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