How many bricks, each of size 25 cm × 13.5 cm × 6 cm, will be required to build a wall 5.4 m long, 5.4 m high and 33 cm thick?

Question

How many bricks, each of size 25 cm × 13.5 cm × 6 cm, will be required to build a wall 5.4 m long, 5.4 m high and 33 cm thick?

Asked on March 26, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-26T06:53:01.703Z

Step 1: Convert all dimensions to centimeters. Length ($l$) = $1.05 \text{ m} = 1.05 \times 100 \text{ cm} = 105 \text{ cm}$ Breadth ($b$) = $70 \text{ cm}$ Height ($h$) = $1.5 \text{ cm}$ Step 2: Calculate the volume of the rectangular piece of iron. The formula for the volume of a cuboid is $V = l \times b \times h$. $$V = 105 \text{ cm} \times 70 \text{ cm} \times 1.5 \text{ cm}$$ $$V = 7350 \text{ cm}^2 \times 1.5 \text{ cm}$$ $$V = 11025 \text{ cm}^3$$ Step 3: Calculate the total weight of the iron piece in grams. Given that $1 \text{ cm}^3$ of iron weighs 8 grams. $$\text{Weight in grams} = \text{Volume} \times \text{Weight per unit volume}$$ $$\text{Weight in grams} = 11025 \text{ cm}^3 \times 8 \frac{\text{g}}{\text{cm}^3}$$ $$\text{Weight in grams} = 88200 \text{ g}$$ Step 4: Convert the total weight from grams to kilograms. We know that $1 \text{ kg} = 1000 \text{ g}$. $$\text{Weight in kilograms} = \frac{88200 \text{ g}}{1000 \frac{\text{g}}{\text{kg}}}$$ $$\text{Weight in kilograms} = 88.2 \text{ kg}$$ The weight of the piece of iron is $\boxed{\text{88.2 kg}}$.

Accepted Answer · 2026-03-26T06:53:01.703Z

Step 1: Convert all dimensions to centimeters. Length ($l$) = $1.05 \text{ m} = 1.05 \times 100 \text{ cm} = 105 \text{ cm}$ Breadth ($b$) = $70 \text{ cm}$ Height ($h$) = $1.5 \text{ cm}$ Step 2: Calculate the volume of the rectangular piece of iron. The formula for the volume of a cuboid is $V = l \times b \times h$. $$V = 105 \text{ cm} \times 70 \text{ cm} \times 1.5 \text{ cm}$$ $$V = 7350 \text{ cm}^2 \times 1.5 \text{ cm}$$ $$V = 11025 \text{ cm}^3$$ Step 3: Calculate the total weight of the iron piece in grams. Given that $1 \text{ cm}^3$ of iron weighs 8 grams. $$\text{Weight in grams} = \text{Volume} \times \text{Weight per unit volume}$$ $$\text{Weight in grams} = 11025 \text{ cm}^3 \times 8 \frac{\text{g}}{\text{cm}^3}$$ $$\text{Weight in grams} = 88200 \text{ g}$$ Step 4: Convert the total weight from grams to kilograms. We know that $1 \text{ kg} = 1000 \text{ g}$. $$\text{Weight in kilograms} = \frac{88200 \text{ g}}{1000 \frac{\text{g}}{\text{kg}}}$$ $$\text{Weight in kilograms} = 88.2 \text{ kg}$$ The weight of the piece of iron is $\boxed{\text{88.2 kg}}$.