A cleaner way to solve this is to work on a per-100-kg-of-dry-leaves basis instead of per kg of wet leaves, which avoids carrying decimals through every step.
Take 100 kg of dry leaf solids as the basis. Nitrogen in this = $0.7\% \times 100 = 0.7$ kg, so carbon in it = $C/N \times N = 50 \times 0.7 = 35$ kg.
Let $y$ = kg of dry sludge solids added to this 100 kg of dry leaves. Nitrogen from this sludge = $0.056y$ kg, and carbon from it = $6.3 \times 0.056y = 0.3528y$ kg.
Setting the combined C/N ratio to 25:
\[ \frac{35 + 0.3528y}{0.7 + 0.056y} = 25 \]
\[ 35 + 0.3528y = 17.5 + 1.4y \]
\[ 17.5 = 1.0472y \implies y = 16.711 \text{ kg dry sludge} \]
Now convert both basis quantities back to wet weights using the moisture contents. Since dry leaves solids = 50% of wet leaves, the 100 kg dry leaves corresponds to $100/0.50 = 200$ kg of wet leaves. Since dry sludge solids = 25% of wet sludge, $y = 16.711$ kg dry sludge corresponds to $16.711/0.25 = 66.844$ kg of wet sludge.
The required ratio of wet sludge to wet leaves is therefore:
\[ x = \frac{66.844}{200} = 0.334 \]
This matches the earlier per-kg-basis calculation exactly, confirming the result.
\[\boxed{x = 0.334 \text{ kg sludge per kg of leaves}}\]