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

Latest revision as of 23:18, 15 July 2024

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 1

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 $3 \times 4 + 1 = 13 \Rightarrow \boxed{A}$

Solution 2

Let $x$ be the number of free throws. Then the number of points scored by two-pointers is $2(x-1)$ and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is $x+4(x-1) = 61 \Rightarrow x=13$ , giving us $\boxed{(A)}$ for an answer.

Solution 3

We let $a$ be the number of $2$ -point shots, $b$ be the number of $3$ -point shots, and $x$ be the number of free throws. We are looking for $x.$ We know that $2a=3b$ , and that $x=a+1$ . Also, $2a+3b+1x=61$ . We can see

$\begin{align*} a&=x-1 \\ 2a &= 2x-2 \\ 3a &= 2x-2. \\ \end{align*}$

Plugging this into $2a+3b+1x=61$ , we see

$\begin{align*} 2x-2+2x-2+x &= 61 \\ 5x-4 &= 61 \\ 5x &= 65 \\ x &= \boxed{\textbf{(A) }13}. \end{align*}$

~Technodoggo

~MrThinker

Video Solution

https://www.youtube.com/watch?v=6tlqpAcmbz4 ~Shreyas S

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

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 == Solution 2 ==
 Let <math>x</math> be the number of free throws. Then the number of points scored by two-pointers is <math>2(x-1)</math> and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is <math>x+4(x-1) = 61 \Rightarrow x=13</math>, giving us <math>\boxed{(A)}</math> for an answer.
+==Solution 3==
+We let <math>a</math> be the number of <math>2</math>-point shots, <math>b</math> be the number of <math>3</math>-point shots, and <math>x</math> be the number of free throws. We are looking for <math>x.</math>
+We know that <math>2a=3b</math>, and that <math>x=a+1</math>. Also, <math>2a+3b+1x=61</math>. We can see
+<cmath>
+\begin{align*}
+a&=x-1 \\
+a &= 2x-2 \\
+a &= 2x-2. \\
+\end{align*}
+</cmath>
+Plugging this into <math>2a+3b+1x=61</math>, we see
+<cmath>
+\begin{align*}
+x-2+2x-2+x &= 61 \\
+x-4 &= 61 \\
+x &= 65 \\
+x &= \boxed{\textbf{(A) }13}.
+\end{align*}
+</cmath>
+~Technodoggo
+~MrThinker
 ==Video Solution ==
@@ Line 24: / Line 51: @@
 == See also ==
 {{AMC12 box|year=2011|num-b=5|num-a=7|ab=A}}
-{{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}
+{{AMC10 box|year=2011|num-b=11|num-a=13|ab=A}}
 {{MAA Notice}}

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