I have this problem with the fraction: Sylvia is among 12 students to have successfully taken successfully taken the challenging General Mathematics exam at the university. To pass it, in fact, it was necessary to succeed in both the first test and the second. She determines the number of students initially enrolled in the exam, knowing that the outcome of the first test was negative for $2/3$ of those enrolled and that the outcome of the second test was turned out negative for the $3/5$ of the remainder.
Let $x$ be the number of initially enrolled students for the exam.
In the first test, $\frac{2}{3}$ of the students failed, so $ \frac{1}{3}$ passed the first test. The number of students who passed the first test is $\frac{1}{3}x $. After the first test, $ \frac{2}{3}x $ students remain to take the second test. For the second test, $ \frac{3}{5} $ of the remaining students failed. So, the number of students who passed both the first and the second tests is $ \frac{2}{5} $ of the remaining $ \frac{2}{3}x $.
$$\frac{2}{5} \times \frac{2}{3}x = \frac{4}{15}x$$
Now, since Sylvia is among the 12 students who passed both tests:
$$ \frac{4}{15}x = 12 $$
To solve for $x$:
$$ x = \frac{12 \times 15}{4} = 45 $$
But the solution is $90$.
The outcome of the second test was negative for $\frac{3}{5}$ of the remainder, i.e. $\frac{3}{5}$ of the people that passed the first test, failed the second one. Recall that we had $\frac{x}{3}$ people who passed the first test. So you have $$ \frac{2}{5} \times \frac{x}{3} = 12 $$ from which you correctly get $x = 90$.
In short: you just mixed up two numbers.