2002 AMC 10A Problem 6

We know where Cindy ended up (43) and what she did, so we can walk backwards to the number she started with.

1. Her last step was dividing by $3$. To undo it, multiply by $3$: $43 \times 3 = 129$.

1. Her first step was subtracting $9$. To undo it, add $9$: $129 + 9 = 138$.

So the starting number is $138$. Check: $138 - 9 = 129$, and $129 \div 3 = 43$. Correct.

3. Now do what the teacher asked: subtract $3$, then divide by $9$:
$$138 - 3 = 135, \qquad 135 \div 9 = 15.$$

The answer is $\boxed{\textbf{(A)}\ 15}$.

A chain of steps can be undone by doing the opposite steps in the opposite order: undo the last thing first. Once the starting number is known, the correct calculation is quick. Watch out: the numbers $3$ and $9$ are swapped on purpose, and $138$ sits among the choices to catch students who stop early.

Let the starting number be $N$. Cindy's wrong procedure gives
$$\frac{N - 9}{3} = 43.$$
Reverse her steps: multiply by $3$ to get $N - 9 = 129$, then add $9$ to get $N = 138$.

Now do what the teacher asked: subtract $3$ and divide by $9$:
$$\frac{138 - 3}{9} = \frac{135}{9} = 15.$$

The answer is $\boxed{\textbf{(A)}\ 15}$.