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.

Answer:

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

2	896
2	448
2	224
2	112
2	56
2	28
2	14
7	7
	1

2	672
2	336
2	168
2	84
2	42
3	21
7	7
	1

\[896=2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\] \[672=2\times 2\times 2\times 2\times 2\times 3\times 7\] 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\[={{(side)}^{2}}\]]                  \[=\frac{896\times 3}{224}\]                  \[=4\times 3=12\]