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. - Mathematics

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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 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.

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Solution

Length (l1) of bigger box = 25 cm

Breadth (b1) of bigger box = 20 cm

Height (h1) of bigger box = 5 cm

Total surface area of bigger box = 2(lb lh + bh)

= [2(25 × 20 + 25 × 5 + 20 × 5)] cm2

= [2(500 + 125 + 100)] cm2

= 1450 cm2

`"Extra area required for overlapping "= ((1450xx5)/100)cm^2 = 72.5 cm^2`

While considering all overlaps, total surface area of 1 bigger box

= (1450 + 72.5) cm2 =1522.5 cm2

Area of cardboard sheet required for 250 such bigger boxes

= (1522.5 × 250) cm2 = 380625 cm2

Similarly, total surface area of smaller box = [2(15 ×12 + 15 × 5 + 12 × 5] cm2

= [2(180 + 75 + 60)] cm2

= (2 × 315) cm2

= 630 cm2

`"Therefore, extra area required for overlapping "= ((630xx5)/100)cm^2 = 31.5cm^2`

Total surface area of 1 smaller box while considering all overlaps

= (630 + 31.5) cm2 = 661.5 cm2

Area of cardboard sheet required for 250 smaller boxes = (250 × 661.5) cm2

= 165375 cm2

Total cardboard sheet required = (380625 + 165375) cm2

= 546000 cm2

Cost of 1000 cm2 cardboard sheet = Rs 4

Cost of 546000 cm2 cardboard sheet`=((546000xx4)/1000) = "Rs. "2184`

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

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Chapter 13: Surface Area and Volumes - Exercise 13.1 [Page 213]

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NCERT Mathematics Class 9
Chapter 13 Surface Area and Volumes
Exercise 13.1 | Q 7 | Page 213

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