II. (Full marks 15 points) Now we have three types of salt water, A, B, and C, with concentrations of $5\%, 8\%$, and $9\%$ respectively. They are 60 grams, 60 grams, and 47 grams each. We need to prepare 100 grams of salt water with a concentration of $7\%$. How many grams of type A salt water can be used at most? How many grams can be used at least?
Let's take $x$ grams, $y$ grams, and $z$ grams of salt water from solutions A, B, and C respectively, to mix a 100-gram salt water solution with a concentration of $7\%$. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=100 \\
5 x+8 y+9 z=700 .
\end{array}\right.
$$

where, $0 \leqslant x \leqslant 60$,
$$
\begin{array}{l}
0 \leqslant y \leqslant 60, \\
0 \leqslant z \leqslant 47
\end{array}
$$

From (1) and (2), we get
$$
y=200+4 x-3 x-100 .
$$
5 is, from (1) we have $0 \leqslant 200-4 x<60$.
Solving this, we get $35<x \leqslant 50$.
From (5) we have $0<3 x-100 \leqslant 47$.
Solving this, we get $33-\frac{1}{3}<x \leqslant 49$.
(i)
Combining (3), (7), and (7), we get $35 \leqslant x \leqslant 49$.
Fact 1: When 35 grams of solution A is taken, 60 grams of solution B, and 5 grams of solution C, it can satisfy the requirements of equations (1) and (2).

When 49 grams of solution A is taken, 4 grams of solution B, and 47 grams of solution C, it can also satisfy the requirements of equations (1) and (2).
Therefore: The maximum amount of solution A that can be used is 49 grams, and the minimum amount is 35 grams.