## Task B-3.7.

It has been established that the number of student absences in the third grade during the school year can be described by the function $f(t)=2^{t+a}+b$, where $t$ is the number of months that have passed since the start of the school year on September 1st, $f(t)$ is the total number of absences in the first $t$ months, and $a$ and $b$ are constants. This year, students were absent for 128 hours in December.

a) After how many months will the students have 4064 hours of absences?

b) How many unexcused absences did the students have in December if the ratio of excused to unexcused absences in December was 29:3?
## Solution.

a) First, let's determine the constants $a$ and $b$ in the function $f(t)=2^{t+a}+b$.

Given that students were absent for 128 hours in December, we write:

$$
f(3)-f(2)=128
$$

According to the given rule, we have:

$$
\begin{gathered}
2^{3+a}+b-\left(2^{2+a}+b\right)=128 \\
8 \cdot 2^{a}-4 \cdot 2^{a}=128 \\
4 \cdot 2^{a}=128 \\
2^{a}=32 \\
a=5
\end{gathered}
$$

Since there were 0 absences at the beginning of the school year, we have $f(0)=2^{a}+b=0$, so $b=-2^{a}$, or $b=-32$.

Thus, $f(t)=2^{t+5}-32$. Now, let's calculate the required number of months. We have:

$$
\begin{gathered}
2^{t+5}-32=4064 \\
2^{t+5}=4096 \\
2^{t+5}=2^{12} \\
t+5=12 \\
t=7
\end{gathered}
$$

After 7 months, they will have 4096 hours of absences.

b) In December, the number of absences is

$$
f(4)-f(3)=2^{4+5}+32-\left(2^{3+5}+32\right)=2^{9}-2^{8}=2^{8}(2-1)=256
$$

The ratio of justified to unjustified absences is $O: N=29: 3$.

$$
\begin{gathered}
O=29 k, \quad N=3 k \\
29 k+3 k=256 \\
32 k=256 \\
k=8
\end{gathered}
$$

The number of unjustified absences is 24.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

4th grade - high school - B variant
January 5, 2023.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.