UPSC CSE Prelims 2026
The correct option is (a) - 20.
[as per provisional answerkey]The total area to be covered is a rectangle of 3' x 100'. Since all available tiles have a width of 3', we only need to focus on the lengths to cover the total length of 100'.
Let the lengths of the tiles be L1 = 3', L2 = 7', and L3 = 11'.
We need to find combinations of these lengths that sum exactly to 100.
1. Finding x (Maximum number of tiles):
To maximize the number of tiles, we must use the smallest possible tile (3') as much as possible.
If we use only 3' tiles: tiles with a remainder of 1'. This is not possible as we cannot break tiles.
We need to find a combination: , where is maximized.
If we use one 7' tile: .
93 is perfectly divisible by 3 ().
Total tiles (x) = 31 (of 3') + 1 (of 7') = 32 tiles.
(Note: Using 11' tiles would decrease the total count because ).
So, .
2. Finding y (Minimum number of tiles):
To minimize the number of tiles, we must use the largest possible tile (11') as much as possible.
If we use nine 11' tiles: . Remainder is 1' (Not possible).
If we use eight 11' tiles: . Remainder is 12'.
12' can be covered by four 3' tiles ().
Total tiles = 8 (of 11') + 4 (of 3') = 12 tiles.
Can we do better? If we use seven 11' tiles: . Remainder is 23'.
23 can be formed by .
Total tiles = 7 (of 11') + 2 (of 7') + 3 (of 3') = 12 tiles.
If we use nine 11' tiles and try to replace one with smaller tiles to fit the remainder, the count only increases.
So, .
3. Calculating x - y:
.
This problem is based on optimization within linear Diophantine equations, where the objective is to find the maximum and minimum integer solutions for the sum of lengths.