The four walls and the ceiling of a rectangular room have to be painted. The room has length 25 m, breadth 12 m and height 10 m. Painter A can paint 200 m^2 in 5 days, while painter B can paint 250 m^2 in 2 days. If painters A and B work together and continue working at their constant individual rates, in how many days will they finish painting the four walls and the ceiling of the room completely?

Introduction / Context:
This question combines the concept of surface area of a room with the idea of combined rates of work. We first need to calculate the total area that must be painted, and then determine how long two painters working together will take to complete that area, based on their individual speeds. Such questions are common in quantitative aptitude tests where geometry and time-and-work concepts are blended.

Given Data / Assumptions:

• Length of the room = 25 m.
• Breadth of the room = 12 m.
• Height of the room = 10 m.
• Four walls and the ceiling are to be painted, the floor is not painted.
• Painter A paints 200 m^2 in 5 days.
• Painter B paints 250 m^2 in 2 days.
• Both painters work together at constant rates until the entire job is finished.

Concept / Approach:
The solution has two main parts. First, we compute the total area to be painted, which is the sum of the areas of the four walls and the ceiling. For a rectangular room, the total area of four walls is 2 * height * (length + breadth), and the area of the ceiling is length * breadth. Second, we convert the work done by each painter into a daily rate of painting in square metres per day and then add these rates to get their combined rate. Finally, total time equals total area divided by the combined daily rate.

Step-by-Step Solution:
Area of four walls = 2 * height * (length + breadth)= 2 * 10 * (25 + 12)= 20 * 37 = 740 m^2.Area of ceiling = length * breadth = 25 * 12 = 300 m^2.Total area to be painted = 740 + 300 = 1040 m^2.Painter A paints 200 m^2 in 5 days, so A's rate = 200 / 5 = 40 m^2 per day.Painter B paints 250 m^2 in 2 days, so B's rate = 250 / 2 = 125 m^2 per day.Combined rate of A and B = 40 + 125 = 165 m^2 per day.Time taken together = total area / combined rate = 1040 / 165 days.Compute 1040 / 165 = 6 + 50/165 = 6 + 10/33.Therefore, time taken = 6 10/33 days.

Verification / Alternative check:
We can check the reasonableness of the answer by estimating. If the combined rate is 165 m^2 per day, in 6 days they would paint 6 * 165 = 990 m^2. The remaining area is 1040 - 990 = 50 m^2. At 165 m^2 per day, the time for the remaining 50 m^2 is 50 / 165 = 10/33 days. Adding 6 and 10/33 gives 6 10/33 days, which matches the exact calculation and confirms that the result is consistent.

Why Other Options Are Wrong:
6 days: At 165 m^2 per day, 6 days cover only 990 m^2, which is less than 1040 m^2, so the painting would not be complete.
7 10/33 days: This is more than required; it would imply extra time after the work is already finished, which is not correct for the minimum completion time.
8 days: Clearly larger than 6 10/33 days, so this would overestimate the actual required time.
5 5/11 days: This is less than 6 days, and the total work finished in such a short time at 165 m^2 per day would be less than 1040 m^2, so it cannot be correct.

Common Pitfalls:
Candidates often forget to include the ceiling area and calculate only the wall area. Another common mistake is miscomputing the daily painting rates, especially for painter B, or adding times instead of rates. Mixing up perimeter formulas or misusing area of walls as 2 * (length + breadth) without multiplying by height is another typical error. Careful attention to the formula for four walls and the distinction between rate and time helps avoid these issues.

Final Answer:
The correct time required for painters A and B working together to paint the four walls and ceiling of the room is 6 10/33 days.