In a calculus course, 65% of the students were males. At the end of the class 80% of the males and 60% of the females passed the course.
What percentage of those who passed the class were females? (round to the nearest percent)
We don't know the total number of students. Using \(x\) as the total number of students, we can write an expression for the number of students that passed.
$$ x \tag*{\tiny total students}$$
$$ x(0.65) \tag*{\tiny male students}$$
$$ x(0.65)(0.80) \tag*{\tiny male students that passed}$$
$$ x(1-0.65) \text{ or } x(0.35) \tag*{\tiny female students} $$
$$ x(0.35)(0.60) \tag*{\tiny female students that passed} $$
To find the percent of female students among those that passed:
$$ \frac{\text{female students that passed}}{\text{all students that passed}}$$
$$ = \frac{x(0.35)(0.60)}{x(0.65)(0.80)+x(0.35)(0.60)} $$
$$ = \frac{0.21x}{.52x+.21x} $$
$$ = \frac{.21\cancel{x}}{.73\cancel{x}} $$
$$ \approx .288 $$
$$ = \boxed{29\%} $$
We can pick an arbitrary number of students for the class. Since we're dealing with percents, we can pick 100 total students. Since 65% of the students are male, there will be 65 male students and 35 female students.
Similarly, among the male students, 80% of them pass:
$$ 65 \cdot (0.80) $$
$$ = 52 \text{ male students pass}$$
Doing the same for females:
$$ 35 \cdot (0.60) $$
$$ = 21 \text{ female students pass}$$
To find the percent of female students among those that passed:
$$ \frac{\text{female students that passed}}{\text{all students that passed}}$$
$$ =\frac{21}{52+21} $$
$$ =\frac{21}{73} $$
$$ \approx .288 $$
$$ = \boxed{29\%} $$