# 2013 AMC 10A Problems/Problem 9

## Problem

In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score? $\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36$

## Solution 1

Let the number of attempted three-point shots be $x$ and the number of attempted two-point shots be $y$. We know that $x+y=30$, and we need to evaluate $3(0.2x) + 2(0.3y)$, as we know that the three-point shots are worth $3$ points and that she made $20$% of them and that the two-point shots are worth $2$ and that she made $30$% of them.

Simplifying, we see that this is equal to $0.6x + 0.6y = 0.6(x+y)$. Plugging in $x+y=30$, we get $0.6(30) = \boxed{\textbf{(B) }18}$

## Solution 2 (cheap)

The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes $0.2 \cdot 30 = 6$ shots, which are worth $6 \cdot 3 = \boxed{\textbf{(B) }18}$ points. If we assume Shenille only attempts two-pointers, then she makes $0.3 \cdot 30 = 9$ shots, which are worth $9 \cdot 2 = 18$ points.

## Video Solution (CREATIVE THINKING)

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