I-ac can of pai m of aea is puintedHow many asof_w need 1 punt

1.	I-ac can of pai m of aea is puintedHow many asof_w need 1 punt
Answer» Solution: Given; Length of wall(l)= 15 m,Breadth of wall(b)= 10 m Height of wall(h)= 7 m Total area to be painted= area of 4 walls + area of ceiling = 2(l+b)h + lb =2(15+10)7 + 15×10 = 2×25×7+ 150 =50× 7+150= 350+ 150=500 Total area to be painted=500m² Given 100m² area can be painted from each can. Number of cans Required= Area of hall/ area of 1 can = 500/100= 5 Hence, 5 cans are required to paint the room.

Answer»

Solution:

Given;

Length of wall(l)= 15 m,Breadth of wall(b)= 10 m Height of wall(h)= 7 m

Total area to be painted= area of 4 walls + area of ceiling

= 2(l+b)h + lb

=2(15+10)7 + 15×10

= 2×25×7+ 150

=50× 7+150= 350+ 150=500

Total area to be painted=500m²

Given 100m² area can be painted from each can.

Number of cans Required=

Area of hall/ area of 1 can

= 500/100= 5

Hence, 5 cans are required to paint the room.

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