Difference between revisions of "2013 AMC 10A Problems/Problem 9"
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==Solution 1== | ==Solution 1== | ||
| − | Let the number of attempted three-point shots be <math>x</math> and the number of attempted two-point shots be <math>y</math>. We know that <math>x+y=30</math>, and we need to evaluate <math>(0. | + | Let the number of attempted three-point shots be <math>x</math> and the number of attempted two-point shots be <math>y</math>. We know that <math>x+y=30</math>, and we need to evaluate <math>3(0.2x) + 2(0.3y)</math>, as we know that the three-point shots are worth <math>3</math> points and that she made <math>20</math>% of them and that the two-point shots are worth <math>2</math> and that she made <math>30</math>% of them. |
Simplifying, we see that this is equal to <math>0.6x + 0.6y = 0.6(x+y)</math>. Plugging in <math>x+y=30</math>, we get <math>0.6(30) = \boxed{\textbf{(B) }18}</math> | Simplifying, we see that this is equal to <math>0.6x + 0.6y = 0.6(x+y)</math>. Plugging in <math>x+y=30</math>, we get <math>0.6(30) = \boxed{\textbf{(B) }18}</math> | ||
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The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes <math>0.2 \cdot 30 = 6</math> shots, which are worth <math>6 \cdot 3 = \boxed{\textbf{(B) }18}</math> points. If we assume Shenille only attempts two-pointers, then she makes <math>0.3 \cdot 30 = 9</math> shots, which are worth <math>9 \cdot 2 = 18</math> points. | The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes <math>0.2 \cdot 30 = 6</math> shots, which are worth <math>6 \cdot 3 = \boxed{\textbf{(B) }18}</math> points. If we assume Shenille only attempts two-pointers, then she makes <math>0.3 \cdot 30 = 9</math> shots, which are worth <math>9 \cdot 2 = 18</math> points. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/kXKTqoJdLTk | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
Latest revision as of 12:07, 1 July 2023
Contents
Problem
In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on 
of her three-point shots and 
of her two-point shots. Shenille attempted 
shots. How many points did she score?
Solution 1
Let the number of attempted three-point shots be 
and the number of attempted two-point shots be 
. We know that 
, and we need to evaluate 
, as we know that the three-point shots are worth 
points and that she made 
% of them and that the two-point shots are worth 
and that she made % of them.
Simplifying, we see that this is equal to 
. Plugging in , we get 
Solution 2 (cheap)
The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes 
shots, which are worth 
points. If we assume Shenille only attempts two-pointers, then she makes 
shots, which are worth 
points.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
| 2013 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
