16) Among the 200 students of a school, 150 participated in a chemistry competition and 130 participated in a physics competition. How many students participated in both competitions?
(A) 70, (B) 80,
(E) it is not possible to determine the

number based on the data in the problem.
16. The answer is $(\mathbf{E})$.

Let $N$ be the number of students who did not participate in any competition, $F$ and $C$ the number of those who participated only in physics and chemistry competitions respectively, and $E$ the number of those who participated in both.

The problem provides the following information:

$$
F+E=130, \quad C+E=150, \quad F+C+E+N=200 .
$$

One might immediately observe that this is a system of 3 linear equations in 4 unknowns, whose solution, if it exists, cannot be unique. This allows us to rule out answers A, $\mathrm{B}, \mathrm{C}, \mathrm{D}$, so the correct answer can only be $(\mathbf{E})$.

Proceeding constructively, we observe instead that:

$$
130+150=(F+E)+(C+E)=(F+C+E)+E=(200-N)+E,
$$

which means $E=80+N$. The maximum value of $N$ is 50, corresponding to the case where all 130 participants in the physics competition also participated in the chemistry competition. $N$ must necessarily be a non-negative number, so it can take any integer value between 0 and 50. For each valid value of $N$, there will be a different solution, with $E$ taking any integer value between 80 and 130 inclusive. The problem therefore admits 51 distinct solutions, so the data provided are not sufficient to determine a specific value of $E$.