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

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\textbf{(E)}\ 17 </math>
 
\textbf{(E)}\ 17 </math>
  
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== Solution ==
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== Solution 1 ==
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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 <math>61</math>, we get <math>x = 4</math>, and therefore the number of free throws they made <math>3 \times 4 + 1 = 13 \Rightarrow \boxed{A}</math>
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 +
== 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.
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 +
==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*}
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a&=x-1 \\
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2a &= 2x-2 \\
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3a &= 2x-2. \\
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\end{align*}
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</cmath>
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 +
Plugging this into <math>2a+3b+1x=61</math>, we see
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 +
<cmath>
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\begin{align*}
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2x-2+2x-2+x &= 61 \\
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5x-4 &= 61 \\
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5x &= 65 \\
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x &= \boxed{\textbf{(A) }13}.
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\end{align*}
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</cmath>
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 +
~Technodoggo
 +
 
 +
~MrThinker
 +
 
 +
==Video Solution ==
 +
 
 +
https://www.youtube.com/watch?v=6tlqpAcmbz4
 +
~Shreyas S
 +
 
 
== 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}}
 +
{{AMC10 box|year=2011|num-b=11|num-a=13|ab=A}}
 +
{{MAA Notice}}

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

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
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

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