Answer

1. Length of plastic box = 1.5 m
Width of plastic box = 1.25 m
Depth of plastic box = 1.25 m
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box = Lateral surface area + Area of the base
= 2(l+b)×h + (l×b)
= 2[(1.5 + 1.25)×1.25] + (1.5 × 1.25) m² 
                                              = (3.575 + 1.875) m² 
= 5.45 m² 
The sheet required required to make the box is 5.45 m² 
(ii) Cost of 1 m² of sheet = Rs 20
∴ Cost of 5.45 m² of sheet = Rs (20 × 5.45) = Rs 109

2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of rs 7.50 per m².

Answer

length of the room = 5m
breadth of the room = 4m
height of the room = 3m
Area of four walls including the ceiling = 2(l+b)×h + (l×b)
= 2(5+4)×3 + (5×4) m² 
 = (54 + 20) m² 
= 74 m² Cost of white washing = ���7.50 per m²
Total cost = rs (74×7.50) = rs 555

Answer

Perimeter of rectangular hall = 2(l + b) = 250 m
Total cost of painting = ���15000
Rate per m² = rs10
Area of four walls = 2(l + b) h m² = (250×h) m²
A/q,
(250×h)×10 = rs15000
⇒ 2500×h = rs15000
⇒ h = 15000/2500 m
⇒ h = 6 m
Thus the height of the hall is 6 m.

Answer

Volume of paint = 9.375 m² = 93750 cm²
Dimension of brick = 22.5 cm×10 cm×7.5 cm
Total surface area of a brick = 2(lb + bh + lh) cm²
= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm²
= 2(225 + 75 + 168.75) cm²
= 2×468.75 cm² = 937.5 cm²
Number of bricks can be painted = 93750/937.5 = 100

(i) Lateral surface area of cubical box of edge 10cm = 4×10² cm² = 400 cm²
Lateral surface area of cuboid box = 2(l+b)×h
= 2×(12.5+10)×8 cm²
= 2×22.5×8 cm² = 360 cm²
Thus, lateral surface area of the cubical box is greater by (400 – 360) cm² = 40 cm²

(ii) Total surface area of cubical box of edge 10 cm =6×102cm²=600cm²
Total surface area of cuboidal box = 2(lb + bh + lh)
= 2(12.5×10 + 10×8 + 8×12.5)cm²
= 2(125+80+100) cm²
= (2×305) cm² = 610 cm²
Thus, total surface area of cubical box is smaller by 10 cm²

(i) Dimensions of greenhouse:
l = 30 cm, b = 25 cm, h = 25 cm
Total surface area of green house = 2(lb + bh + lh)
= 2(30×25 + 25×25 + 25×30) cm²
= 2(750 + 625 + 750) cm²
= 4250 cm²
(ii) Length of the tape needed = 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4×80 cm = 320 cm

Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
= 2(25×20 + 20×5 + 25×5) cm²
= 2(500 + 100 + 125) cm²
= 1450 cm²
Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
= 2(15×12 + 12×5 + 15×5) cm²
= 2(180 + 60 + 75) cm²
= 630 cm²
Total surface area of 250 boxes of each type = 250(1450 + 630)cm²
= 250×2080 cm² = 520000 cm²
Extra area required = 5/100(1450 + 630) × 250 cm² = 26000 cm²

Total Cardboard required = 520000 + 26000 cm² = 546000 cm²
Total cost of cardboard sheet = rs (546000 × 4)/1000 = rs 2184

Dimensions of the box- like structure = 4m × 3m × 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(l+b)×h + lb
= [2(4+3)×2.5 + 4×3] m²
= (35×12) m²
= 47 m²