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Square tiles of side 20 cm are to be lai...

Square tiles of side 20 cm are to be laid on the floor of a room 10 m by 4.5 m . How many tiles will be needed ? Find the cost of putting the tiles at Rs. 131 .40 per tile .

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To solve the problem step by step, we will follow these calculations: ### Step 1: Convert the dimensions of the room from meters to centimeters. - The dimensions of the room are given as 10 m by 4.5 m. - We know that 1 meter = 100 centimeters. - Therefore, we convert: - Length: \( 10 \, \text{m} = 10 \times 100 = 1000 \, \text{cm} \) - Breadth: \( 4.5 \, \text{m} = 4.5 \times 100 = 450 \, \text{cm} \) ### Step 2: Calculate the area of the room. - The area of a rectangle is calculated using the formula: \[ \text{Area} = \text{Length} \times \text{Breadth} \] - Substituting the values: \[ \text{Area of the room} = 1000 \, \text{cm} \times 450 \, \text{cm} = 450000 \, \text{cm}^2 \] ### Step 3: Calculate the area of one tile. - The side of the square tile is given as 20 cm. - The area of a square is calculated using the formula: \[ \text{Area} = \text{side}^2 \] - Substituting the value: \[ \text{Area of one tile} = 20 \, \text{cm} \times 20 \, \text{cm} = 400 \, \text{cm}^2 \] ### Step 4: Calculate the number of tiles needed. - The number of tiles required can be calculated by dividing the area of the room by the area of one tile: \[ \text{Number of tiles} = \frac{\text{Area of the room}}{\text{Area of one tile}} = \frac{450000 \, \text{cm}^2}{400 \, \text{cm}^2} \] - Performing the division: \[ \text{Number of tiles} = 1125 \] ### Step 5: Calculate the total cost of the tiles. - The cost of one tile is given as Rs. 131.40. - To find the total cost, we multiply the number of tiles by the cost per tile: \[ \text{Total cost} = \text{Number of tiles} \times \text{Cost per tile} = 1125 \times 131.40 \] - Performing the multiplication: \[ \text{Total cost} = 147825 \] ### Final Answer: - The total number of tiles needed is **1125**. - The total cost of putting the tiles is **Rs. 147825**. ---
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ICSE-MENSURATION-EXERCISE 23 C
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