Now we have three types of salt water, A, B, and C, with concentrations of $5\%$, $8\%$, and $9\%$ respectively, and weights of 60 grams, 60 grams, and 47 grams. We need to prepare 100 grams of salt water with a concentration of $7\%$. Try to find out the maximum and minimum amount of salt water A that can be used. Please write down the sum of these two numbers as the answer to this question.
Concentration problem.
Let the amount of salt water A be $x$ grams.
(1) To make the amount of salt water A as much as possible, then the amount of salt water C should be as much as possible; thus, 47 grams of salt water C is used, $x$ grams of salt water A is used, and $100-47-x=53-x$ grams of salt water B is used;
$$
x \times 5 \%+(53-x) \times 8 \%+47 \times 9 \%=100 \times 7 \% \text {; }
$$

Solving for $x$ gives $x=49$; that is, the maximum amount of salt water A that can be used is 49 grams.
(2) To make the amount of salt water A as little as possible, then the amount of salt water B should be as much as possible; thus, 60 grams of salt water B is used, $x$ grams of salt water A is used, and $100-60-x=40-x$ grams of salt water C is used;
$$
x \times 5 \%+60 \times 8 \%+(40-x) \times 9 \%=100 \times 7 \% \text {; }
$$

Solving for $x$ gives $x=35$; that is, the minimum amount of salt water A that can be used is 35 grams.
Therefore, the answer to this problem is $49+35=84$.