2007 AMC 10B Problem 12

Try a small case first to see what happens. Say Tom is $30$ and his children are $8$, $10$ and $12$ (sum $30$). One year ago Tom was $29$, but the children's ages added to $7 + 9 + 11 = 27$. The children's total dropped by $3$, one for each child.

1. Now: Tom is $T$ and the children's ages add to $T$.
2. $N$ years ago: Tom was $T - N$. Each of the three children was $N$ years younger, so their total was $T - 3N$.
3. Back then Tom's age was twice the children's total:
$$T - N = 2(T - 3N) = 2T - 6N.$$
4. Move the $T$'s to one side and the $N$'s to the other: $6N - N = 2T - T$, so $5N = T$.
5. Therefore
$$\frac{T}{N} = 5.$$

The answer is $\boxed{\textbf{(D)}\ 5}$.

Everyone gets older at the same speed, so age problems turn into simple equations. The one thing to watch is a total of several people's ages: going back $N$ years, a total of $3$ ages drops by $3N$, not by $N$.

Plug in a number. Take $N = 1$. Then $T - 1 = 2(T - 3)$, so $T - 1 = 2T - 6$ and $T = 5$. That gives $T/N = 5$. (A real family fitting these numbers would need $N$ larger, for example $N = 8$ and $T = 40$, but the ratio is the same.)