Each unit of $C$ contains $\frac15$ unit of $A$ and $\frac45$ unit of $B$. Each unit of $D$ contains $\frac47$ unit of $A$ and $\frac37$ unit of $B$. Thus, $c$ units of $C$ and $d$ units of $D$ contain $\frac{c}5+\frac{4d}7=\frac{7c+20d}{35}$ units of $A$ and $\frac{4c}5+\frac{3d}7=\frac{28c+15d}{35}$ units of $B$. (Sanity check: these two quantities do in fact sum to $c+d$.)
You want these two quantities to be in the proportion $5:6$, so you want
$\frac{7c+20d}{35}=\frac56\cdot\frac{28c+15d}{35}\;,$
or $6(7c+20d)=5(28c+15d)$. Simplifying yields $45d=98c$, or $\dfrac{c}d=\dfrac{45}{98}$: the correct proportion of $C$ to $D$ is $45:98$.
The calculations below resulted from a misreading of the problem; they answer a different question from the one actually asked, but I’ll leave them up, since someone may at some point find them useful.
From $1$ unit of $A$ and $4$ units of $B$ you get $5$ units of $C$. From $4$ units of $A$ and $3$ units of $B$ you get $7$ units of $D$. If you scale that second combination down by a factor of $\frac67$, you find that $\frac67\cdot4=\frac{24}7$ units of $A$ and $\frac67\cdot3=\frac{18}7$ units of $B$ will give you $\frac67\cdot7=6$ units of $D$. These data are summarized in first four rows of the table below.
$\begin{array}{cccc|l} A&B&C&D\\ \hline 1&4&5&-&*\\ 4&3&-&7\\ 24/7&18/7&-&6&*\\ \hline 31/7&46/7&5&6&** \end{array}$
Now add the starred rows to get the bottom row: $\frac{31}7$ units of $A$ and $\frac{46}7$ units of $B$ give you $5$ units of $C$ and $6$ of $D$. That’s $11$ units altogether, so for a single unit of a $5:6$ $CD$ mixture divide the quantities of $A$ and $B$ by $11$: you need $\frac{31}{11\cdot7}=\frac{31}{77}\approx0.4026$ units of $A$ and $\frac{46}{11\cdot7}=\frac{46}{77}\approx0.5974$ units of $B$.