2021 AMC 10A Problem 5

Means cannot be subtracted from each other, but totals can. Since mean $=$ total $\div$ count, we have total $=$ mean $\times$ count. So turn each mean into a total first.

1. The whole class has $k$ students with mean $8$, so all the scores add up to $8k$.
2. The chosen $12$ scores have mean $14$, so they add up to $12 \times 14 = 168$.
3. The other students number $k - 12$, and their scores add up to what is left: $8k - 168$.
4. Their mean is their total divided by how many there are:
$$\frac{8k - 168}{k - 12}.$$

Quick check with a real number: if $k = 30$, the class total is $240$, the remaining $18$ students have $240 - 168 = 72$ points, and their mean is $72 \div 18 = 4$. The formula gives $\frac{240 - 168}{18} = 4$ too.

The answer is $\boxed{\textbf{(B)}\ \frac{8k-168}{k-12}}$.

For any "mean of the rest" question, the safe routine is: total of everyone, minus total of the group you know, divided by the number of people left. Working with totals avoids the tempting but wrong step of subtracting one mean from another.

Pick a number for $k$ and test the choices. With $k = 30$ the true answer is $4$ (worked out above). Choice (A) gives $\frac{6}{18}$, (C) gives about $0.9$, (D) gives $\frac{14 \times 18}{900} = 0.28$, and (E) gives $\frac{14 \times 18}{240} = 1.05$. Only (B) gives $4$.