Question

Question 195

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 1

Given, length of the floor = 8 m 96 cm

= 8 $\times$ 100 cm + 96 cm [ $\because$ 1 m = 100 cm]

= (800 +96) cm = 896 cm

and breadth of the floor = 6 m 72 cm

= 6 $\times$ 100 cm + 72 cm [ $\because$ 1 m = 100 cm]

= (600 + 72) cm = 672 cm

Area of the floor = Length $\times$ Breadth $=896\times 672c{m}^2$

Now, length of the square tile = HCF of 896 and 672

Prime factorization of 896 and 672

$\begin{array}{cc}2& 896\\ 2& 448\\ 2& 224\\ 2& 112\\ 2& 56\\ 2& 28\\ 2& 14\\ 7& 7\\ & 1\end{array}\begin{array}{cc}2& 672\\ 2& 336\\ 2& 168\\ 2& 84\\ 2& 42\\ 3& 21\\ 7& 7\\ & 1\end{array}$
$896=2\times 2\times 2\times 2\times 2\times 2\times 2\times 7$
$672=2\times 2\times 2\times 2\times 2\times 7\times 3$
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
$\therefore$ Minimum number of square tiles $=\frac{\text{Area of the floor}}{\text{Area of a square tile}}$
$=\frac{896\times 672}{224\times 224}=12$
Hence, the minimum number of square tiles of the same size that is required to cover the entire floor is 12.