Surface Area Increase of Melted Cubes

CAT 2017 Slot 1 · QA · Medium · Mensuration and Geometry

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A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1:1:8:27:271 : 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. A.

    10

  2. B.

    50

  3. C.

    60

  4. D.

    20

Answer

B

Explanation

Total volume ratio = 1+1+8+27+27=641 + 1 + 8 + 27 + 27 = 64. Original cube volume = 64, side = 4, surface area = 6×42=966 \times 4^2 = 96. Sides of the 5 smaller cubes = 13,13,83,273,273=1,1,2,3,3\sqrt[3]{1}, \sqrt[3]{1}, \sqrt[3]{8}, \sqrt[3]{27}, \sqrt[3]{27} = 1, 1, 2, 3, 3. Sum of surface areas = 6(12+12+22+32+32)=6(1+1+4+9+9)=6(24)=1446(1^2 + 1^2 + 2^2 + 3^2 + 3^2) = 6(1 + 1 + 4 + 9 + 9) = 6(24) = 144. Percentage excess = 144−9696×100=4896×100=50%\frac{144 - 96}{96} \times 100 = \frac{48}{96} \times 100 = 50\%.

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