If $p, q$ and $r$ are positive integers and $p+\frac{1}{q+\frac{1}{r}}=\frac{25}{19}$, then $q$ equals
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Solution 1

We know $p+\frac{1}{q+\frac{1}{r}}=\frac{25}{19}=1+\frac{6}{19}=1+\frac{1}{\frac{19}{6}}=1+\frac{1}{3+\frac{1}{6}}$.

Therefore, comparing the two fractions, $p=1, q=3$ and $r=6$.

## Solution 2

Since $p, q$ and $r$ are positive integers, then $q+\frac{1}{r}$ is at least 1 , so $\frac{1}{q+\frac{1}{r}}$ is between 0 and 1 .

Since $p+\frac{1}{q+\frac{1}{r}}$ is equal to $\frac{25}{19}$ which is between 1 and 2 , then $p$ must be equal to 1 .

Therefore, $\frac{1^{r}}{q+\frac{1}{r}}=\frac{25}{19}-1=\frac{6}{19}$ or $q+\frac{1}{r}=\frac{19}{6}$.

Since $r$ is a positive integer, then $\frac{1}{r}$ is between 0 and 1 , so since $\frac{19}{6}$ is between 3 and 4 , then $q=3$.

(We are not asked to determine what the value of $r$, but we can check that $r=6$.)

ANSWER: (C)