4. A rope, the first time cut off $\frac{1}{3}$ of its total length, the second time cut off $30\%$ of the remaining part. If the parts cut off twice are 0.4 meters more than the remaining part, then the original length of the rope is $\qquad$ meters.
【Solution】Solution: The second cut represents the following fraction of the total length:
$$
\begin{array}{l}
\left(1-\frac{1}{3}\right) \times 30 \% \\
= \frac{2}{3} \times \frac{3}{10} \\
= \frac{1}{5}, \\
0.4 \div\left[\frac{1}{3}+\frac{1}{5}-\left(1-\frac{1}{3}-\frac{1}{5}\right)\right] \\
= 0.4 \div\left[\frac{8}{15}-\frac{7}{15}\right] \\
= 0.4 \div \frac{1}{15} \\
= 0.4 \times 15 \\
= 6 \text { (meters); }
\end{array}
$$

Answer: The original length of the rope is 6 meters.
Therefore, the answer is: 6.