Difference between revisions of "2013 AMC 12A Problems/Problem 6"

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== Problem==
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In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on <math>20\%</math> of her three-point shots and <math>30\%</math> of her two-point shots. Shenille attempted <math>30</math> shots. How many points did she score?
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<math> \textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36 </math>
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==Solution==
 
Let the number of 3-point shots attempted be <math>x</math>. Since she attempted 30 shots, the number of 2-point shots attempted must be <math>30 - x</math>.
 
Let the number of 3-point shots attempted be <math>x</math>. Since she attempted 30 shots, the number of 2-point shots attempted must be <math>30 - x</math>.
  
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<math>18</math>, which is <math>B</math>
 
<math>18</math>, which is <math>B</math>
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}

Revision as of 17:34, 22 February 2013

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 $B$

See also

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