want 20%, then p is 0.20:

x = (6800(0.20) - 1156) / (1 - 0.20) = 255 gallons

You can solve this more generally. Let A be the quantity of the first liquid and a be the concentration of it. We will use up all of this first liquid.

Now let B be the quantity of the second liquid whose concentration is b. We will use as much of it as we need to make a mixture whose concentration is c (in between a and b, of course).

The total active ingredient in the mixture is
Aa + Bb
and the total liquid is A + B.

So c = (Aa + Ba) / (A + B)

And the unknown quantity B is therefore:
Ac + Bc = Aa + Ba
Bc - Ba = Aa - Ac
B = A(c-a)/(b-c)

You can see that this is the same as your specific question because
when A = 6800, a = 0.17, b = 1.00 and c = 0.20, you have:
B = 6800(0.20 - 0.17) / (1 - 0.20) = [ 6800(0.20) - 1156 ] / (1 - 0.20)

the same as we arrived at earlier.

Hope this helps,
Stephen La Rocque.

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.