Let the total number of students in the class be represented by \( x \).
According to the problem:
- One-third of the students scored 90% or better, which is \( \frac{1}{3} x \).
- Two-fifths of the students scored between 80% and 89%, which is \( \frac{2}{5} x \).
To determine the fraction of students who scored below 80%, we first need to add the fractions of students who scored 90% or better and those who scored 80% - 89%.
The total fraction of students scoring 90% or better and 80% - 89% is: \[ \frac{1}{3} + \frac{2}{5} \]
To add these fractions, we need a common denominator, which is 15. We will convert each fraction:
\[ \frac{1}{3} = \frac{5}{15} \] \[ \frac{2}{5} = \frac{6}{15} \]
Now, we can add them:
\[ \frac{5}{15} + \frac{6}{15} = \frac{11}{15} \]
This means that \( \frac{11}{15} \) of the students scored either 90% or better or between 80% and 89%. Thus, the fraction of students who scored below 80% can be found by subtracting this from 1:
\[ 1 - \frac{11}{15} = \frac{15}{15} - \frac{11}{15} = \frac{4}{15} \]
Therefore, the fraction of students who scored below 80% on the test is \( \frac{4}{15} \).
Thus, the answer is:
\[ \boxed{\frac{4}{15}} \]