Math, asked by chandanab73, 3 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
4 for 1000 cm?, find the cost of cardboard required for supplying 250 boxes of each
kind.

Answers

Answered by MalikramJangde
1

Answer:

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

2

cardboard sheet

= Rs.

1000

546000×4

= Rs. 2184

Answered by kumarineelam39806
0

Answer:

bigger box t.s.a=2(lb+bh+hl)

2(25×20+20×5+5×25)

2(500+100+125)

2(725)

1450cm2

5%overlap 1450×5/100

72.5

total area =1450+72.5

1522.5cm2

1 box =1522.5cm2

250 box =1522.5×250

380625

smaller box t.s.a =2(lb+bh+hl)

2(15×12+12×5+5×15)

2(180+60+75)

2(315)

630cm2

5%overlap =530×5/100

31.5cm2

total area =630+31.5

661.5cm2

1 box=661.5cm2

250 box=661.5×250

165375

total cardboard =380625+165375

546000cm2

total cost=546000×4/1000

2184rupees

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