To pave the floor of a ship of 12.3 m long by 9 m wide, square tiles have been used, they have been fair, without cutting any. As you will have the side of each tile, knowing they have the largest used that there was in stock.
+118608 That is a very good question. 
then you do this you should draw the rectangular floor - It will make it easier to follow.
Let the tiles be x metres by x metres in size
9=c*x and 12.3=k*x
9/c=x and 12.3/k=x
where c and k are whole numbers (cardinal numbers or positive integers if you like)
so
$$\frac{9}{c}=\frac{12.3}{k}$$
$$\\\frac{9}{12.3}=\frac{c}{k}\\\\
\frac{c}{k}=\frac{9}{12.3}\\\\
\frac{c}{k}=\frac{90}{123}\\\\
\frac{c}{k}=\frac{30}{41}\\\\$$
this fraction cannot be simplified further.
So 30 tiles fit on the short side and 41 tiles fit along the longer side.
9/30=0.3 mtres 12.3/41=0.3metres
The tiles will be 30cm by 30cm.
+118608 That is a very good question.
then you do this you should draw the rectangular floor - It will make it easier to follow.
Let the tiles be x metres by x metres in size
9=c*x and 12.3=k*x
9/c=x and 12.3/k=x
where c and k are whole numbers (cardinal numbers or positive integers if you like)
so
$$\frac{9}{c}=\frac{12.3}{k}$$
$$\\\frac{9}{12.3}=\frac{c}{k}\\\\
\frac{c}{k}=\frac{9}{12.3}\\\\
\frac{c}{k}=\frac{90}{123}\\\\
\frac{c}{k}=\frac{30}{41}\\\\$$
this fraction cannot be simplified further.
So 30 tiles fit on the short side and 41 tiles fit along the longer side.
9/30=0.3 mtres 12.3/41=0.3metres
The tiles will be 30cm by 30cm.
