Question

Shanti Sweets stall was placing an order for making cardboard boxes for packing their sweets. 2 sizes of Boxers wear required the bigger of dimensions 25 cm × 20 cm × 5 cm and smaller of dimension 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 rupees 4 for 1000 CM sq., find the cost of cardboard required for supplying 250 boxes of each kind.

Answer 1

length of bigger box = 25 cm
breadth of bigger box = 20 cm
height of bigger box = 5cm

Total surface area of bigger box = 2(lb+bh+hl)
= 2( 25× 20 + 20 × 5 + 5 ×25)
= 2( 500 + 100 + 125)
1450 cm^2

extra area required = 5% × surface area of bigger box
= 5/100 × 1450
= 72.5 cm^2

Total area required for 1 bigger box
= total surface area + extra area required
= 1450+ 72.5
= 1522.5 cm^2
Total area required for 250 box
= area of 1 box × 250
= 1522.2 × 250
= 380625cm^2

FOR SMALLER BOX
length of smaller box = 15 cm
breadth of smaller box = 12 cm
height of smaller box = 5 cm

Total surface area of smaller box = 2 (lb+bh+hl)
= 2(15×12+ 12×5 + 5×15)
= 2(180+75+60)
= 2×315
= 630cm^2
extra area required = 5% of surface area of smaller box
= 5/100 × 630
=31.5cm^2
Total area required for 1 smaller box = surface area + extra area
630+31.5 cm^2
= 661.5cm^2
Area of cardboard sheet required for 250 smaller boxes
= Total area × 250
= 250 × 661.5
= 165375cm^2
Total cardboard sheet required
= area of 250 big boxes + area of 250 smaller boxes
= 380625+ 165375
= 546000cm^2

Cost of 1000cm^2 cardboard sheet = rs 4
cost of 1 cm^2 cardboard sheet = rs 1/1000× 4
cost of 546000cm^2 cardboard sheet =
rs 546000/1000×4
= rs 2184
THANKYOU
hope it will help u..

Answer 2

$ANSWER:$

Case 1

Given:

In case of the bigger box,

L=25cm,

B=20cm

H=5cm

To find:

Total surface area.

$Total \: Surface \: Area \: of \: the \: Bigger \: Box = 2(lb + bh + hl) \\ \\ Total \: Surface \: Area \: of \: the \: Bigger \: Box = 2(25 \times 20 + 20 \times 5 + 5 \times 25) \\ \\ Total \: Surface \: Area \: of \: the \: Bigger \: Box \: = 2(500 + 100 + 125) \\ \\ Total \: Surface \: Area \: of \: the \: Bigger \: Box = 2(725) \\ \\ Total \: Surface \: Area \: of \: the \: Bigger \: Box = 1450 {cm}^{2}$

Case 2:

Given:

In case of smaller box,

L=15cm

B=12cm

H=5cm.

To find:

Total surface area of the smaller box.

$Total \: Surface \: Area \: of \: the \: Smaller \: Box = 2(lb + bh + hl) \\ \\ Total \: Surface \: Area \: of \: the \: Smaller \: Box = 2(15 \times 12 + 12 \times 5 + 5 \times 15) \\ \\ Total \: Surface \: Area \: of \: the \: Smaller \: Box = 2(180 + 60 + 75) \\ \\ Total \: Surface \: Area \: of \: the \: Smaller \: Box = 2 \times 315 \\ \\ Total \: Surface \: Area \: of \: the \: Smaller \:Box = 630 {cm}^{2}$

Now, calculate the total surface area of 250 boxes of each type.

$= No.of \: boxes \: (surface \: area \: of \: big \: box+ surface \: area \: of \: small \: box) \\ \\ = 250(1450 + 630) \\ \\ = 250 \times 2080 \\ \\ = 520000 {cm}^{2}$

Therefore,

total surface area of 250 boxes of each type =520000cm^2.

Now, cardboard required (i.e including 5% for overlaps,etc)

$= Total \: Surface \: Area \: of \: 250 \: boxes \: of \: each \: type \: \times \frac{105}{100} \\ \\ = (520000 \times \frac{105 }{100}) \\ \\ = 546000 {cm}^{2}$

$\\ Cost \: of \: 1000 {cm}^{2} \: of \: cardboard = rs \: 4 \\ \\ Total \: Cost \: of \: Cardboard = ( \frac{54600}{1000} \times 4) \\ \\ Cost \: of \: Cardboard = 546 \times 4 = Rs\: 2184$

Therefore,

cost of cardboard required for supplying boxes is Rs 2,184.