4. (7 points) Master Li produced a batch of parts one day, and he divided them into two piles, A and B. If 15 parts are moved from pile A to pile B, then the number of parts in both piles will be equal; if 15 parts are moved from pile B to pile A, then the number of parts in pile A will be three times the number of parts in pile B. Therefore, pile A originally had $\qquad$ parts, and Master Li produced a total of $\qquad$ parts that day. $\qquad$
【Solution】Solution: The number of parts in pile B later:
$$
\begin{aligned}
& (15 \times 2+30) \div(3-1) \\
= & 60 \div 2 \\
= & 30 \text { (pieces) }
\end{aligned}
$$

The number of parts in pile A later:
$$
30 \times 3=90 \text { (pieces) }
$$

The original number of parts in pile A:
$$
90-15=75 \text { (pieces) }
$$

The total number of parts produced by Master Li on this day:
$$
30+90=120 \text { (pieces) }
$$

Answer: Pile A originally had 75 parts, and Master Li produced a total of 120 parts on this day. Therefore, the answers are: $75, 120$.