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Difference between revisions of "2011 AMC 12A Problems/Problem 6"

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== Solution ==
 
== Solution ==
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For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a <math>3:2</math> ratio. Therefore, assume they made <math>3x</math> and <math>2x</math> two- and three- point shots, respectively, and thus <math>3x+1</math> free throws. The total number of points is <cmath>2 \times (3x) + 3 \times (2x) + 1 \times (3x+1) = 15x+1</cmath>
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Set that equal to 61, we get <math>x = 4</math>, and therefore the number of free throws they made <math>3 \times 4 + 1 = 13 \Rightarrow \boxed{C}</math>
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=5|num-a=7|ab=A}}
 
{{AMC12 box|year=2011|num-b=5|num-a=7|ab=A}}

Revision as of 17:49, 10 February 2011

Problem

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?

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

Solution

For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a $3:2$ ratio. Therefore, assume they made $3x$ and $2x$ two- and three- point shots, respectively, and thus $3x+1$ free throws. The total number of points is \[2 \times (3x) + 3 \times (2x) + 1 \times (3x+1) = 15x+1\]

Set that equal to 61, we get $x = 4$, and therefore the number of free throws they made $3 \times 4 + 1 = 13 \Rightarrow \boxed{C}$

See also

2011 AMC 12A (Problems • Answer Key • Resources)
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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