Ans: B
Let $l_0$, $w_0$ and $A_0$ be the original length, width and area of the rectangle respectively. Let $l_1$, $w_1$ and $A_1$ be the new length, width and area of the rectangle respectively. Hence we have
Let $l_0$, $w_0$ and $A_0$ be the original length, width and area of the rectangle respectively. Let $l_1$, $w_1$ and $A_1$ be the new length, width and area of the rectangle respectively. Hence we have
The original area $A_0=l_0w_0$.
The new area $A_1=l_1w_1$.
Note that
$\begin{array}{rcl}
l_1 & = & l_0(1+20\%) \\
w_1 & = & w_0(1+x\%)
\end{array}$
Therefore, $A_1=(1.2)(1+x\%)l_0w_0$.
Since the area is increased by $50\%$, we have
$\begin{array}{rcl}
\dfrac{A_1-A_0}{A_0} \times 100\% & = & 50\% \\
\dfrac{(1.2)(1+x\%)l_0w_0-l_0w_0}{l_0w_0} & = & 50\% \\
1.2+1.2x\%-1 & = & 50\% \\
1.2x\% +20\% & = & 50\% \\
1.2x & = & 30 \\
x & = & 25
\end{array}$
