Question

If each edge of a cube is increased by 50%, then the percentage increase in its surface area is
(a) 50%
(b) 75%
(c) 100%
(d) 12%

Answer 1

Let a be the edge of the cube.
Then the surface area is $6a^2=S\left(\mathrm{say}\right)$

Now, increased edge = $\left(a+\frac{50}{100}a\right)$ = $\frac{150}{100}a=\frac{3}{2}a$

$$\begin{gathered} \mathrm{Then},\mathrm{new}\mathrm{surface}\mathrm{area}=6{\left(\frac{3}{2}a\right)}^2=6\times \frac{9}{4}{a}^2=\frac{9}{4}S \\ \mathrm{In}\mathrm{crease}\mathrm{in}\mathrm{surface}\mathrm{area}=\frac{9}{4}S-S=\frac{5}{4}S \\ \therefore \mathrm{Percentage}\mathrm{increase}\mathrm{in}\mathrm{surface}\mathrm{area}=\frac{\frac{5}{4}S}{S}=\frac{5}{4}\times 100\%=125\% \end{gathered}$$