If the length and breadth of a rectangle is increased by $100\% $,then what will be the percentage increase in the area?
$\begin{gathered}
  A - 100 \\
  B - 200 \\
  C - 300 \\
  D - 400 \\
\end{gathered} $

Hint: Though this sum might look easy to solve, it becomes slightly difficult for those who are not well versed with percentage. In this Particular sum, the student has to first assume variables for the area of rectangle, length and breadth of the rectangle. Then increase its length and breadth by $100\% $. After this step the student has to apply the percentage formula once again to come to the final answer. A student is advised to go step by step in this sum otherwise he may get confused if he misses one of the steps.

Complete answer:
Let us assume the Area of the given Rectangle be $A sq.units$
Similarly , let us assume length of the Rectangle as $L$ units
 let us assume Breadth of the Rectangle as$B$units
Area of the Rectangle before increase in its dimensions is given be
$A = L \times B........(1)$
Now ,we have to find the length of the rectangle if its length is increased by $100\% $. Let us assume the New length of the rectangle as ${L_1}$.
For $100\% $ increase we have to find the value of ${L_1}$ in terms of $L$.
Formula to be used is $(\dfrac{{New - Original}}{{Original}}) \times 100 = \% Increase$
$\therefore \dfrac{{{L_1} - L}}{L} \times 100 = 100...........(2)$
$\therefore {L_1} = 2L......(3)$
Similarly we have to follow same steps for Breadth
We have to find the Breadth of the rectangle if its breadth is increased by $100\% $. Let us assume the New breadth of the rectangle as ${B_1}$.
For $100\% $ increase we have to find the value of ${B_1}$ in terms of $B$.
$\therefore \dfrac{{{B_1} - B}}{B} \times 100 = 100...........(4)$
\[\therefore {B_1} = 2B......(5)\]
Let's say that the New area is ${A_1}$.
New area of the rectangle is ${A_1} = {L_1} \times {B_1}$
But from equation $3\& 5$, we can also say that
${A_1} = 2L \times 2B$
$ \Rightarrow {A_1} = 4LB$
But from equation $1$ we can say that
${A_1} = 4A$
Therefore Percentage increase in the Area is
$\therefore \dfrac{{4A - A}}{A} \times 100 = 300\% $

Thus the answer for this numerical is option $C - 300$

Note: In order to solve this sum, the student should have knowledge of percentage increase and the area of the rectangle. There can be similar sums where it is asked to find percentage increase but the formulae for the volumes and perimeter might be complicated. In such cases the student has to carefully apply the formula and follow a step by step approach.