Difference between revisions of "2011 AMC 12A Problems/Problem 6"
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\textbf{(E)}\ 17 </math> | \textbf{(E)}\ 17 </math> | ||
− | == Solution == | + | == 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 <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> | 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> | ||
− | 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{ | + | 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> |
+ | |||
+ | == 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 \\ | ||
+ | 2a &= 2x-2 \\ | ||
+ | 3a &= 2x-2. \\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Plugging this into <math>2a+3b+1x=61</math>, we see | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 2x-2+2x-2+x &= 61 \\ | ||
+ | 5x-4 &= 61 \\ | ||
+ | 5x &= 65 \\ | ||
+ | x &= \boxed{\textbf{(A) }13}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | ~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 00:18, 16 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 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
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
. Also,
. We can see
Plugging this into , we see
~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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