2013 AMC 10B Problem 10

Everything in the problem depends on the number of three-point tries, so give that a name and build everything else from it.

1. Let $x$ be the number of three-point tries.
2. "$50\%$ more" two-point tries means $x$ plus half of $x$, which is $1.5x$ tries.
3. Points from two-pointers: half of the $1.5x$ tries go in, which is $0.75x$ baskets, and each is worth $2$ points: $2 \times 0.75x = 1.5x$ points.
4. Points from three-pointers: $40\%$ of $x$ go in, which is $0.4x$ baskets, each worth $3$ points: $3 \times 0.4x = 1.2x$ points.
5. The total is $54$:
$$1.5x + 1.2x = 2.7x = 54, \qquad x = \frac{54}{2.7} = 20.$$
6. Check: $20$ three-point tries means $30$ two-point tries. Then $15$ two-pointers go in for $30$ points and $8$ three-pointers go in for $24$ points. $30 + 24 = 54$. It works.

The answer is $\boxed{\textbf{(C)}\ 20}$.

Each percent is a multiplier: tries become baskets, and baskets become points. Writing that chain out step by step with one letter avoids mixing up "50% of" and "50% more than".

Try the answer choices. With $10$ three-point tries there are $15$ two-point tries; that gives $7.5$ two-pointers made, which is not a whole number, so (A) is out. With $20$ tries you get the check in step 6 above: exactly $54$ points. The total grows as the number of tries grows, so no other choice can also give $54$.