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 $100 m^{2}$ of area is painted. How many cans of paint will she need to paint the room?

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Solution

Here, the length of wall l = 15 m, the breadth of wall b = 10 m, and the height of wall h = 7 m.
(0.5 mark)

Surface area of the classroom to be painted = lb + 2(bh + hl)
$= 15 \times 10 + 2 (10 \times 7 + 7 \times 15) = 150 + 2 (70 + 105) = 150 + 350 = 500 m^{2}$
(2 marks)

$Now, the required number of cans = \frac{A r e a o f h a l l}{A r e a o f o n e c a n} = \frac{500}{100} = 5 c a n s$
Hence, 5 cans are required to paint the room.
(0.5 mark)

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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 100m2 of area is painted. How many cans of paint will she need to paint the room?

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 $100 m^{2}$ of area is painted. How many cans of paint will she need to paint the room?