jmc

Let $S$ be the sum of all of Brian's test scores up to this point, and let $n$ be the number of tests Brian has taken up to this point. Thus the arithmetic mean of his scores now is $\frac{S}{n}$ and the arithmetic mean of his scores after getting a 98 on the last test will be $\frac{S+98}{n+1}$. This gives the system of equations: $$\begin{array}{rlrl}\frac{S}{n}& =91& \frac{S+98}{n+1}& =92\end{array}$$ From the first equation we have $S=91n$. Substituting this into the second equation gives: $$\begin{array}{rl}\frac{S+98}{n+1}& =92\\ S+98& =92(n+1)\\ 91n+98& =92n+92\\ 92n-91n& =98-92\\ n& =6\end{array}$$ So Brian has to take $n+1=\boxed{7}$ tests.