Question

A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to
1. 10
2. 50
3. 60
4. 20

Answer 1

Correct Answer - Option 2 : 50

Calculation:

Let be assume the volumes of the cube is x³, x³, 8x³, 27x³, 27x³ respectively.

⇒ Volume of cube = x³ × (1 + 1 + 8 + 27 + 27) = 64x³

⇒ Side of the original cube = ³√(64x3) = 4x

⇒ Surface area of original cube = 6 × (4x) × (4x) = 96 x²

⇒ The sides of the smaller cube are x, x, 2x, 3x, 3x respectively.

⇒  The surface area of smaller cubes = 6 × (x² + x² + 4x² + 9x² + 9x²) = 144 x²

⇒ Required % change = (144 x² - 96 x²)/(96 x2) × (100) = 50%

∴ The required result will be 50%.