If the [length](https://artofproblemsolving.com/wiki/index.php/Length) and width of a [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle) are each increased by $10\%$, then the [perimeter](https://artofproblemsolving.com/wiki/index.php/Perimeter) of the rectangle is increased by
$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$
Let the width be $w$ and the length be $l$. Then, the original perimeter is $2(w+l)$. 
After the increase, the new width and new length are $1.1w$ and $1.1l$, so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$.
Therefore, the percent change is 
\begin{align*} \frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \frac{.2(w+l)}{2(w+l)} \\ &= \frac{.2}{2} \\ &= 10\% \\ \end{align*}
$\boxed{\text{B}}$