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A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio $1: 1:8: 27: 27$. 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. $20$
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Given that, the ratio of the volume of five cubes $= 1:1:8:27:27$

Total volume of smaller cubes $ = 1 + 1 + 8 + 27 + 27 = 64$

Side of original cube $a = \sqrt[3]{64} = 4$

Surface area of original cube $=6a^{2}=6(4^{2}) = 96$

The ratio of the sides of smaller cubes $= \sqrt[3]{1}: \sqrt[3]{1}: \sqrt[3]{8}:  \sqrt[3]{27}: \sqrt[3]{27}=1:1:2:3:3$

The surface area of smaller cubes $=6(1^{2}+1^{2}+2^{2}+3^{2}+3^{2})$

$\qquad \qquad = 6(1+1+4+9+9) =6(24) = 144$

$\therefore$  The required percentage change $ = \left(\frac{144-96}{96}\right)\times100 \% = \left(\frac{48}{96}\right)\times 100 \% = 50 \%.$

Correct Answer $:\text{B}$
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