In one class in the school, number of abscent students is $\frac{1}{6}$ of number of students who were present. When teacher sent one student to bring chalk, number of abscent students was $\frac{1}{5}$ of number of students who were present. How many students are in that class?
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 \(\boxed{7}\)