Difference between revisions of "2011 AMC 12A Problems/Problem 6"
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== Solution 2 == | == 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. | 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> We know that <math>2a+3b+1x=61.</math> We can see: | ||
| + | \begin{align*} | ||
| + | a&=x-1 \\ | ||
| + | 2a&=2x-2 \\ | ||
| + | 3b&=2x-2 \\ | ||
| + | 2a+3b+x&=2x-2+2x-2+x \\ | ||
| + | &=5x-4 \\ | ||
| + | &=61 \\ | ||
| + | 5x&=65 \\ | ||
| + | x&=\boxed{\text{(A)}~13}. \\ | ||
| + | \end{align*} | ||
==Video Solution == | ==Video Solution == | ||
Revision as of 18:33, 10 August 2023
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 
points. How many free throws did they make?
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 
ratio. Therefore, assume they made 
and 
two- and three- point shots, respectively, and thus 
free throws. The total number of points is ![\[2 \times (3x) + 3 \times (2x) + 1 \times (3x+1) = 15x+1\]](http://latex.artofproblemsolving.com/2/3/2/23266fed552512114a8d3bf8b07a2ec846b2faec.png)
Set that equal to , we get 
, and therefore the number of free throws they made 
Solution 2
Let 
be the number of free throws. Then the number of points scored by two-pointers is 
and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is 
, giving us 
for an answer.
Solution 3
We let 
be the number of 
-point shots, 
be the number of 
-point shots, and be the number of free throws. We are looking for 
We know that 
and that 
We know that 
We can see:
\begin{align*}
a&=x-1 \\
2a&=2x-2 \\
3b&=2x-2 \\
2a+3b+x&=2x-2+2x-2+x \\
&=5x-4 \\
&=61 \\
5x&=65 \\
x&=\boxed{\text{(A)}~13}. \\
\end{align*}
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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