There are three types of rectangular tiles : 3' × 3', 3' × 7' and 3' × 11'. An area of rectangular shape of dimensions 3' × 100' is to be covered using these tiles without breaking them. If x and y are the maximum and minimum numbers of tiles of various sizes, respectively, that can be used to cover the area exactly, then x - y is

Since the width of the area and all tile types is a uniform 3', this problem simplifies to a one-dimensional linear Diophantine equation. We need to find non-negative integers a, b, and c representing the number of 3', 7', and 11' length tiles such that their sum equals the 100' length:

3a + 7b + 11c = 100

Finding the Maximum (x): To maximize the total number of tiles, we must use the smallest tile (3') as much as possible.

• Max possible a is 100 / 3 = 33 (remainder 1, which cannot be filled by 7' or 11' tiles).
• Try a = 32: sum is 96, remainder is 4 (cannot be filled).
• Try a = 31: sum is 93, remainder is 7. This is exactly covered by one 7' tile (b = 1, c = 0).
• Therefore, the maximum number of tiles is x = 31 + 1 + 0 = 32 tiles.

Finding the Minimum (y): To minimize the total number of tiles, we maximize the use of the largest tile (11').

• Max possible c is 100 / 11 = 9 (remainder 1, cannot be filled).
• Try c = 8: sum is 88, remainder is 12. This is exactly covered by four 3' tiles (a = 4, b = 0).
• Therefore, the minimum number of tiles is y = 4 + 0 + 8 = 12 tiles.

Finally, the requested difference is x - y = 32 - 12 = 20.

Why the other options are incorrect:

• Option B (12): This is incorrect because 12 is the minimum number of tiles (y) itself, rather than the difference x - y.
• Option C (10): This is incorrect and would only result from arithmetic errors, such as mistakenly finding a sub-optimal maximum of 22 tiles (22 - 12 = 10).
• Option D (7): This is incorrect; it is a distractor likely stemming from falsely assuming a different combination constraint or incorrectly subtracting the tile lengths rather than tile counts.

Takeaway: In optimization problems involving lengths or denominations (like the Coin Change problem), apply a Greedy Approach. Maximize the count by favoring the smallest unit, and minimize the count by favoring the largest unit, stepping back iteratively until the remainder is perfectly satisfied.