IAS 2026 — CSAT Maths

The correct option is (a) - 20.

Solution

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: $100 / 3 = 33$100/3=33 tiles with a remainder of 1'. This is not possible as we cannot break tiles.
We need to find a combination: $3a + 7b + 11c = 100$3a+7b+11c=100, where $a + b + c$a+b+c is maximized.
If we use one 7' tile: $100 - 7 = 93$100−7=93.
93 is perfectly divisible by 3 ($93 / 3 = 31$93/3=31).
Total tiles (x) = 31 (of 3') + 1 (of 7') = 32 tiles.
(Note: Using 11' tiles would decrease the total count because $11 > 3+3+3$11>3+3+3).
So, $x = 32$x=32.

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: $11 \times 9 = 99$11×9=99. Remainder is 1' (Not possible).
If we use eight 11' tiles: $11 \times 8 = 88$11×8=88. Remainder is 12'.
12' can be covered by four 3' tiles ($3 \times 4 = 12$3×4=12).
Total tiles = 8 (of 11') + 4 (of 3') = 12 tiles.
Can we do better? If we use seven 11' tiles: $11 \times 7 = 77$11×7=77. Remainder is 23'.
23 can be formed by $(2 \times 7) + (3 \times 3) = 14 + 9 = 23$(2×7)+(3×3)=14+9=23.
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, $y = 12$y=12.

3. Calculating x - y:
$x - y = 32 - 12 = 20$x−y=32−12=20.

Why the other options are incorrect

• Option (b) - 12: This value represents the minimum number of tiles (y) required, not the difference between the maximum and minimum.
• Option (c) - 10: This is a result of a calculation error, likely by incorrectly identifying x as 22 or y as 22, or failing to find the most efficient combination for the 100' length.
• Option (d) - 7: This does not correspond to any logical derivation of the tile counts or their difference; it may be a distractor based on one of the tile dimensions.

Key Concept

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.