Question

Chapter - Mensuration
Class - 8

★ Question -

The paint in a can is sufficient to paint 8.5 sq m of surface . How many cans of paint are required to paint the walls of a hall which is 12 m long , 5 m wide and 4.25 m high ? Find the cost of the paint if each can cost ₨ 510 .

Note -

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Answer 1

Answer:

The total cost of painting the hall is Rs 12240.

Step-by-step explanation:

Surface area that can be painted by 1 can = 8.5 m²

★ Dimension of hall

• Length = 12m
• Breadth = 5m
• Height = 4.25m

In order to know the area to be painted we need to find the lateral surface area (area of 4 walls) and the area of ceiling.

Area to be painted

= Lateral Surface Area + Area of ceiling

= 2h(l + b) + lb

= [{2×4.25(12 + 5)} + {12 × 5}]m²

= {(2 × 4.25 × 17) + 60}m²

= (144.5 + 60)m²

= 204.5 m²

Number of cans required to paint 8.5 m² = 1

$\implies \textsf{Number of cans required to paint 1m}^{2} \: = \: \dfrac{1}{8.5}$

$\implies \: \textsf{Number of cans required to paint 204.5m}^{2} \: = \: \dfrac{1}{8.5} \times 204.5$

⇒ Number of cans required to paint 204.5 m² = 24.05

⇒ Number of cans required to paint 204.5 m² ≈ 24

Cost of 1 can of paint = Rs 510

⇒ Cost of 24 cans of paint = 24 × Rs 510

⇒ Cost of 24 cans of paint = Rs 12240

Therefore the total cost of painting the hall is Rs 12240.

Thanks !

Answer 2

Answer:

area of the walls of a hall (or)

lateral surface area of the hall = 2 × (l+b) × h

= 2 × 12+5 × 170

= 5780sq.m

now number of cans = total area of the hall ÷ area that each can paints

= 5780÷ 8.5 = 680

now , total cost of cans =

Number of cans × rate of cost per can

= 680 × 510 = Rs 346800