Question

Find the least number of square tiles required to pave the ceiling of a room $15\mathrm{m}17\mathrm{cm}$ long and $9\mathrm{m}2\mathrm{cm}$ broad.

Answer 1

Solution:

Step 1: Convert the units of length and breadth into centimeters:

$$\begin{gathered} \mathrm{l}=15\mathrm{m}17\mathrm{cm} \\ =1500\mathrm{cm}+17\mathrm{cm}\mathbf{[}\mathbf{\because }\mathbf{1}\mathbf{}\mathbf{m}\mathbf{=}\mathbf{100}\mathbf{}\mathbf{cm}\mathbf{]} \\ =1517\mathrm{cm} \end{gathered}$$

And,

$$\begin{gathered} \mathrm{b}=9\mathrm{m}2\mathrm{cm} \\ =900\mathrm{cm}+2\mathrm{cm} \\ =902\mathrm{cm} \end{gathered}$$

Step 2: Finding the maximum size of the square tile by finding the HCF:

Prime factors of $1517=37\times 41$

Prime factors of $902=2\times 11\times 41$

$\mathrm{HCF}(1517,902)=41$

Hence, the size of tile is $41cm$.

Step 3: Find the number of tiles (n):

$n=\frac{Areaofroom}{Areaoftile}$

$=\frac{1517\times 902}{41\times 41}\mathbf{}\left[\because Area=length\times breadth\right]$

$=37\times 22$

$=814$

Final answer: Hence, the least number of square tiles required to pave the ceiling of the room is $\mathbf{814}$.