Difference between revisions of "2013 AMC 10A Problems/Problem 9"
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| − | Let the number of attempted three-point shots made 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>(0.2\cdot3)x +(0.3\cdot2)y</math>, as we know that the | + | Let the number of attempted three-point shots made 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>(0.2\cdot3)x +(0.3\cdot2)y</math>, as we know that the three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them. |
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| − | three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them. | ||
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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> | ||
Revision as of 12:24, 27 January 2015
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
Let the number of attempted three-point shots made 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 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them.
Simplifying, we see that this is equal to 
. Plugging in , we get 
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
