Have we lost our marbles yet ?

Let the numbers of red and black marbles be $4n$ and $7n$ respectively, so they are in the ratio $4:7$ .

The probability of picking two black marbles is $\underbrace{\frac {7n}{11n}}_{\text{first marble}} \cdot \underbrace{\frac {7n-1}{11n-1}}_{\text{second marble}} = \frac {49n-7}{121n-11}$ .

Setting this equal to $\frac {35}{88}$ gives

$\begin{aligned} \frac {49n-7}{121n-11} &= \frac {35}{88} \\ 88(49n-7) &= 35(121n-11) \\ 4312n-616 &= 4235n-385 \\ 77n-231 &= 0 \\ 77n &= 231 \\ n &= 3 \end{aligned}$

Since we defined the numbers of red and black marbles as $4n$ and $7n$ the total number of marbles will be $11n = \boxed{33}$