2012 AMC 10B Problem 7

1. Let the chipmunk dig $h$ holes. With $3$ acorns per hole, it hid $3h$ acorns.
2. The squirrel dug $4$ fewer holes, so it dug $h - 4$ holes. With $4$ acorns per hole, it hid $4(h - 4) = 4h - 16$ acorns.
3. They hid the same number of acorns, so
$$3h = 4h - 16.$$
4. Take $3h$ away from both sides: $0 = h - 16$, so $h = 16$.
5. The question asks for acorns, not holes. The chipmunk hid $3 \times 16 = 48$ acorns.
6. Check: the squirrel dug $16 - 4 = 12$ holes and hid $4 \times 12 = 48$ acorns. The totals match.

The answer is $\boxed{\textbf{(D)}\ 48}$.

"Same total, different amounts per group" gives an equation: amount per group times number of groups, on each side. Another way to see it: the squirrel puts $1$ more acorn in each of its holes than the chipmunk does, and that has to make up for the $4 \times 4 = 16$ acorns in the $4$ holes it skipped. Always reread the question at the end, since here the easy mistake is to answer with the number of holes.

The total is a multiple of $3$ (chipmunk) and of $4$ (squirrel), so it is a multiple of $12$. Among the choices, only $36$ and $48$ are.

• $36$ acorns: the chipmunk digs $12$ holes and the squirrel digs $9$. The difference is $3$, not $4$.
• $48$ acorns: the chipmunk digs $16$ holes and the squirrel digs $12$. The difference is $4$. This one works.

Let the chipmunk dig $h$ holes. It hides $3h$ acorns. The squirrel digs $h - 4$ holes with $4$ acorns in each, hiding $4(h - 4)$ acorns. The totals are equal:
$$3h = 4(h - 4) = 4h - 16 \quad\Longrightarrow\quad h = 16.$$
The chipmunk hid $3 \cdot 16 = 48$ acorns. Check: the squirrel used $12$ holes, and $4 \cdot 12 = 48$.

The answer is $\boxed{\textbf{(D)}\ 48}$.