The following question was asked in a competitive exam
Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. No tile should overshoot any edge of the floor.What is the maximum number of tiles that can be accommodated on the floor ?
Solution obtained by trial and error, showing an answer of $6$:
But is there a equation or inequality based approach (because I cannot rely on trial and error) like $$130\cdot{110} - 70\cdot{30\cdot{n}} < 70\cdot{30} $$ $$ n >5.8 $$ So, $n$ is the next integer, so $6$, so we might get the answer right, but is the approach right?
For example, if it is some other shape which is also of area $130\times 110 = 14300$ but not a rectangle, there is no guarantee that we can place inside it $6$ smaller tiles. So this approach is not applicable at all places.
So guide me in knowing the right approach. Please help. I am stuck.
The area of the entire floor is $110*130=14300$. The area of each smaller tile is $70*30=2100$. Therefore, the most possible tiles to fit would be 6, as $7*2100>14300$. While this can not be applied to every possible floor layout, it works in the given one.
Edit: When generalized, given $A*B$ as the dimensions of the floor, and $c*d$ as the dimensions of each tile, the maximum number of tiles would theoretically be $\lfloor AB/cd \rfloor$. (Again, this will not always give the maximum number in practice.)