Math, asked by ahana71, 9 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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