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Difference between revisions of "2013 AMC 10A Problems/Problem 9"

(Created page with "==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-po...")
 
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==Problem==
 
==Problem==
  
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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?
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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?
  
  

Revision as of 20:49, 7 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 attempted three-point shots made be $x$ and the number of attempted two-point shots be $y$. We know that $x+y=30$, and we need to evaluate $(0.2\cdot3)x +(0.3\cdot2)y$, 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 $0.6(30) = 18$, $\textbf{(B)}$.

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