To celebrate her birthday, Ana is going to prepare pear pies and apple pies. At the market, an apple weighs $300 \mathrm{~g}$ and a pear $200 \mathrm{~g}$. Ana's bag can hold a maximum weight of $7 \mathrm{~kg}$. What is the maximum number of fruits she can buy to be able to make pies of both types?
Let $m$ denote the number of apples and $p$ the number of pears that Ana bought, so the weight she carries in the bag is $300 m+200 p$ grams. Since the bag can hold a maximum of 7000 grams, we have

$$
300 m+200 p \leq 7000, \text { which is equivalent to } 3 m+2 p \leq 70
$$

Since pears weigh less, Ana has to carry the maximum number of pears, and therefore, the minimum number of apples. Thus, if she carries 1 apple, we have:

$$
2 p \leq 70-3=67 \Longrightarrow p \leq 33.5
$$

Therefore, by carrying 1 apple, she can carry 33 pears. So, the maximum number of fruits is 34.

In the table below, we see that Ana can also carry 2 apples and 32 pears.

| $p$ | $m$ | $300 m+200 p$ | $p+m$ |
| :---: | :---: | :---: | :---: |
| 34 | 0 | 6800 | 34 |
| 33 | 1 | 6900 | 34 |
| 32 | 2 | 7000 | 34 |
| 31 | 2 | 6800 | 33 |