# 1950 AHSME Problems/Problem 34

## Problem

When the circumference of a toy balloon is increased from $20$ inches to $25$ inches, the radius is increased by:

$\textbf{(A)}\ 5\text{ in} \qquad \textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad \textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad \textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad \textbf{(E)}\ \dfrac{\pi}{5}\text{ in}$

## Solutions

### 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 $\frac{circumference}{2 \pi}$

$$2 \pi r = 20$$ $$r = \frac{10}{\pi}$$

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 $$\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}$$ $$4r_2=5r_1$$ $$4(r_2-r_1)=r_1$$ $$r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$$