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Size of a tile is 9 inches by 9 inches. The number of tiles required to cover the floor of 12 ft by 18 ft is
[a] 384
[b] 32
[c] 24
[d] 216

Answer
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635.4k+ views
Hint: Use the fact that the area of rectangle = lb where l is the length of the rectangle and b is the breadth of the rectangle.
The number of tiles required will be the ratio of the area of the floor to the area of the tile.

Complete step-by-step answer:
We have the length of the tile = 9 inches.
The breadth of the tile = 9 inches.
Hence the area of the tile $=9\times 9$ = 81 square inches.
Now the length of the floor = 12 ft.
We know that 1 ft = 12 inches.
Hence the length of the floor = $12\times 12$ =144 inches.
Similarly, the breadth of the floor $=18\times 12$ =216 inches.
Hence the area of the floor $=216\times 144$ square inches.
Now we have 81 square inches and require one tile.
Hence 1 square inch requires $\dfrac{1}{81}$ tiles
Hence $216\times 144$ square inch requires = $\dfrac{216\times 144}{81}=384$ tiles.
Hence the number of tiles required to cover the floor = 384.
Hence option [a] is correct.

Note: We have assumed that in the whole process of covering the floor, no part of a tile is wasted. If some part of the tile gets wasted, then the required number of tiles is more than 384. Hence 384 is the minimum number of tiles required.
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