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 cm 2 , find the cost of cardboard required for supplying 250 boxes of each kind.

Answer should be with explanition. ​

Answer 1

Answer:

May be 2332

Step-by-step explanation:

first we have to multiply 25×20×5 =2500

then 15×12×5 = 900

then add there answer 2500+900=3400

then percentage

=5×34

=170

then mines 4 from 1000

=996

then 170 plus 996

=1166

1166×2

2332

Answer 2

$\large{\bf{\underline{\underline{\blue{Given-]}}}}}$

• OVERLAP = 5%

• COST = 4 Rs 1000 Cm²

• BIGGER DIMENSIONS - = L = 25 Cm

= B = 20 Cm

= H = 5 Cm

• SMALLER DIMENSIONS = L = 15 Cm

= B = 12Cm

= H = 5 Cm

[ TOTAL SURFACE AREA ] =

• = 2 [ LB + BH + HL ]

• 2 [ (25 × 20) + (20 × 5) + (5 × 25) ]

• 2 [ 500 + 100 + 125]

• 2 [ 725 ]

• 1450 CM²

OVERLAP = 5%

$\frac{1450 \times 5}{100} \\ = \frac{145}{2} \\ = 72.5$

SO,

.

FOR AREA OF CARDBOARD SHEET =

.

${\large{\bf{\underline{\underline{\ \: {(1522.5 \times 250)Cm²}}}}}} \\ {\small{\bf{\underline{\underline{\ { = 380625 cm²}}}}}}$

.

FOR SMALLER DIMENSIONS =

• TOTAL SURFACE AREA = 2 [ LB + BH+HL ]

$2[(15\times12)+(12\times5)+(5\times15)] \\ 2[180 + 60 + 75] \\ 2[315] \\ = 630$

.

OVERLAP = 5%

.

$\frac{630 \times 5}{100}\\= \frac{63}{2} \\ 31.5$

SO,

.

[630 + 31.5] Cm² = 661.5 m²

.

COST OF CARDBOARD SHEET = 1000 Cm² = 4 Rs

.

FOR AREA OF CARDBOARD

.

${\large{\bf{\underline{\underline{\red{ 2[661.5 \times 250]}}}}}} \\ {\small{\bf{\underline{\underline{\blue{165375 cm²}}}}}}$

.

TOTAL AREA OF CARDBOARD SHEET

.

${\huge{\bf{\underline{\underline{\blue{ \frac{546000×4}{1000} }}}}}}$

.

FINAL ANSWER

2184.

.

KNOW MORE FORMULAS

• LATERAL SUFRACE AREA

• = 2(L+B) × H

.

• AREA OF BASE OF CUBOID

• L×B

.

• DIAGONAL OF THE CUBOID =

• √( l² + b² +h²)

.

• VOLUME OF CUBOID =

• = AREA OF AREA OF BASE × HEIGHT × L×B×H