Percent of a percent

Question

I'm trying to calculate the correct percentages of both red and blue paint from a starting mix... It's difficult for me to explain, but the example should be easy enough...

Let's say I have the following recipe:

Starting Mix: 22.20% (already mixed at 90% Red, 10% Blue)

Added Red: ??%
Added Blue: ??%

Extras: 9% (These stay as they are, without any Red or Blue)

The starting mix has 90% Red in it, 10% Blue in it, and at the end I want 60% Red and 40% Blue in the whole mix (excluding the "Extras"), so I want 60% Red in the total of 91%, and 40% Blue in the total of 91%.

How do I calculate the Added Blue and Added Red percentages for this? Is there a simple formula I can follow, as the starting mix may change, for example 100% Red, and the output may be different (for example 20% Red, 80% Blue).

Thanks in advance!

EDIT This is not a homework question, it's actually to help me work out e-liquid PG/VG splits when making my own vape juices, but I simplified the question to "paint" to try to make it easier to understand.

Answer 1

First we have:

0.9A + m = 0.6 * 0.91T
0.1A + n = 0.4 * 0.91T

where A is the initial amount, m is how much red to add, n is how much blue to add and T is the final total amount. If you specify the final total you can then solve them for m and n. For the first equation:

m = 0.6 * 0.91T - 0.9A

Answer 2

This is the balance for a volume $V_1$ of 90-10 mix and a volume $V_2$ of 60-40 mix.

$$ \frac{V_1}{9+1} \left[ \begin{matrix} 9 \\ 1 \end{matrix} \right] + V_r \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right] + V_b \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] = \frac{V_2}{6 + 4} \left[ \begin{matrix} 6 \\ 4 \end{matrix} \right] \\ $$

Here we can solve for the parts $V_r$ of red paint and $V_b$ of blue paint.

$$ \begin{align} \left[ \begin{matrix} V_r \\ V_b \end{matrix} \right] &= \frac{V_2}{6 + 4} \left[ \begin{matrix} 6 \\ 4 \end{matrix} \right] - \frac{V_1}{9+1} \left[ \begin{matrix} 9 \\ 1 \end{matrix} \right] \\ &= \left[ \begin{matrix} \frac{6}{10} V_2 - \frac{9}{10} V_1 \\ \frac{4}{10} V_2 - \frac{1}{10} V_1 \end{matrix} \right] \end{align} $$

Relative to $V_2$, we have $$ \begin{align} p_r &= \frac{V_r}{V_2} \cdot 100 \, \% = 60 \% - 90 \% \cdot \frac{V_1}{V_2} \\ p_b &= \frac{V_b}{V_2} \cdot 100 \, \% = 40 \% - 10 \% \cdot \frac{V_1}{V_2} \end{align} $$

Note:

$$ V_r + V_b = V_2 - V_1 $$ like it should be. And to get positive parts the condition $$ \frac{V_1}{V_2} \le \frac{2}{3} = 0.\bar{6} $$ must hold.

Example:

$V_1 = 100$, $V_2 = 200$ of whatever volume units, then we need to add $V_r = 30$ and $V_b = 70$ volume units.

The percentages are: $p_r = 15 \%$ and $p_b = 35 \%$ of $V_2 = 200$ volume units. ($V_1 / V_2 = 1 / 2$)

Now we want $V_3$ volume units of the 60-40-mix with extras $x$, by adding $V_x$ to $V_2$. We have $$ V_3 = V_2 + V_x = V_2 + 9\% \cdot V_3 $$

this gives $$ V_3 = \frac{100}{91} \cdot V_2 $$

In the example you end up with $V_3 = 219.78$ volume units of 60-40 mix with extras.

Example 2:

If we want to end up with $V_3 = 150$ volume units of the 60-40 mix with extras, then we need $V_2 = 91\% \cdot V_3 = 136.5$ volume units of the 60-40 mix and $V_x = 9\% \cdot V_3 = 13.5$ volume units of extras.

If we start with $V_1 = 50$ volume units of 90-10 mix, then we need $V_r = 36.9$ volume units of red paint and $V_b = 49.6$ volume units of blue paint.

The percentages (relative to $V_2$) are $p_r = 27 \%$ and $p_b = 36 \%$. ($V_1 / V_2 = 0.3663$)