Question

Q18. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height 15m,
10m and 7m respectively. From each can of paint 100m of area is painted. How many cans of
paint will be required to paint the room? Find the cost of paint if each can cost 238.
​

Answer 1

Answer:

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

=2(hl+hb)+lb

=[2×(7×15+7×10)+(15×10)]=500 sq m

No. of cans required = 500/100= 5

cost of each can = 238

=5*238= 1190

Answer 2

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Given

• Length = 15m
• Breadth = 10m
• Height = 7m

To Calculate

• Cost of painting the hall

Solution

Area to be painted = L.S.A of hall + Area of ceiling

$\bf \implies [2(l + b)h]+ [l \times b]$

$\bf \implies [2(15 + 10) \times 7]+ [15\times 10]$

$\bf \implies [2 \times 25 \times 7]+ [150]$

$\bf \implies 350 + 150$

$\bf \implies {500m}^{2}$

Given That,

1 can paint 100m sq.

So, Cans required = 500/ 100 = 5 Cans

Now, Cost of 1 can = Rs.238

Cost of 5 cans = Rs. 238 × 5

= Rs. 1175

Therefore, Cost of painting the hall is Rs. 1175.