Question

If each edge of a cube is increased by $50\%$, the percentage increase in its surface area is:

Answer 1

Let the edge of a cube be $a$

$\therefore$ Surface area of a cube $=6a^2$

Now, if side is increased by $50\%$, then

New side $\left(a^{\prime }\right)=a+50\%$ of $a$
$=a+\frac{a}{2}=\frac{3a}{2}$

So, new surface area
$=6{\left(\frac{3a}{2}\right)}^2$

$=\frac{6\times 9\times a^2}{4}$

$=13.5a^2$

$\therefore$ Increase in area
$=13.5a^2-6a^2$
$=7.5a^2$

$\%\text{increase}=\frac{\text{Increase in area}}{\text{original area}}\times 100$

$=\frac{7.5a^2}{6a^2}\times 100\%=125\%$

Hence, (d) is the correct option.