CBSE Notes, Lectures

CBSE - Mathematics - Surface Areas and Volumes

Surface Areas and Volumes

NCERT Exercise Exercise 13.1

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 rs4 for 1000 cm2 , find the cost of cardboard required for supplying 250 boxes of each kind.

Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
= 2(25×20 + 20×5 + 25×5) cm2
= 2(500 + 100 + 125) cm2
= 1450 cm2
Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
= 2(15×12 + 12×5 + 15×5) cm2
= 2(180 + 60 + 75) cm2
= 630 cm2
Total surface area of 250 boxes of each type = 250(1450 + 630)cm2
= 250×2080 cm= 520000 cm2
Extra area required = 5/100(1450 + 630) × 250 cm= 26000 cm2

Total Cardboard required = 520000 + 26000 cm2 = 546000 cm2
Total cost of cardboard sheet = rs (546000 × 4)/1000 = rs 2184

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