Question 15 Marks
The floor of a room is $8\ m\ 96\ cm$ long and $6\ m\ 72\ cm$ broad. Find the minimum number of square tiles of the same size needed to cover the entire floor.
Answer
View full question & answer→Given, length of the floor $= 8\ m \ 96\ cm = 8 \times 100\ cm + 96 \ cm$
$[\therefore 1\text{m}=100\text{cm}]$
$= (800 + 96)cm = 896\ cm$ and
breadth of the floor $= 6\ m\ 72\ cm = 6 \times 100\ cm + 72\ cm $
$[\therefore 1\text{m}=100\text{cm}]$
$= (600 + 72)cm = 672\ cm$
Area of the floor $= ($Length $\times $ Breadth$) = 896 \times 672\ cm^2$
Now, length of the square tile $= HCF$ of $896$ and $672$ Prime factorization of $896$ and $672.$
The $HCF$ of $896.$
$\begin{array}{c|c}2&896\\ \hline2&448\\ \hline2&224\\ \hline2&112\\ \hline2&56\\ \hline2&28\\ \hline2&14\\ \hline7&7\\ \hline&1\end{array}$
Prime factorization of $896 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7$ The $HCF$ of $672.$
$\begin{array}{c|c}2&672\\ \hline2&336\\ \hline2&168\\ \hline2&84\\ \hline2&42\\ \hline3&21\\ \hline7&7\\ \hline&1\end{array}$
Prime factorization of $672 = 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 3$
Common factors of $896$ and $672 = 2 \times 2 \times 2 \times 2 \times 2 \times 7 = 224$
So, length of a square tile $= HCF$ of $896$ and $672 = 224\ cm$
Area of a square tile $= (224 \times 224)cm^2$
$\therefore$ Minimum number of square tiles $=\frac{\text{Area of the floor}}{\text{Area of a square tile}}$
$=\frac{896\times672}{224\times224}=12$
Hence, the minimum number of square tiles of the same size need to cover the entire floor is $12.$
$[\therefore 1\text{m}=100\text{cm}]$
$= (800 + 96)cm = 896\ cm$ and
breadth of the floor $= 6\ m\ 72\ cm = 6 \times 100\ cm + 72\ cm $
$[\therefore 1\text{m}=100\text{cm}]$
$= (600 + 72)cm = 672\ cm$
Area of the floor $= ($Length $\times $ Breadth$) = 896 \times 672\ cm^2$
Now, length of the square tile $= HCF$ of $896$ and $672$ Prime factorization of $896$ and $672.$
The $HCF$ of $896.$
$\begin{array}{c|c}2&896\\ \hline2&448\\ \hline2&224\\ \hline2&112\\ \hline2&56\\ \hline2&28\\ \hline2&14\\ \hline7&7\\ \hline&1\end{array}$
Prime factorization of $896 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7$ The $HCF$ of $672.$
$\begin{array}{c|c}2&672\\ \hline2&336\\ \hline2&168\\ \hline2&84\\ \hline2&42\\ \hline3&21\\ \hline7&7\\ \hline&1\end{array}$
Prime factorization of $672 = 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 3$
Common factors of $896$ and $672 = 2 \times 2 \times 2 \times 2 \times 2 \times 7 = 224$
So, length of a square tile $= HCF$ of $896$ and $672 = 224\ cm$
Area of a square tile $= (224 \times 224)cm^2$
$\therefore$ Minimum number of square tiles $=\frac{\text{Area of the floor}}{\text{Area of a square tile}}$
$=\frac{896\times672}{224\times224}=12$
Hence, the minimum number of square tiles of the same size need to cover the entire floor is $12.$