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)

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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:

$$\frac{\text{female students that passed}}{\text{all students that passed}}$$ $$=\frac{21}{52+21}$$ $$=\frac{21}{73}$$ $$\approx .288$$ $$= \boxed{29\%}$$