At a national math contest, students are being housed in single rooms and double rooms; it is known that $75\%$ of the students are housed in double rooms. What percentage of the rooms occupied are double rooms?

[i]Proposed by Evan Chen[/i]
1. Let the total number of students be \( S \). According to the problem, \( 75\% \) of the students are housed in double rooms. Therefore, the number of students in double rooms is:
   \[
   0.75S
   \]
2. Since each double room houses 2 students, the number of double rooms \( D \) is:
   \[
   D = \frac{0.75S}{2} = 0.375S
   \]
3. The remaining \( 25\% \) of the students are housed in single rooms. Therefore, the number of students in single rooms is:
   \[
   0.25S
   \]
4. Since each single room houses 1 student, the number of single rooms \( S_r \) is:
   \[
   S_r = 0.25S
   \]
5. The total number of rooms \( R \) is the sum of the number of double rooms and single rooms:
   \[
   R = D + S_r = 0.375S + 0.25S = 0.625S
   \]
6. The percentage of rooms that are double rooms is given by:
   \[
   \frac{D}{R} \times 100 = \frac{0.375S}{0.625S} \times 100 = \frac{0.375}{0.625} \times 100 = 0.6 \times 100 = 60\%
   \]

The final answer is \( \boxed{ 60\% } \)