2013 AMC 12A Problems/Problem 6

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

Let the number of 3-point shots attempted be $x$. Since she attempted 30 shots, the number of 2-point shots attempted must be $30 - x$.

Since she was successful on $20\%$, or $\frac{1}{5}$, of her 3-pointers, and $30\%$, or $\frac{3}{10}$, of her 2-pointers, then her score must be

$\frac{1}{5}*3x + \frac{3}{10}*2(30-x)$


$\frac{3}{5}*x + \frac{3}{5}(30-x)$


$\frac{3}{5}(x+30-x)$


$\frac{3}{5}*30$


$18$, which is $B$

Alternative Solution

Since the problem doesn't specify the number of 3-point shots she attempted, it can be assumed that number doesn't matter, so let it be $0$. Then, she must have attempted $30$ 2-point shots. So, her score must be:

$\frac{3}{10}*30*2$,which is $B$.

Additional note

It is also reasonably easy to find all possibilities for the number of two-point and three-point shots she made. Just note that both numbers of successful throws have to be integers. For "$30\%$ of her two-point shots" to be an integer we need the number of two-point shots to be divisible by 10. This only leaves four possibilities for the number of two-point shots: 0, 10, 20, or 30. Each of them also works for the three-point shots, and as shown above, for each of them the total number of points scored is the same.

See also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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