Question

a Courtyard is 20 m long and 15 metre broad and it is to be paved with bricks of length 25 cm and breadth is 12 cm find the number of bricks required ​

Answer 1

AnswEr :

$\bf{ Courtyard}\begin{cases}\sf{Length = 20m=(20 \times 100)cm}\\ \qquad \:\quad\sf{= 2000 \:cm}\\\sf{Breadth=15m=(15 \times 100)cm} \\ \qquad \quad\: \:\:\sf{= 1500 \:cm}\end{cases}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{A}}\put(6.9,2){\mathsf{\large{1500 cm}}}\put(7.7,1){\large{B}}\put(9.2,0.7){\matsf{\large{2000 cm}}}\put(11.1,1){\large{C}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11.1,3){\large{D}}\end{picture}$

$\rule{100}{2}$

$\bf{Brick}\begin{cases}\sf{Length=25 \:cm}\\\sf{Breadth=12 \:cm}\end{cases}$

$\setlength{\unitlength}{1cm}\begin{picture}(8,2)\thicklines\put(7.6,3){\large{A}}\put(6.8,2){\mathsf{\large{12 cm}}}\put(7.6,1){\large{B}}\put(9.2,0.6){\matsf{\large{25 cm}}}\put(11.1,1){\large{C}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11.1,3){\large{D}}\end{picture}$

$\rule{200}{1}$

• According to the Question Now :

$\longrightarrow \textsf{Area of Courtyard = Area of Brick \times Number of Bricks}\\\\\\\longrightarrow \textsf{LENGTH \times BREADTH = (length \times breadth) \times Number} \\\\\\\longrightarrow \sf(2000cm \times 1500cm) = (25cm \times 12cm)\times Number \\ \\ \\\longrightarrow \sf \cancel\dfrac{(2000cm \times 1500cm)}{(25cm \times 12cm)} = Number \\ \\ \\\longrightarrow \sf80 \times 125 = Number \\ \\ \\\longrightarrow \large\boxed{\sf Number = 10000}$

⠀

∴ 10000 Bricks are required for courtyard.

Answer 2

$\bf{\Huge{\boxed{\sf{\green{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}}}$

A courtyard is 20m long and 15m broad and it is to be paved with bricks of length 25cm and breadth is 12cm.

$\bf{\Large{\underline{\bf{To\:find\::}}}}}}$

The number of bricks required.

$\bf{\Large{\underline{\sf{\blue{Explanation\::}}}}}$

$\bf{We\:have\begin{cases}\sf{Courtyard\:of\:length\:=\:20m}\\ \sf{Courtyard\:of\:new\:length\:=(20*100)cm=2000cm}\\ \sf{Coutyard\:of\:breadth\:=\:15m}\\ \sf{Courtyard\:of\:new\:breadth\:=\:(15*100)=1500cm}\end{cases}}$

Therefore,

$\leadsto\bf{Area\;of\:courtyard\:=\:Length*Breadth}$

$\longmapsto\tt{Area\:of\:cortyard\:=\:(2000*1500)cm^{2} }$

$\longmapsto\tt{Area\:of\:courtyard\:=\:3000000cm^{2} }$

&

$\bf{We\:have}\begin{cases}\sf{Bricks\:of\:length\:=\:25cm}\\ \sf{Bricks \:of\:breadth\:=\:12cm}\end{cases}}$

$\leadsto\bf{Area\:of\:bricks\:=\:Length*Breadth}$

$\longmapsto\tt{Area\:of\:bricks\:=\:(25*12)cm^{2} }$

$\longmapsto\tt{Area\:of\:bricks\:=\:300cm^{2} }$

Now,

$\mapsto\sf{Number\:of\:bricks\:=\:\frac{Area\:of\:Courtyard}{Area\:of\:Bricks} }$

$\mapsto\sf{Nuber\:of\:bricks\:=\:\cancel{\frac{3000000cm^{2} }{300cm^{2} }} }$

$\mapsto\sf{\red{Number\:of\:bricks\:=\:10000\:bricks}}$

Thus,

$\bf{\large{\boxed{\tt{\orange{The\:number\:of\:bricks\:required\:is\:10000.}}}}}}$