4. Barbara scored 98 points on the second-to-last test of the school year, thus increasing the average of the points she had achieved so far by 1 point. On the last test, she scored 70 points and lowered the new average by 2 points. How many tests did she take in the entire school year?
II/4. Namely, if Barbara had written $n$ tests and had an average of $m$ points before the last 2 tests, then

$$
\begin{aligned}
& \frac{n \cdot m+98}{n+1}=m+1 \\
\text { and } \quad & \frac{n \cdot m+98+70}{n+2}=m+1-2=m-1
\end{aligned}
$$

From the first equation, we get $n \cdot m+98=n \cdot m+n+m+1$ or $m+n=97$, and from the second, $n \cdot m+168=$ $n \cdot m-n+2 m-2$ or $2 m-n=170$. Thus, we arrive at $3 m=267$ or $m=89$ and $n=8$. This means that Barbara wrote 10 tests in total during the school year.

The equations (1) and (2) (or equivalent) based on the problem statement: 2+2 points. Solution of the system: 3 points.