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(C) Length $(l_1)$ of bigger box $= 25 \,cm$,Breadth $(b_1) = 20 \,cm$,Height $(h_1) = 5 \,cm$.Total surface area of bigger box $= 2(l_1b_1 + l_1h_1 +

Vedclass · Accepted Answer

$Rs. 2184$

Vedclass · Answer

(C) Length $(l_1)$ of bigger box $= 25 \,cm$,Breadth $(b_1) = 20 \,cm$,Height $(h_1) = 5 \,cm$.Total surface area of bigger box $= 2(l_1b_1 + l_1h_1 + b_1h_1) = 2(25 	imes 20 + 25 	imes 5 + 20 	imes 5) \,cm^2 = 2(500 + 125 + 100) \,cm^2 = 1450 \,cm^2$.Extra area for overlaps $= 5\% 	ext{ of } 1450 \,cm^2 = (1450 	imes 0.05) \,cm^2 = 72.5 \,cm^2$.Total area for $1$ bigger box $= 1450 + 72.5 = 1522.5 \,cm^2$.Area for $250$ bigger boxes $= 250 	imes 1522.5 = 380625 \,cm^2$.Length $(l_2) = 15 \,cm$,Breadth $(b_2) = 12 \,cm$,Height $(h_2) = 5 \,cm$.Total surface area of smaller box $= 2(15 	imes 12 + 15 	imes 5 + 12 	imes 5) \,cm^2 = 2(180 + 75 + 60) \,cm^2 = 630 \,cm^2$.Extra area for overlaps $= 5\% 	ext{ of } 630 \,cm^2 = (630 	imes 0.05) \,cm^2 = 31.5 \,cm^2$.Total area for $1$ smaller box $= 630 + 31.5 = 661.5 \,cm^2$.Area for $250$ smaller boxes $= 250 	imes 661.5 = 165375 \,cm^2$.Total cardboard area $= 380625 + 165375 = 546000 \,cm^2$.Cost $= (546000 / 1000) 	imes 4 = 546 	imes 4 = Rs. 2184$.

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