2007 AMC 10A Problem 10

Averages are hard to combine, but totals just add. So turn each average into a total.

1. Name the unknown. Say there are $n$ children. Then the whole family has $n + 2$ people, and the mother with the children is $n + 1$ people.
2. Total of the mother and children. Average $16$ over $n + 1$ people: total $16(n + 1)$.
3. Total of the whole family. Average $20$ over $n + 2$ people: total $20(n + 2)$.
4. The difference is the father. The whole family is the mother and children plus the father, so
$$20(n + 2) - 16(n + 1) = 48.$$
5. Solve. $20n + 40 - 16n - 16 = 48$, so $4n + 24 = 48$, $4n = 24$ and $n = 6$.
6. Check. Mother and $6$ children: $7$ people, total $16 \times 7 = 112$. Add the father: $112 + 48 = 160$ over $8$ people, average $20$. Correct.

The answer is $\boxed{\textbf{(E)}\ 6}$.

When a problem gives the average of a group and the average of part of that group, multiply each average by the number of people to get totals. The person who is missing from the smaller group is just the difference of the two totals.

Think of balancing around the family average of $20$. The father is $28$ years above it. The mother and children average $16$, so each of them is $4$ below it on average. For the family average to be exactly $20$, the amounts above and below must cancel: $4 \times (\text{number of mother and children}) = 28$. So there are $7$ of them: the mother and $6$ children.