Math, asked by ahana71, 10 months ago

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
Rs 4 for 1000 cm, find the cost of cardboard required for supplying 250 boxes of each
kind.​

Answers

Answered by Anonymous
51

\huge\star{\green{\underline{\mathfrak{Answer: -}}}}

Dimension of bigger box = 25 cm × 20 cm × 5 cm

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

= 2(25×20 + 20×5 + 25×5) cm2

= 2(500 + 100 + 125) cm2

= 1450 cm2

Dimension of smaller box = 15 cm × 12 cm × 5 cm

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

= 2(15×12 + 12×5 + 15×5) cm2

= 2(180 + 60 + 75) cm2

= 630 cm2

Total surface area of 250 boxes of each type = 250(1450 + 630) cm2

= 250×2080 cm2 = 520000 cm2

Extra area required = 5/100(1450 + 630) × 250 cm2 = 26000 cm2

Total Cardboard required = 520000 + 26000 cm2 = 546000 cm2

Total cost of cardboard sheet = ₹ (546000 × 4)/1000 = ₹ 2184.

Answered by saanvigrover2007
9

 \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}

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