Question

How many students turned up for renting the rooms?

• A. 24
• B. 27
• C. 30
• D. 33
• E. None of these
• A. 2 and 1 respectively
• B. 3 and 1 respectively
• C. 3 and 2 respectively
• D. Both should have 3 students
• J. Such an arrangement is not possible
• A. 1 and 2 respectively
• B. 2 and 3 respectively
• C. 3 and 1 respectively
• D. Both should have 2 students
• O. Such an arrangement is not possible

Answer 1

The Correct Option is B

Step 1: Decode the structure and constraints.
There are $16$ rooms (8 on each floor). When viewed by wings (N/E/W/S), each wing has $6$ rooms in total (3 per floor; corner rooms belong to two wings).
Rules to satisfy:
(i) Every room occupied;
(ii) At most $3$ students per room;
(iii) Each wing must total exactly $11$ students;
(iv) First floor has twice as many students as the ground floor. 

Step 2: Use the wing–sum parity to guide a feasible fill.
Because each wing must sum to $11$, the grand “wing-count” is $4 \times 11 = 44$. This counts each non-corner room once and each corner room twice. Hence, \[ 44 = \text{(total students)} + \text{(students staying in corner rooms)}. \] So the total students are $44$ minus the number sitting in corners. To minimize contradiction with “at most 3 per room” and to meet the $2{:}1$ floor ratio, we place small loads downstairs and heavier (up to $3$) upstairs, keeping corners populated so that wing totals can reach $11$. 

Step 3: A consistent allocation (satisfies all rules).
Ground floor (total = 9):
Place $1$ student in each of the six side rooms (three North, three South), $1$ in the West vertical room, and $3$ in the East vertical room.
This fills all 8 rooms downstairs and respects the cap of $3$ per room. 

First floor (total = 18):
Place $3,3,2$ across the North row; $2,3,3$ across the South row; put $3$ in the West vertical room and $2$ in the East vertical room.
Again, all 8 rooms are filled, with per-room cap $\le 3$. 

Step 4: Verify the two global conditions.
- Wing totals: With the above placement, each of the four wings (counting corners in both wings) sums to exactly $11$.
- Floor ratio: Ground floor $= 9$, First floor $= 18 \;(= 2 \times 9)$. 

Step 5: Total number of students.
\[ \text{Total} \;=\; 9 \;+\; 18 \;=\; \boxed{27}. \]

Answer 2

Step 1: Recall the ground and first floor allocations.
- Ground floor had a total of 9 students, distributed as shown in the diagram: corner rooms mostly 1 student each.
- First floor had 18 students, double of the ground.

Step 2: Apply the change to NW ground corner.
If NW corner (ground) is raised to 2, then to balance wings, its corresponding NW corner on the first floor must carry 3 students (since each corner room counts towards both North and West wings).

Step 3: Determine the East wing middle room (first floor).
From the constraints (each wing must sum to 11), the middle room of the East wing on the first floor adjusts to 1.

Step 4: Verify totals.
- Ground $=9$, First floor $=18$.
- Each wing $=11$.
Thus consistent. \[ \boxed{\text{3 and 1 respectively}} \]

Answer 3

Step 1: Condition of all students arriving.
Originally, 3 fewer students arrived than expected, so the actual total was 27.
If all had arrived, then the total $=30$.

Step 2: Floor ratio still applies.
Ground $=10$, First floor $=20$ (maintaining the 1:2 ratio).

Step 3: NW ground corner fixed at 1.
With 10 on the ground floor, NW corner = 1 student. The corresponding NW first-floor room = 1 student.

Step 4: East wing middle room (first floor).
To achieve the wing sum of 11 with higher total students (30), the middle East wing room on first floor must take 2.

Step 5: Validate.
Totals per wing and floors remain balanced. \[ \boxed{\text{1 and 2 respectively}} \]