Shanti Sweets Stall was placing an order for making cardboard boxes for packing
their sweets. Two sizes of boxes were required. The bigger of dimensions
25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the
overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is
4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each
kind.
Answers
Answer:
Cost of 250 bigger boxes = Rs. 1522.5
Cost of 250 smaller boxes = Rs. 346.5
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Question :-
Shanti Sweets Stalll was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Answer :-
For bigger box:
Length (l) = 25 cm,
Breadth (b) = 20 cm,
Height (h) = 5 cm
Total surface area of a box = 2(lb + bh + hl)
= 2[(25 x 20) + (20 x 5) + (5 x t25)] cm^2
= 2 [500 + 100 + 125] cm^2
= 2[725] cm^2
= 1450 cm^2
Total surface area of 250 boxes = (250 x 1450) cm^2 = 362500 cm^2
For smaller box:
l = 15 cm, b = 12 cm, h = 5 cm
Total surface area of a box = 2 [lb + bh + hl]
= 2[(15 x 12) + (12 x 5) + (5 x 15)] cm^2
= 2[180 + 60 + 75] cm^2 = 2[315] cm^2 = 630 cm^2
∴ Total surface area of 250 boxes = (250 x 630) cm^2 = 157500 cm^2
Now, total surface area of both type of boxes = 362500 cm^2 +157500 cm^2 = 520000 cm^2 Area for overlaps = 5% of [total surface area]
= 5/100 x 520000 cm^2 = 26000 cm^2
∴ Total surface area of the cardboard required = [Total surface area of 250 boxes of each type] + [Area for overlaps]
= 520000 cm^2 + 26000 cm^2 = 546000 cm^2
∵ Cost of 1000 cm^2 cardboard = Rs. 4
∴ Cost of 546000 cm^2 cardboard
= Rs.4×546000/1000 = Rs. 2184
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