Problem 1370

1. Let $x$ be the number of students who were present initially.
2. Let $y$ be the number of students who were absent initially.

From the problem, we know:
$$y = \frac{1}{6}x$$

3. When the teacher sent one student to bring chalk, the number of present students becomes $x - 1$, and the number of absent students remains $y$.

4. At this point, the number of absent students is $\frac{1}{5}$ of the number of present students:
$$y = \frac{1}{5}(x - 1)$$

5. We now have two equations:
$$y = \frac{1}{6}x$$
$$y = \frac{1}{5}(x - 1)$$

6. Substitute $y$ from the first equation into the second equation:
$$\frac{1}{6}x = \frac{1}{5}(x - 1)$$

7. To eliminate the fractions, multiply both sides by 30 (the least common multiple of 6 and 5):
$$30 \cdot \frac{1}{6}x = 30 \cdot \frac{1}{5}(x - 1)$$
$$5x = 6(x - 1)$$

8. Simplify and solve for $x$:
$$5x = 6x - 6$$
$$5x - 6x = -6$$
$$-x = -6$$
$$x = 6$$

9. Substitute $x = 6$ back into the first equation to find $y$:
$$y = \frac{1}{6} \cdot 6$$
$$y = 1$$

10. The total number of students in the class is the sum of the present and absent students:
$$x + y = 6 + 1 = 7$$

The final answer is $\boxed{7}$