Question

Each side of a cube is increased by 50%. Then the surface area of the cube is increased by:

• A. 100%
• B. 50%
• C. 150%
• D. 125%

Answer 1

The correct option is D 125%

Let original side be L units in length.
New side length = 1.5L (after increase).
Original surface area = $6\times L^2$
Increased surface area = $6\times {\left(1.5L\right)}^2=13.5L^2$
Then, % increase in surface area
$=\frac{Increasedsurfacearea-Originalsurfacearea}{Originalsurfacearea}\times 100$
$=\frac{13.5L^2-6L^2}{6L^2}\times 100=125\%$

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

TSA of the original cube $=6a^2$

TSA of new cube
$=6\times {\left(\frac{3a}{2}\right)}^2=6\times \frac{9a^2}{4}=2.25\times 6a^2$

Increase in area $=[2.25\times 6a^2]-\left[6a^2\right]=(2.25-1)\times 6a^2=1.25\times 6a^2=1.25\times \text{TSA of the original cube}$

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