• # question_answer 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.

Given, length of the floor = 8 m 96 cm $=8\times 100\text{ }cm+96\text{ }cm$ [$\therefore$$1\,m=100\text{ }cm$] $=800+96\text{ }cm$ $=896\text{ }cm$ and breadth of the floor $=6m\text{ }72cm$ $=6\times 100\text{ }cm+72\text{ }cm$ [$\therefore$  $1\,m=100\text{ }cm$] $=672\text{ }cm$ Now, size of the square tile = HCF of 896 and 672 Prime factorization of 896 and 672
$896=2\times 2\times 2\times 2\times 2\times 2\times 2\times 7$ $672=2\times 2\times 2\times 2\times 2\times 3\times 7$ Common factors of 896 and 672 $=2\times 2\times 2\times 2\times 2\times 7$ $=224$ $\therefore$ Minimum no. of square tiles                  $\text{=}\frac{\text{Area}\,\,\text{of}\,\,\text{floor}}{\text{Area}\,\,\text{of}\,\,\text{square}\,\,\text{tile}}$                  $=\frac{896\times 672}{224\times 224}$ [$\therefore$  area of square$={{(side)}^{2}}$]                  $=\frac{896\times 3}{224}$                  $=4\times 3=12$