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 4 for 1000 cm² find the cost of cardboard required for supplying 250 boxes of each kind. ​

Answer 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 ......hii can you give me brilliant list

Answer 2

Step-by-step explanation:

Total S.A of bigger box = 2(lb+bh+lh)

= 2(25×20+25×5+20×5) cm²

= 2(500+125+100) cm²

= 1450 cm²,

⇒For overlapping extra area required =

$\frac{450 \times 5}{100}$

= 72.5 cm²

∴ Total S.A (including overlaps),

= 1450 + 72.5 = 1522.5 cm²

Area of cardboard sheet for 250 such boxes = (1522.5×250) cm²

Total S.A of smaller box = 2(15×12+15×5+12×5)cm²

= 630 cm²

For overlapping area required =

$\frac{630 \times 5}{100}$

= 31.5 cm²

Total S.A (including overlaps) = 630 + 31.5 = 661.5 cm²

Area of cardboard sheet required for 250 such boxes

= 250×661.5cm²

= 165375 cm²

Total cardboard sheet required = 380625+165375

= 546000 cm²

Cost of 1000 cm² cardboard sheet = Rs.4

Cost of 546000 cm² cardboard sheet = Rs.

$\frac{546000 \times 4}{1000}$

= Rs. 2184