One frequency $x$ is missing from the table below and the median is 55.
0-20 -> 4, 20-40 -> 6, 40-60 -> $x$, 60-80 -> 15, 80-100 -> 10.
Find $x$.

Here $n=35+x$ so $\frac{n}{2}=17.5+\frac{x}{2}$, and cf = 10 for the class before 40-60. Since the median 55 lies in 40-60, $l=40,f=x,h=20:55=40+\left(\frac{17.5+x/2-10}{x}\right)\times 20$, so $15x=20\left(7.5+\frac{x}{2}\right)=150+10x$ and $x=30$. Check: 4, 6, 30, 15, 10 gives $n=65$, cumulative 4, 10, 40, 55, 65, and $40+\frac{32.5-10}{30}\times 20=55$. Each wrong option is the exact result of one classic slip: $x=32$ uses $\frac{n+1}{2},x=10$ forgets that $x$ is part of $n$, and $x=6$ uses the cumulative frequency of the median class itself instead of the class before it.