142 views

1 vote

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 $:\text{B}$

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 $:\text{B}$