2011 AMC 10A Problem 12

Both conditions compare something with the two-point shots, so let the number of two-point shots be the unknown. Call it $y$.

1. Two-point shots: $y$ shots, worth $2y$ points.
2. Three-point shots: the team scored the same number of points with these as with two-pointers, so they are worth $2y$ points too. (We never need to know how many shots that was.)
3. Free throws: there were one more of these than two-point shots, so $y + 1$ free throws. Each is worth $1$ point, so they give $y + 1$ points.
4. Total: add the points and set them equal to $61$:
$$2y + 2y + (y + 1) = 61.$$
That is $5y + 1 = 61$, so $5y = 60$ and $y = 12$.
5. The number of free throws is $y + 1 = 13$.

Check: $12$ two-pointers make $24$ points, $8$ three-pointers make $24$ points, and $13$ free throws make $13$ points. $24 + 24 + 13 = 61$. ✓

The answer is $\boxed{\textbf{(A)}\ 13}$.

Pick the unknown that every condition refers to, and each other amount can be written in terms of it. Read each condition carefully: one compares points, the other compares number of shots. Mixing those up is the usual mistake.

Let $y$ be the number of two-point shots. They scored $2y$ points from these, and the three-point shots produced the same number of points, $2y$. The free throws numbered $y + 1$, worth $y + 1$ points.

Total score:
$$2y + 2y + (y + 1) = 5y + 1 = 61,$$
so $y = 12$. The number of free throws is $y + 1 = 13$.

(Check: $12$ two-pointers give $24$ points, $8$ three-pointers give $24$ points, $13$ free throws give $13$; total $61$.)

The answer is $\boxed{\textbf{(A)}\ 13}$.