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Question

A retaining wall with measurements 30 m × 12 m × 6 m was constructed with bricks of dimensions 8 cm × 6 cm × 6 cm. If 60% of the wall consists of bricks, the number of bricks used for the construction is _______ lakhs.

The correct answer is

45

To determine the total number of bricks used in the construction of the retaining wall, we need to calculate the volumes involved and then consider the percentage of the wall that consists of bricks. It's essential to ensure all measurements are in the same units before performing calculations.

Retaining Wall Volume Calculation

First, let's determine the dimensions of the retaining wall and convert them to centimeters (cm) to match the dimensions of the individual bricks. This unit consistency is vital for accurate volume calculation.

  • Length of the retaining wall = 30 m = $\text{30} \times \text{100 cm} = \text{3000 cm}$
  • Width of the retaining wall = 12 m = $\text{12} \times \text{100 cm} = \text{1200 cm}$
  • Height of the retaining wall = 6 m = $\text{6} \times \text{100 cm} = \text{600 cm}$

Now, we can calculate the total volume of the retaining wall using the formula for the volume of a rectangular prism:

Volume of the wall ($\text{V}_{\text{wall}}$) = Length $\times$ Width $\times$ Height

$\text{V}_{\text{wall}} = \text{3000 cm} \times \text{1200 cm} \times \text{600 cm}$

$\text{V}_{\text{wall}} = \text{2,160,000,000 cm}^3$

Brick Volume Calculation

Next, we need to calculate the volume of a single brick using its given dimensions. The dimensions of one brick are:

  • Length of one brick = 8 cm
  • Width of one brick = 6 cm
  • Height of one brick = 6 cm

The volume of one brick is calculated as:

Volume of one brick ($\text{V}_{\text{brick}}$) = Length $\times$ Width $\times$ Height

$\text{V}_{\text{brick}} = \text{8 cm} \times \text{6 cm} \times \text{6 cm}$

$\text{V}_{\text{brick}} = \text{288 cm}^3$

Volume of Bricks within the Wall

The problem states that 60% of the retaining wall consists of bricks. This means we only consider 60% of the total wall volume when calculating the space occupied by the bricks themselves. The remaining 40% would typically account for mortar, gaps, or other materials.

Volume occupied by bricks ($\text{V}_{\text{bricks used}}$) = 60% of $\text{V}_{\text{wall}}$

$\text{V}_{\text{bricks used}} = \frac{\text{60}}{\text{100}} \times \text{2,160,000,000 cm}^3$

$\text{V}_{\text{bricks used}} = \text{0.60} \times \text{2,160,000,000 cm}^3$

$\text{V}_{\text{bricks used}} = \text{1,296,000,000 cm}^3$

Number of Bricks Calculation

To find the total number of bricks required for the construction, we divide the total volume occupied by the bricks by the volume of a single brick.

Number of bricks = $\frac{\text{Volume occupied by bricks}}{\text{Volume of one brick}}$

Number of bricks = $\frac{\text{1,296,000,000 cm}^3}{\text{288 cm}^3}$

Number of bricks = $\text{4,500,000}$

Converting Bricks to Lakhs

The question asks for the number of bricks in lakhs. One lakh is a unit in the Indian numbering system equal to 100,000 (one hundred thousand). To convert the total number of bricks to lakhs, we divide by 1,00,000.

Number of bricks in lakhs = $\frac{\text{Total number of bricks}}{\text{1,00,000}}$

Number of bricks in lakhs = $\frac{\text{4,500,000}}{\text{1,00,000}}$

Number of bricks in lakhs = $\text{45}$

Therefore, 45 lakhs of bricks were used for the construction of the retaining wall.

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Important Questions from Numerical Computation

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  3. Leila aspires to buy a car worth Rs. 10,00,000 after 5 years. What is the minimum amount in Rupees that she should deposit now in a bank which offers 10% annual rate of interest, if the interest was compounded annually?

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  5. There are nine bags of rice looking alike, eight of which have equal and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance the minimum number of weighting required to identify the heavier bag is

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