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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.

Guest Sep 28, 2014

Best Answer 

 #1
avatar+92781 
+5

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.                     

Melody  Sep 28, 2014
 #1
avatar+92781 
+5
Best Answer

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.                     

Melody  Sep 28, 2014

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