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  • 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.

Q: 7. 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 \small 25\hspace {1mm}cm \times 20\hspace {1mm}cm \times 5 \hspace {1mm}cm, and the smaller of dimensions \small 15\hspace {1mm}cm \times 12\hspace {1mm}cm \times 5 \hspace {1mm}cm. For all the overlaps, \small 5\% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for \small 1000\hspace {1mm}cm^2,  find the cost of cardboard required for supplying 250 boxes of each kind.

Answers (1)

best_answer

Given,

Dimensions of the bigger box = \small 25\hspace {1mm}cm \times 20\hspace {1mm}cm \times 5 \hspace {1mm}cm,

Dimensions of smaller box =  \small 15\hspace {1mm}cm \times 12\hspace {1mm}cm \times 5 \hspace {1mm}cm

We know,

total surface area of a cuboid = 2(lb+bh+hl)

\therefore Total surface area of the bigger box = 2(25\times20+20\times5+5\times25)

= 2(500+100+125) = 1450\ cm^2

\therefore Area of the overlap for the bigger box = 5\%\ of\ 1450\ cm^2 = \frac{5}{100}\times1450 = 72.5\ cm^2

Similarly,

 

total surface area of the smaller box = 2(15\times12+12\times5+5\times15)

= 2(180+60+75) = 630\ cm^2

\therefore Area of overlap for the smaller box = 5\%\ of\ 630\ cm^2 = \frac{5}{100}\times630 = 31.5\ cm^2

Since 250 of each box are required,

\therefore Total area of cardboard required = 250[(1450+72.5)+(630 + 31.5)] = 546000\ cm^2

Cost of \small 1000\hspace {1mm}cm^2 of the cardboard = Rs 4 

\therefore Cost of \small 546000\hspace {1mm}cm^2 of the cardboard = \small Rs. (\frac{4}{1000}\times546000) = Rs. 2184

Therefore, the cost of the cardboard sheet required for 250 such boxes of each kind is \small Rs.\ 2184

Posted by

HARSH KANKARIA

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