To celebrate her birthday, Ana is going to prepare pear 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 fruits?
Let $m$ denote the number of apples and $p$ the number of pears that Ana buys, 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 $300 m + 200 p \leq 7000$, which is equivalent to $3 m + 2 p \leq 70$. As pears weigh less, Ana should take a larger number of pears and, therefore, a smaller number of apples. Since Ana wants to make pies of both fruits, she needs to take at least 1 apple.

If she takes one apple, we have $2 p \leq 70 - 3 = 67$, so $p \leq 33.5$, which means Ana can take up to 33 more pears, for a total of 34 fruits. If she takes two apples, we have $2 p \leq 70 - 6 = 64$, so $p \leq 32$, which means Ana can take up to 32 more pears, again for a total of 34 fruits. If she takes three apples, we have $2 p \leq 70 - 9 = 61$, so $p \leq 30.5$, which means Ana can only take up to 30 more pears, for a total of 33 fruits.

From the calculations, we see that for each additional apple Ana takes, she needs to take 1.5 fewer pears. Thus, if she takes more than two apples, she will never be able to take more than 30 pears, for a total always less than 34 fruits.

Conclusion: the maximum number of fruits Ana can take is 34 fruits (either one apple and 33 pears, or two apples and 32 pears).