The mean of a list of numbers is 42. When one more number, 66, is written at the end of the list, the mean becomes 44. How many numbers were in the list to begin with?

Let the list hold $n$ numbers, so its total is $42n$. Afterwards the total is $42n+66$ spread over $n+1$ numbers and equals 44, so $42n+66=44n+44$, giving $2n=22$ and $n=11$. Check: 11 numbers total 462, and $(462+66)÷12=528÷12=44$. A neater route: the new number is 66 − 44 = 22 above the new mean, and that surplus is exactly what lifts each of the other $n$ numbers by the 2 from 42 to 44, so $2n=22$. The value 12 is the new count and 22 is $2n$ left un-halved.