The ratio of the number of white balls in a bag to that of black balls is $1:2$. If $9$ grey balls are added, the ratio of the numbers of white, black and grey balls becomes $2:4:3$. How many black balls were originally in the bag?

Let the original number of white balls be $w$ and black balls be $b$.
Given original ratio:

$$\frac{w}{b}=\frac{1}{2}⟹b=2w$$

When $9$ grey balls are added, the new ratio of white $:$ black $:$ grey becomes $2:4:3$.
Let the common multiplier for the new ratio be $k$.
Then:

* Number of white balls = $2k$
* Number of black balls = $4k$
* Number of grey balls = $3k$

Since no white or black balls were added or removed, the counts remain unchanged:

$$w=2k$$

$$b=4k$$

We are given that the number of grey balls added is $9$, which corresponds to the grey share:

$$3k=9⟹k=3$$

Now, we can find the original number of black balls ($b$):

$$b=4k=4\times 3=12$$