Volume of the original box = 18 X 15 X 10 = 2700 cm3 New breadth of the box = 15 X 1.2 = 18 cm New height of the box = 10 X 0.8 = 8 cm So, new volume of the box = 18 X 18 X 8 = 2592 cm3 So, percentage reduction in volume = {(2700 - 2592) ÷ 2700} X 100 = 4% Alternate solution Let the length, breadth and height of the cuboid originally be 'l' units, 'b' units and 'h' units, respectively Therefore, original volume of the cuboid = 'lbh' cubic units New breadth of the cuboid = '1.2b' units New height of the cuboid = '0.8h' units New volume of the cuboid = 1.2b X 0.8h X l = '0.96lbh' cubic units Required percentage = {(lbh - 0.96lbh)/lbh} X 100 = 4%
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