Let's solve the problem step by step:

1. Define variables:
- Let $R$ be the number of red balls originally in the basket.
- Let $B$ be the number of blue balls originally in the basket.

2. Initial proportions:
- 40% of the balls are red, so $R=0.4(R+B)$.
- 60% of the balls are blue, so $B=0.6(R+B)$.

3. Simplify:
- Substitute $R$ into the equation: $R=0.4(R+B)$ implies $R=0.4R+0.4B$.
- Rearrange to find $R$: $0.6R=0.4B$ or $R=\frac{2}{3}B$.

4. Add two red balls:
- New number of red balls = $R+2$.
- New total number of balls = $R+B+2$.

5. New proportions:
- The new proportion of red balls is 50%. Thus:
$$\frac{R+2}{R+B+2}=0.5$$

6. Solve for $R$ and $B$:
$$R+2=0.5(R+B+2)$$
$$R+2=0.5R+0.5B+1$$
$$R+1=0.5R+0.5B$$
$$0.5R=0.5B-1$$
$$R=B-2$$

7. Substitute $R=B-2$ into $R=\frac{2}{3}B$:
$$B-2=\frac{2}{3}B$$
$$B-\frac{2}{3}B=2$$
$$\frac{1}{3}B=2$$
$$B=6$$

8. Find $R$:
$$R=B-2=6-2=4$$

9. Total number of balls now:
$$\text{New total number of balls}=R+B+2=4+6+2=12$$

Thus, the number of balls in the basket now is:

C. 12