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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%
(56.04% of 550.06 + 19.92 × 18.13) – 121.97 = ?
9214.39 - 6843.57 + 8435.22 + ? = 17620.47
20.11 × 6.98 + 21.03 × 6.12 – 37.95 + 92.9 × 5.02 =?
? × 32.91 – 847.95 ÷ √16.4 – 13.982 = √24.7 × 24.04
8.15 of 124.95 ÷ 40.13 + 249.84 X 14.18 - √325 X 149.87 = ? X 10.85
189.23 + 18.11² + ?³ = √841.76 * 28.94
31.98% of 224.99 = 24.98% of ? + 9.91% of 499.99
56.05 2 – 24.24 2 + (63.98) 3/2 – 32.28% of 1500 = ? 2 + 113.03 × 5.09
456 x 99.999 + 654 = ?