Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of $15 m$ , $10 m$ and $7 m$ respectively. From each can of paint $100 s q m e t r e$ of area is painted. How many cans of paint will she need to paint the room ?

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Solution

Area to be painted $=$ area of the walls $+$ area of ceiling

$= 2 (h l + h b) + l b$

$= [2 \times (7 \times 15 + 7 \times 10) + (15 \times 10)] = 500$ sq m

No. of cans required $=$ $\frac{500}{100}$ $= 5$

Hence, $5$ cans are required to paint the walls and the ceiling of the cuboidal hall.

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Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m , 10 m and 7 m respectively. From each can of paint 100 sq metre of area is painted. How many cans of paint will she need to paint the room ?

Area to be painted = area of the walls + area of ceiling =2(hl+hb)+lb=[2×(7×15+7×10)+(15×10)]=500 sq mNo. of cans required = 500100 =5Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.

Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of $15 m$ , $10 m$ and $7 m$ respectively. From each can of paint $100 s q m e t r e$ of area is painted. How many cans of paint will she need to paint the room ?

Area to be painted $=$ area of the walls $+$ area of ceiling

$= 2 (h l + h b) + l b$

$= [2 \times (7 \times 15 + 7 \times 10) + (15 \times 10)] = 500$ sq m

No. of cans required $=$ $\frac{500}{100}$ $= 5$

Hence, $5$ cans are required to paint the walls and the ceiling of the cuboidal hall.