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 is Rs. 4 for 1000 \({cm}^2\), find the cost of cardboard required for supplying 250 boxes of each kind.

Given:
Length (l) of bigger box = 25 cm
Breadth (b) of bigger box = 20 cm
Height (h) of bigger box = 5 cm

we know that,

Total surface area of bigger box
= 2(lb + lh + bh)
= [2 (25 x 20 + 25 x 5 + 20 x 5)] \({cm}^2\)
= [2(500 + 125 + 100)] \({cm}^2\)
= 1450 \({cm}^2\)

Therefore, Extra area required for overlapping
= \(\frac{1450 × 5}{100}\) \({cm}^2\) = 72.5 \({cm}^2\)

While considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5) \({cm}^2\) = 1522.5 \({cm}^2\)

So, Area of cardboard sheet required for 250 such bigger boxes = (1522.5 x 250) \({cm}^2\) = 380625 \({cm}^2\)

Similarly, total surface area of smaller box = [2 (15 + 15 x 5 + 12 x 5)]
= [2 (180 + 75 + 60)] \({cm}^2\)
= (2 x 315) \({cm}^2\)
= 630 \({cm}^2\)

Therefore, extra area required for overlapping
= \(\frac{630 × 5}{100}\) \({cm}^2\) = 31.5 \({cm}^2\)

So, Total surface area of 1 smaller box while considering all overlaps
= (630 + 31.5) \({cm}^2\)
= 661.5 \({cm}^2\)

So, Area of cardboard sheet required for 250 smaller boxes
= (250 x 661.5) \({cm}^2\) = 165375 \({cm}^2\)

Thus, Total cardboard sheet required
= (380625 + 165375) \({cm}^2\)
= 546000 \({cm}^2\)

Cost of 1000 \({cm}^2\) cardboard sheet = Rs. 4

Therefore, Cost of 546000 \({cm}^2\) cardboard sheet will be:
\(Rs. \frac{546000 × 4}{100}\) \({cm}^2\) = Rs. 2184

Therefore, the cost of cardboard sheet required for 250 such boxes of each kind will be Rs 2184.