Question

7. Shanti Sweets Stall was placing an order for making
cardboard boxes for paking their sweets. Two sizes of boxes were
required. The bigger of dimensions 25 cm by 20 cm by 5 cm and
the smaller of dimensions 15 cm by 12 cm by 5 cm. 5% of the total
surface area is required extra, for all the overlaps. If the cost of
the cardboard is * 4 for 1000 cm?, find the cost of cardboard
required for supplying 250 boxes of each kind.
​

Answer 1

Explanation:

Total S.A of bigger box

=2(lb+bh+lh)

=2(25×20+25×5+20×5) cm

2

=2(500+125+100)

=1450 cm

2

⇒For overlapping extra area required =

100

450×5

=72.5 cm

2

∴ Total S.A (including overlaps)

=1450+72.5=1522.5 cm

2

Area of cardboard sheet for 250 such boxes

=(1522.5×250) cm

2

Total S.A of smaller box

=2(15×12+15×5+12×5)cm

2

=630 cm

2

For overlapping area required =

100

630×5

=31.5 cm

2

Total S.A (including overlaps)=630+31.5=661.5 cm

2

Area of cardboard sheet required for 250 such boxes

=250×661.5cm

2

=165375 cm

2

Total cardboard sheet required =380625+165375

=54000 cm

2

⇒Cost of 1000 cm

2

cardboard sheet = Rs.4

⇒Cost of 546000 cm

cardboard sheet

= Rs.

1000

546000×4

= Rs. 2184

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

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

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