Question

What is the percentage increase in the lateral surface area of a cube, if each of its side is increased by 50% ?

• A. 125%
• B. 25%
• C. 100%
• D. 135%

Answer 1

The correct option is A 125%

Let original side be L units in length.
New side length = 1.5L (after increase).
Original lateral surface area = $4\times L^2$
Increased lateral surface area = $4\times {\left(1.5L\right)}^2=9L^2$
Then, % increase in surface area
$=\frac{\text{Increased lateral surface area}-\text{Original lateral surfac area}}{\text{Original lateral surface area}}\times 100$
$=\frac{9L^2-4L^2}{4L^2}\times 100=125\%$

[Alternate method:
Let the edge be $a\text{cm}$.
New edge (after increase)
$=a+50\%a=a+\frac{a}{2}=\frac{3a}{2}$

LSA of the original cube $=4a^2$

LSA of new cube
$=4\times {\left(\frac{3a}{2}\right)}^2=4\times \frac{9a^2}{4}=9a^2$

Increase in area $=9a^2-4a^2=5a^2=\frac{5}{4}\times 4a^2=1.25\times 4a^2=1.25\times \text{LSA of the original cube}$

$\Rightarrow$ Increase $\%=125\%$]