Section A has 30 students with mean marks 52. Section B has n students with mean marks 68. Taken together, the mean marks of both sections is 58. Find n.

$\frac{30(52)+68n}{30+n}=58$ gives $1560+68n=1740+58n$, so $10n=180$ and $n=18$. (By alligation: 58 - 52 = 6 and 68 - 58 = 10, so $n$ : 30 = 6 : 10, i.e. $n=18$.) "48" reports the combined strength $30+n$; "50" inverts the alligation ratio to 10 : 6; "6" stops at the distance 58 - 52 without finishing.