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Question

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×20 cm×5 cm and the smaller of dimensions 15 cm×12 cm×5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard isRs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.

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Solution

Since the cardboard boxes are cuboidal in shape, the total area of the cardboard is the same as the total surface area of the cuboid added to the overlap area that is, 5% of the total surface area is required extra.

Hence, the area of each box can be obtained by adding 5% of the total surface area to the total surface area of the cuboid.

For bigger boxes:

Dimensions of bigger boxes: L=25 cm, B=20 cm, H=5 cm

Total surface area of the bigger box (T.S.A)=2(lb+bh+lh)

T.S.A=2{(25 cm×20 cm)+(20 cm×5 cm)+(25 cm×5 cm)}

T.S.A=2{500 cm2+100 cm2+125 cm2}

T.S.A=2×725 cm2
T.S.A=1450 cm2
With 5% extra we get,
Requires area for each box =1450 cm2+0.05×1450 cm2=1522.5 cm2

For 250 boxes , required are is =1522.5 cm2×250=380625 cm2 ---(1)

For smaller boxes:
Dimensions of smaller boxes: l=15 cm, b=12 cm, h=5 cm
Total surface area of the bigger box (t.s.a)=2(lb+bh+lh)
(t.s.a)=2{(15 cm×12 cm)+(12 cm×5 cm)+(15 cm×5 cm)}

(t.s.a)=2{180 cm2+60 cm2+75 cm2}

(t.s.a)=2×315 cm2

T.S.A=630 cm2

With 5% extra we get,

Required area for the box =630 cm2+0.05×630 cm2=661.5 cm2

For 250 boxes , required are is =661.5 cm2×250=165375 cm2 ---(2)

So, Total T.S.A of the boxes =380625 cm2+165375 cm2=546000 cm2

Rs. 4 is charged or 1000 cm2.

Total Cost =546000 cm21000 cm2×Rs. 4=Rs. 2184

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