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 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 1

Answer:

TSA of big box = 2(lb+bh+hl)

=> 2 (25*20 + 20*5 + 5*25)

=> 2 (500 + 100 + 125)

=> 2 (725)

=> 1450cm2

therefore for 250 such boxes = 1450*250

=> 362500cm2

TSA of small box = 2(lb+bh+hl)

=> 2 (12*15 + 5*12 + 5*15)

=> 2 (180 + 60 + 75)

=> 2 (315)

=> 630 cm2

therefore for 250 such boxes = 630*250

=> 157500 cm2

Cardboard required for boxes = 362500+157500

=> 520000cm2

total = 520000 + 5% of 520000.......................(Overlaps)

=> 520000+26000

=> 586000cm2

total Cardboard required is 586000cm2

therefore Cost = 586000*4/1000

=> rs 2344 is required

Hope it helps

Please Mark brainliest because it took a lot of time and effort for the answer

Answer 2

$\pmb{ \sf{Question :}}$

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 is Rs. 4 for 1000 cm², find the cost of cardboard required for supplying 300 boxes of each kind.

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$\pmb{ \sf { Formula \: to \: be \: used: }}$

$:\mapsto{\small{\underline{\boxed{\sf{TSA \: of \: cuboid =2(lb +bh + hl)}}}}}$

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$\pmb{\sf{Solution : }}$

$\red \bigstar\textsf{ \: For bigger box,}$

$\textsf{ l= 25 cm b = 20 cm h = 5 cm}$

$\textsf{Total surface area of the bigger box} \\ \textsf{= 2 (lb + bh + hl)} \\ \textsf{= 2[(25)(20) + (20)(5) + (5)(25)]} \\ \textsf{= 2(500 + 100 + 125) = 1450 cm²}$

$\sf{Cardboard \: required \: for \: all \: the \: overlap }\\ \sf{ = 1450 \times \frac{5}{100} = 72.5 \: {cm}^{2} }$

$\sf{\therefore Net \: surface \: area \: of \: the \: bigger \: box} \\ \sf{= 1450 cm² + 72.5 cm² = 1522.5 cm²}$

$\therefore \textsf {Net surface area of 250 bigger boxes} \\ \sf{= 1522.5 \times 250 = 380625 cm²}$

$\sf{Cost \: of \: cardboard = \frac{4}{1000} \times 38065 = ₹1522.50}$

$\red \bigstar \textsf { \: For smaller box,}$

$\textsf{I = 15 cm b = 12 cm h = 5 cm}$

$\textsf{Total surface area of the smaller box} \\\textsf{= 2 (lb + bh + hl)} \\\textsf{= 2[(15)(12) + (12)(5) + (5)( 15)] }\\\textsf{= 2[ 180 + 60 + 75] }\textsf{= 630cm²}$

$\textsf{Cardboard required for all the overlaps} \\ \sf{ = 630 \times \frac{5}{100} = 31.5 {cm}^{2} }$

$\therefore \textsf{Net surface area of the smaller box} \\\sf{= 630 cm^2 + 31.5 cm^2 = 661.5 cm^2}$

$\therefore \textsf{Net surface area of 250 smaller boxes} \\ \sf{ = 661.5 \times 250 = 165375 cm^2}$

$\sf{Cost \: of \: cardboard = \frac{4}{1000} \times 165375 = ₹661.50}$

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$\pmb{\sf{Final \: Answer :}}$

$\textsf{Total Cost = ₹ 1522.50 + ₹661.50} \\ = \boxed{ \underline{\large{\sf \pink{ ₹2184}} }}$

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$\pmb{ \sf{Note : }}$

$\footnotesize \rm{ : \mapsto For \: diagrams, \: refer \: to \: the \: attachment}$

$\footnotesize \textrm{:↦ Do thanks and rate the answer if it was helpful}$