# Alligation and Percentage

What is the formula used to calculate how many mL should be used from a higher concentration solution that results in an ordered % if added to a fixed volume of the solution of a lower concentration?

For example:

• How many mL of 50% dextrose that should be added to 500 mL of 5% dextrose to prepare 8% dextrose?

• How many mL of 50% dextrose that should be added to 500 mL of 5% dextrose to prepare 12.5% dextrose?

• Not to be confused with crocodilation. Commented Jun 29, 2013 at 22:05

The way I'm doing it is simpler and translates English into math which is more straightforward than the above example.

But the way that I basically added in sort of a English language way the math terms and then figured this out and made a substitution.

X ML's of 50% solution +500 ml of 5%solution = 8% of total solution Translation in to math

X(50%) + 500(5%) = 8%(total solution in ml)

Total solution is 500+ x

x(50/100) + 500(5/100) = (8/100)(x+500)

Then solve for x and you come up with the answer of 35.7. You can keep the units on the equation if you'd like and they will cancel each other out but basically this is a simpler way to think of it by putting the English parameters into a math equations and then figure out what each of the individual math parts really mean.

Let $V$ be the fixed volume of solution $A$. Let $100\alpha$ be the percentage of dextrose in $A$. Let $100\beta$ be the percentage of dextrose in the higher concentration solution $B$. And let $100\delta$ be the desired percentage of dextrose. So for example if we start with $700$ ml of $10\%$ dextrose, and we wish to pour in enough $40\%$ dextrose to make $15\%$ dextrose, then $V=700$, $\alpha=0.10$, $\beta=0.40$, and $\delta=0.15$.

Let $x$ be the amount of $B$ we should add to $A$. We will end up with volume $V+x$. The amount of dextrose in this quantity will be $(V+x)\delta$.

Let's find the amount of dextrose another way. An amount $V\alpha$ will come from $A$, and $x\beta$ will come from $B$, for a total of $V\alpha+x\beta$. So we have $$(V+x)\delta=V\alpha+x\beta.$$ Rewrite this as $$V(\delta-\alpha)=x(\beta-\delta).$$ Now solve for $x$. We get $$x=V\frac{\delta-\alpha}{\beta-\delta}.$$

Remark: Because the multiplier $\frac{\delta-\alpha}{\beta-\delta}$ does not change if we multiply each of $\alpha$, $\beta$, and $\delta$ by a constant, we can think of, say, $25\%$ as $25$, as long as we are consistent.

Let us take the first of your examples. There $V=500$, $\alpha=0.05$, $\beta=0.50$, and $\delta=0.08$. But if we wish we can use $5$, $50$, and $8$. Then $$x=500\frac{8-5}{50-8}.$$ Calculate. We get $x\approx 35.71$.

• Thank you Andre, I have spent the last day to come to that result! I appreciate your answer. Commented Apr 13, 2013 at 0:38
• x=500*(8−5)/(50−8) = 35.714 :) Commented Apr 13, 2013 at 0:48
• @user72343: Thanks. My arithmetic is weak. Actually, keying error. Commented Apr 13, 2013 at 1:07