There are three types of rectangular tiles : 3' x 3', 3' x 7' and 3' x 11'. An area of rectangular shape of dimensions 3' x 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

• The total area to be covered is 3' x 100' = 300 square feet.

• To find the maximum number of tiles (x), use the smallest tile area, which is 3' x 3' = 9 sq ft. Since 300 is divisible by 9, x = 300 / 9 = 33.33, but tiles must be whole numbers. Wait, the problem implies covering the 3x100 strip. The 3' dimension matches the width. So we are tiling a 1x100 strip with lengths 3, 7, and 11.

• Let the number of 3' tiles be $n_3$, 7' tiles be $n_7$, and 11' tiles be $n_{11}$. The equation is $3n_3 + 7n_7 + 11n_{11} = 100$.

• The total number of tiles is $N = n_3 + n_7 + n_{11}$.

• To maximize N (x), we should use as many small tiles (3') as possible. $100 / 3 = 33$ remainder 1. We cannot have a remainder of 1. We need to adjust. If we use 32 tiles of 3', length is 96, remaining 4 (impossible). If we use 31 tiles of 3', length is 93, remaining 7. This can be covered by one 7' tile. So, $n_3=31, n_7=1, n_{11}=0$. Total tiles $x = 31 + 1 = 32$.

• To minimize N (y), we should use as many large tiles (11') as possible. $100 / 11 = 9$ remainder 1. If we use 9 tiles of 11', length is 99, remaining 1 (impossible). If we use 8 tiles of 11', length is 88, remaining 12. 12 can be covered by one 3' and one... wait, 12 is not 7 or 11. 12 = 3+3+3+3 (4 tiles) or 3+... no. 12 cannot be formed by 7 and 11 alone? 12 = 3*4. So $n_{11}=8, n_3=4, n_7=0$. Total tiles $y = 8 + 4 = 12$.

• Let's check if a smaller y is possible. Try 7 tiles of 11' (77), remaining 23. 23 = 11 + 11 + 1 (no), 11 + 7 + 5 (no), 7+7+7+2 (no). 23 = 11 + 3*4 (11+12). So $n_{11}=7+1=8$? No, we started with 7. $7*11=77$. Rem 23. $23 = 11 + 12$. So add one 11' and four 3's. Total 11' tiles = 8, 3' tiles = 4. Same as before. Total 12.

• Try 6 tiles of 11' (66), remaining 34. 34 = 11*3 + 1 (no). 34 = 11*2 + 12 (11+11+3*4). Total 11' tiles = 6+2=8, 3' tiles=4. Total 12.

• Try 5 tiles of 11' (55), remaining 45. 45 = 11*4 + 1 (no). 45 = 11*3 + 12 (11*3 + 3*4). Total 11' tiles = 5+3=8, 3' tiles=4. Total 12.

• It seems the minimum is 12. Let's re-evaluate max. Max tiles: Use 3's. $3n_3 + 7n_7 + 11n_{11} = 100$. Maximize $n_3+n_7+n_{11}$. This is equivalent to minimizing the 'waste' or using smallest units. Since 3 is the smallest, we want max $n_3$. $100 = 3(33) + 1$. Not possible. $100 = 3(32) + 4$. Not possible. $100 = 3(31) + 7$. Possible ($31 imes 3 + 1 imes 7$). Total tiles = 32. Can we get 33? No, because 33*3=99, rem 1. 34*3=102 > 100. So max is 32.

• $x = 32, y = 12$. $x - y = 32 - 12 = 20$.