AHSME 1950 Problem 34

When the circumference of a toy balloon is increased from $20$ inches to $25$ inches, the radius is increased by: $\textbf{(A)}\ 5\text{ in} \textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \textbf{(E)}\ \dfrac{\pi}{5}\text{ in}$