How to calculate mixing of two ratios to get desired ratio?

Question

For the purpose of this example lets say I have two ingredients:

1. water ($100\%$)
2. a chocolate mix ($80\%$ chocolate, $20\%$ water)

The goal is to mix them together to create a mix that is exactly $50\%$ water and $50\%$ chocolate.

How can I calculate exactly how much of the 2 ingredients to combine to create 500ml without waste?

Answer 1

If you take $s$ liters (or quarts, etc.) number 1, and $(1-s)$ liters of #2, you'll end up with $1$ liter of mix. That liter will contain $$ (1-s) 0.8 = 0.8 - 0.8s $$ liters of chocolate; you want that to be $0.5$ liters. So set \begin{align} 0.8 - 0.8s &= 0.5 \\ -0.8s &= -0.3 \\ s &= \frac{-0.3}{-0.8} = \frac{3}{8} \end{align} So: take 3/8 pure water, and 5/8 chocolate mix.

If you have, say, 7 liters of the 80% mix (what I've called "#2"), you'll need $\frac{3}{5} \times 7$ of the water to make your 50-50 mix.

In short: let's say the amount of chocolate mix that you have (measured in cups, gallons, liters, whatever) is $A$. You'll need to add $0.6 \times A$ of water to it. You'll end up with $1.6 \times A$ of mix.

For instance, if you have 9 cups of the 80% mix, you'll need to add $0.6 \times 9 = 5.4 cups$ of water to it to get to a 50% mix; when you're done, you'll have the original 9 cups of mix plus 5.4 cups of water to get 14.4 cups of mix, whic is $1.6 \times 9$.

(I got $s$ and $1-s$ screwed up in my earlier response, which I've now fixed up.)

Final Edit

Now that the problem's been clarified -- you want to get some amount $C$ of final 50-50 mix -- the solution is this:

quantity of pure water: (3/8) * C

quantity of 80% choc: (5/8) * C.

In your example where $C = 500$ml, you get

water: (3/8) * 500 = 187.5 ml

mix: (5/8) * 500 = 312.5 ml

Answer 2

Suppose, that you have $x$ liters of the mixture. Then you have $0,2x$ water. Now you add $y$ liters of pure water. The sum of these two liquids has to have the average water proportion. It is $0.5\cdot (x+y)$

Thus the equation is $0.2x+y=0.5x+0.5y$

Solve this equation for y. This gives you the amount of pure water, which you have to add-in relation to x.

Answer 3

There are two related problems of this sort that come up, and it is worth distinguishing them carefully. They involve solving the same equations but with different givens.

Problem 1:

You have $V_2$ liters of mixture 2, which has concentration $0.8$, and you want to dilute it with $V_1$ liters of mixture 1, which has concentration $0$, in such a way that the result has concentration $0.5$. So here $V_2$ is given, $V_1$ is unknown, and you want

$$\frac{0.8V_2}{V_1+V_2}=0.5.$$ Solving this you find $0.5V_1=0.3V_2$ so $V_1=0.6V_2$.

Problem 2:

You have as much of mixtures 1 and 2 as is required, and you want a total volume of $V$ which has concentration $0.5$. So here $V$ is given and $V_1,V_2$ are both unknown. Then you want to solve the two equations

$$V_1+V_2=V \\ \frac{0.8V_2}{V_1+V_2}=0.5.$$

I will solve this case in detail. Using the first equation, replace $V_1+V_2$ by $V$ in the second equation:

$$\frac{0.8V_2}{V}=0.5.$$

Multiply both sides by $V$:

$$0.8V_2=0.5V.$$

Divide both sides by $0.8$:

$$V_2=\frac{5}{8} V.$$

Now substitute into the first equation to get $V_1=V-\frac{5}{8} V = \frac{3}{8} V.$

One thing that simplifies the situation in practice is that with these types of equations, $V_1,V_2$ will always be directly proportional to $V$, if all the other numbers are held constant. That means you can solve the problem for $V=1$ and then multiply by $V$. This is done in John Hughes' answer.