# 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.

H Harsh Kankaria

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

Since, 250 of each box is required,

$\therefore$ Total area of carboard 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 $\small Rs.\ 2184$

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