Question

A courtyard is 18m long and 12m broad. What will be the total cost of the tiles at the rate ₹450 for 100 tiles. Each tile is of length 40 cm and breadth 30 cm.


Please solve this math as I have my math exam tomorrow. ​

Answer 1

Answer:

Step-by-step explanation:

Area of Courtyard = 18 * 12

                              =  216 m^2

Area of tile = 40 * 30

                  = 1200 cm ^2

                  = 12m^2

No.of tile required = 216/ 12

                              = 18

Cost of 1 tile = 450 /100

                    =  45 / 10

                    =  ₹ 4.5

Cost of 18 tile = 18 * 4.5

                      =  ₹ 81

Answer 2

$\large{{ \underline{\underline\red{\sf{Given:-}}}}}$

• ● A courtyard is 18m long and 12m broad.

• ● Cost of the tiles at the rate ₹450 for 100 tiles

• ● Each tile is of length 40 cm and breadth 30 cm.

$\large{{ \underline{\underline\orange{\sf{To \: Find:-}}}}}$

• ● The total cost of tiles.

$\large{{ \underline{\underline\orange{\sf{Solution:-}}}}}$

$\red\star\large{{ \underline{\blue{\mathfrak{Finding \: the\: area \: of \: courtyard :-}}}}}$

We know that,

$\sf \bigg[Area_{(Rectangle)}=L×B \bigg]$

$\sf \mapsto Area = \ell \times b$

$\sf \mapsto Area = 18 \times 12 \: m$

$\sf \mapsto Area = 216 m$

${\underline{\boxed{\sf\pink{∴ The \: area \: of \: courtyard = 216m ^{2}}}}}$

Now,

$\red\star\large{{ \underline{\blue{\mathfrak{Finding \: the\: area \: of \: tiles :-}}}}}$

• Length = 40 cm

• Breadth = 3p cm

$\sf \mapsto Area = \ell \times b$

$\sf \mapsto Area = 40 \times 30$

$\sf \mapsto Area = 1200 cm$

Area of tiles = $\sf 1200cm {}^{2} = 12m^2$

Number of tiles required $\sf \mapsto \frac{ \cancel{216}}{ \cancel{12}} = 18$

Cost of 1 tile $\sf \mapsto \frac{ \cancel{450}}{ \cancel{100}}= ₹4.5$

Cost of 18 tiles $\sf \mapsto 18×4.5 = ₹ 81$

$\large{{ \underline{\boxed{\sf\pink{∴ The \: cost \: of \: tiles = ₹ 81}}}}}$