Let the amount of water to be added be \( x \) metric tons.
Initially, the mass of water in the mixture is:
\[
\text{Mass of water} = 30 \times 0.10 = 3\ \text{metric tons}
\]
The final moisture content should be 50% of the total mass after adding \( x \) metric tons of water. Thus, the total mass of the mixture after adding water is \( 30 + x \) metric tons, and the mass of water becomes \( 3 + x \). The desired moisture content is 50%, so we set up the equation:
\[
\frac{3 + x}{30 + x} = 0.50
\]
Solving for \( x \):
\[
3 + x = 0.50 \times (30 + x)
\]
\[
3 + x = 15 + 0.50x
\]
\[
3 + 0.50x = 15
\]
\[
0.50x = 12 \quad \Rightarrow \quad x = 24
\]
Thus, the amount of water to be added is:
\[
\boxed{24}\ \text{metric tons}
\]