The mathlete team is exactly 37.5% female. If the number of boys remains unchanged what is the smallest number of girls that can leave the team to lower the percentage of girls to exactly 20%. How many boys and how many girls are originally on the team?
$f$ - females on the team
$m$ - males on the team
$x$ - dropped females
This is what I did
$$\frac f{f+m}= .375$$
$$\begin{align} f&= .375(f+m)\\ &= .375f+ .375m\\ .625f&= .375m\\ f&= .6m\\ &= 3/5 m \end{align}$$
$$\frac{f-x}{f-x+m}= .20$$
$$\begin{align} f-x&= .2(f-x+m)\\ &= .2f-.2x+.2m\\ .8f-.8x&= .2m\\ .8\left(\frac35m\right)-.8x&= .2m\\ .48m-.8x&= .2m\\ .28m&= .8x \end{align}$$
$$\begin{align} x&=\frac{.28}{.8} m\\ &=\frac{28}8 m\\ &=\frac{28}8(8)\\ &=28\text{ dropped females} \end{align}$$
This doesn't make sense though because if $f=\frac35m$ and $m=8$ then the number of females originally on the team is $4.8$ which can't be right and thus $28$ females can't drop. What did I do wrong?
It may help to know that $37.5 \% = 3/8$ and $20 \% = 1/5$.
So, the initial composition of the team must be (exactly) three girls for every five boys, and the final composition must be (exactly) one girl for every four boys.
Initially, $3m = 5f$, and afterwords, $m = 4(f-x)$.
We can eliminate $m$ to see that $7f = 12x$. The least common multiple of $7$ and $12$ is $84$, so $x = 7$ is the smallest number that can leave.
Finding the other quantities to check, $f=12$ and $m = 5f/3 = 20$. With $7$ females leaving, we're left with $f-x=5$ females and $20$ males, which gives the $20 \%$ final composition.
So ... what did you do wrong? The answer you got (28 females) is an answer, but not the smallest answer. At some point you needed to look at all of the numbers to see if there was a common factor. That was what I did along the way with the LCM observation.
That, and I'm not sure where $m=8$ came from. You had $x = 28m/8$, which is correct; it leads to $2x=7m$ and once again $x=7$. But you inserted $m=8$ for some reason, and things went awry.
Anyway, hope this helps!