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A solid cube is melted to form three solid cubes whose volumes are in the ratio of 27 : 64 : 125. Then, the sum of the total surface areas of these three solid cubes how much percentage more than the total surface area of the original cube?

A solid cube is melted to form three solid cubes whose volumes are in the ratio of 27 : 64 : 125. Then, the sum of the total surface areas of these three solid cubes how much percentage more than the total surface area of the original cube? Correct Answer 38.88%

Given:

Ratio of volumes of three cubes = 27 : 64 : 125

Formula used:

Volume of cube = (side)3

Area of cube = 6 × (side)2

Calculation:

The volume of the original cube will be 27 + 64 + 125 = 216 unit3

⇒ The edge of the original cube = ∛(216) = 6 units

The edge of the three solid cubes will be ∛27, ∛64 and ∛125 or 3, 4 and 5 units, respectively.

The total surface areas of the three solid cube = 6 × (3)2 + 6 × (4)2 + 6 × (5)2 = 300 sq. units

Now,

The total surface area of the original cube = 6 × (6)2 = 216 sq. units

∴ Required Percentage = (300 – 216)/216 × 100 = 84/216 × 100 = 38.88%

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