Difference between revisions of "2011 AMC 10A Problems/Problem 12"

(Solution)
Line 7: Line 7:
  
 
Suppose there were <math>x</math> three-point shots, <math>y</math> two-point shots, and <math>z</math> one-point shots. Then we get the following system of equations:
 
Suppose there were <math>x</math> three-point shots, <math>y</math> two-point shots, and <math>z</math> one-point shots. Then we get the following system of equations:
<cmath>\begin{array}
+
<cmath>\begin{align}
 
3x=2y\\ z=y+1\\ 3x+2y+z=61
 
3x=2y\\ z=y+1\\ 3x+2y+z=61
\end{array}</cmath>
+
\end{align}</cmath>
  
 
The value we are looking for is <math>z</math>, which is easily found to be <math>z=\boxed{13 \ \mathbf{(A)}}</math>.
 
The value we are looking for is <math>z</math>, which is easily found to be <math>z=\boxed{13 \ \mathbf{(A)}}</math>.

Revision as of 00:13, 16 February 2011

Problem 12

The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?

$\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17$

Solution

Suppose there were $x$ three-point shots, $y$ two-point shots, and $z$ one-point shots. Then we get the following system of equations: \begin{align} 3x=2y\\ z=y+1\\ 3x+2y+z=61 \end{align}

The value we are looking for is $z$, which is easily found to be $z=\boxed{13 \ \mathbf{(A)}}$.

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