2. Two math teachers are conducting a geometry test, checking each 10th-grade student's ability to solve problems and their knowledge of theory. The first teacher spends 5 and 7 minutes per student, respectively, while the second teacher spends 3 and 4 minutes per student. What is the minimum time they will need to interview 25 students?
# Answer: 110 minutes.

## Solution: (estimation + example)

Let the first teacher accept the test on problems from $X$ students, and on theory from $Y$ students. Then the second teacher accepts the test on problems from (25-X) students, and on theory from (25-Y) students. Let $T$ be the minimum time required for them to interview 25 students. Then, considering the condition of the problem, we can set up the following system of inequalities:

$$
\left\{\begin{array} { c } 
{ 5 X + 7 Y \leq T , } \\
{ 3 ( 2 5 - X ) + 4 ( 2 5 - Y ) \leq T . }
\end{array} \Rightarrow \left\{\begin{array} { c } 
{ 5 X + 7 Y \leq T } \\
{ - 3 X - 4 Y \leq T - 1 7 5 . }
\end{array} \Rightarrow \left\{\begin{array}{c}
15 X+21 Y \leq 3 T \\
-15 X-20 Y \leq 5 T-875
\end{array}\right.\right.\right.
$$

$$
\text { Hence } 0 \leq Y \leq 8 T-875 \Rightarrow 8 T \geq 875 \Rightarrow T \geq 110 \text {. }
$$

Therefore, it is impossible to accept the test from 25 students in less than 110 minutes.

Let's provide an example (of achieving the boundary):

The first teacher accepts tests on problems from 22 students, spending 110 minutes on this; the second teacher accepts tests on problems from the remaining three students, spending 9 minutes on this, and accepts tests on theory from all 25 students, spending 100 minutes on this, thus spending 109 minutes.