Solution:
For the least number of tiles, the size of the tiles must be the maximum.
Maximum size of the tile = HCF of 1517 cm and 902 cm = 41 cm.
Hence, required number of tiles
$$\eqalign{ {\text{ = }}\frac{{{\text{Area of ceiling}}}}{{{\text{Area of each tile}}}} \cr = \left( {\frac{{1517 \times 902}}{{41 \times 41}}} \right) \cr = 814 \cr} $$

Solution:
HCF of 132, 204 and 228
132 = 2 × 2 × 3 × 11
204 = 2 × 2 × 3 × 17
228 = 2 × 2 × 3 × 19
∴ HCF of 132, 204 and 228 is 2 × 2 × 3 i.e. 12
∴ Required HCF = 12

Solution:
75 litres, 45 litres
For maximum capacity take HCF (75, 45) = 15

Solution:
15(x + y) = 240
x + y = 16
1 + 15 (Right)
2 + 14 (Wrong)
3 + 13 (Right)
4 + 12 (Wrong)
5 + 11 (Right)
6 + 10 (Wrong)
7 + 9 (Right)
8 + 8
Total pairs = 4

Solution:
HCF = 8
Now LCM should have a factors 8
So check also the option we have only 60 which does not have a factor 8. So it will never be the LCM.

Solution:
Let the numbers be 2x and 3x
Then, their L.C.M. = 6x
So, 6x = 48 or x = 8
∴ The numbers are 16 and 24
Hence, required sum = (16 + 24) = 40

Solution:
$$\eqalign{ {\text{116690151}}\overline {){\text{427863887(}}} {\text{3}} \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{350070453}} \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {{\text{77793434)116690151(}}} 1 \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{77793434}} \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {{\text{38896717)77793434(}}} {\text{2}} \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{77793434}}{ \times } \cr {\text{So, H}}{\text{.C}}{\text{.F of 116690151 and 427863887}} \cr {\text{ = 38896717}} \cr \therefore \frac{{116690151}}{{427863887}} = \frac{{116690151 \div 38896717}}{{427863887 \div 38896717}} \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{11}} \cr} $$