Problem 195

Solution 1
When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be $\pi$ anymore)
We see that the circumference was increased by $25\%$. This means the radius was also increased by $25\%$. The radius of the original balloon is $\frac{20}{2\pi}=\frac{10}{\pi}$. With the $25\%$ increase, it becomes $\frac{12.5}{\pi}$. The increase is $\frac{12.5-10}{\pi}=\frac{2.5}{\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$.

Solution 2
Circumference of a circle is $2 \pi r$ so the radius is $\frac{circumference}{2 \pi}$
So radius of first circle
$$2 \pi r = 20$$
$$r = \frac{10}{\pi}$$
Radius of second circle
$$2 \pi r = 25$$
$$r = \frac{25}{2 \pi}$$
The difference of these radii is
$$\frac{25}{2 \pi} - \frac{10}{\pi} = \frac{5}{2 \pi}$$
So the answer is $\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$.

Solution 3
Let the radius of the circle with the larger circumference be $r_2$ and the circle with the smaller circumference be $r_1$. Calculating the ratio of the two
$$\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}$$
$$4r_2=5r_1$$
$$4(r_2-r_1)=r_1$$
$$r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$$