Mrs. Siroka was expecting guests in the evening. First, she prepared 25 sandwiches for them. Then she calculated that each guest could take two, but three would not be enough for everyone. She thought that if she made another 10 sandwiches, each guest could take three, but not four. This still seemed too little to her. In the end, she prepared a total of 52 sandwiches. Each guest could then take four sandwiches, but five would not be enough for everyone. How many guests was Mrs. Siroka expecting? She herself is on a diet and never eats in the evening.
First, let's work with the part of the problem where 25 sandwiches are considered. According to this, Mrs. Siroka expected no more than 12 guests, because $25: 2=12$, remainder 1, which means that 12 people could take two sandwiches each, but only one would be left. Here we also find out that Mrs. Siroka expected more than 8 guests, because $25: 3=8$, remainder 1, which means that with 8 guests, everyone could take three sandwiches. So far, it is plausible that she expected 9, 10, 11, or 12 guests.

Now let's consider only the part of the problem that talks about 35 sandwiches. We determine that Mrs. Siroka expected at most 11 guests, since $35: 3=11$, remainder 2, and more than 8 guests, since $35: 4=8$, remainder 3. Therefore, Mrs. Siroka could have expected 9, 10, or 11 guests.

Next, let's work only with the consideration of 52 sandwiches. According to this, Mrs. Siroka expected at most 13 guests, because $52: 4=13$, and more than 10 guests, because $52: 5=10$, remainder 2. She therefore expected 11, 12, or 13 guests.

We see that the only number of guests that fits all the data in the problem is 11.

Another solution. Just as in the first paragraph of the previous solution, we determine that Mrs. Siroka could have expected 9, 10, 11, or 12 guests. For each number, we will check whether it corresponds to the other data in the problem.

9 guests: With 35 sandwiches, everyone could take three sandwiches, not four, because $9 \cdot 3 < 35$. With 52 sandwiches, everyone could take four sandwiches, but even five, because $9 \cdot 4 < 52$. With 52 sandwiches, everyone could take four sandwiches, but even five, because $10 \cdot 4 < 52$. With 52 sandwiches, everyone could take four sandwiches, but not five, because $11 \cdot 4 \leq 52$. This number of guests fits the entire problem.

12 guests: With 35 sandwiches, everyone could not take three sandwiches, because $12 \cdot 3 > 35$. This number of guests is rejected.

Mrs. Siroka expected 11 guests.