How many bricks,each measuring 25, cm times 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) First,convert all dimensions to centimeters $(cm)$:Wall dimensions: $8\, m = 800\, cm$,$6\, m = 600\, cm$,$22.5\, cm$.Volume of the wall $= 800\,

Vedclass · Accepted Answer

$6400$

Vedclass · Answer

(C) First,convert all dimensions to centimeters $(cm)$:Wall dimensions: $8\, m = 800\, cm$,$6\, m = 600\, cm$,$22.5\, cm$.Volume of the wall $= 800\, cm 	imes 600\, cm 	imes 22.5\, cm = 10,800,000\, cm^3$.Volume of one brick $= 25\, cm 	imes 11.25\, cm 	imes 6\, cm = 1,687.5\, cm^3$.Number of bricks $= \frac{	ext{Volume of wall}}{	ext{Volume of brick}} = \frac{800 	imes 600 	imes 22.5}{25 	imes 11.25 	imes 6}$.Simplifying the expression: $\frac{800 	imes 600 	imes 22.5}{1687.5} = \frac{10,800,000}{1,687.5} = 6400$.Thus,$6400$ bricks are required.

How many bricks,each measuring $25\, cm \times 11.25\, cm \times 6\, cm$,will be needed to build a wall $8\, m \times 6\, m \times 22.5\, cm$?

Similar Questions

$A$ well of $11.2 \, m$ diameter is dug $8 \, m$ deep. The earth taken out has been spread all around it to a width of $7 \, m$ to form a circular embankment. Find the height of the embankment in meters.

If each side of a cube is increased by $10 \%$,the volume of the cube will increase by (in $\%$).

$A$ hemisphere is placed on top of a cube. Its front view is shown in the given figure. The total height of the figure is $221 \ cm$. The ratio of the curved surface area of the hemisphere to the total surface area of the cube is $11:42$. What is the total volume (in $cm^3$) of the figure?

If the volume of a sphere is numerically equal to its surface area,then its diameter is ......... $cm$.

$A$ cylinder and a cone have their heights in the ratio $2:3$ and the radii of their bases in the ratio $3:4$. Find the ratio of their volumes.

Vedclass Test Series

Exam Paper Generator

Online Exam Module