ABC26GN4881 · Volume and Surface Area

Let original edge $=a$. The, surface area $=6 a^{2}$. New edge $=\frac{125}{100} a=\frac{5 a}{4}$. New surface area $=6 \times\left(\frac{5 a}{4}\right)^{2}=\frac{75 a^{2}}{8}$. Increase in surface area $=\left(\frac{75 a^{2}}{8}-6 a^{2}\right)=\frac{27 a^{2}}{8}$. $$\text { ∴ Increase } \%=\left(\frac{27 a^{2}}{8} \times \frac{1}{6 a^{2}} \times 100\right) \%=56.25 \% \text {. }$$