2013 AMC 10A Problem 9

The problem does not say how many shots of each kind she took. That is a hint that the answer does not depend on it. Let us see why.

1. Average points for one three-point try. She makes $20\%$ of them, and each make is worth $3$. On average one try gives $0.2 \times 3 = 0.6$ points. (For example, $10$ tries give $2$ makes, which is $6$ points, or $0.6$ per try.)
2. Average points for one two-point try. She makes $30\%$ of them, each worth $2$. On average one try gives $0.3 \times 2 = 0.6$ points. (Ten tries give $3$ makes, which is $6$ points.)
3. Every try is worth the same. Both kinds of shot give $0.6$ points per try. So it does not matter how the $30$ tries are split.
4. Total. $30 \times 0.6 = 18$ points.

Shenille scored $\boxed{\textbf{(B)}\ 18}$ points.

If a question leaves out a detail, the answer probably does not depend on it. Finding the "value per try" for each kind of shot shows the reason: both are $0.6$. Then the total is simply the number of tries times $0.6$.

Since the split does not matter, pick an easy one: all $30$ shots are three-pointers. She makes $20\%$ of $30$, which is $6$ shots, worth $6 \times 3 = 18$ points. With algebra: if $x$ shots are three-pointers, the points are $0.6x + 0.6(30 - x) = 18$, and the $x$ disappears.

Let $x$ be the number of three-point attempts, so there were $30 - x$ two-point attempts.

Points from three-pointers: $20\%$ of $x$ attempts succeed, each worth $3$, giving $0.2 \cdot 3x = 0.6x$.

Points from two-pointers: $30\%$ of $30 - x$ attempts succeed, each worth $2$, giving $0.3 \cdot 2(30 - x) = 0.6(30 - x)$.

Total:
$$0.6x + 0.6(30 - x) = 0.6 \cdot 30 = 18.$$

The $x$ cancels, so the split does not matter. Shenille scored $\boxed{\textbf{(B)}\ 18}$ points.