Difference between revisions of "2013 AMC 10A Problems/Problem 9"
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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> | ||
− | ==Solution 2== | + | ==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 <math>0.2 \cdot 30 = 6</math> shots, which are worth <math>6 \cdot 3 = \boxed{\textbf{(B) }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. | ||
+ | |||
+ | -- I-_-I | ||
==See Also== | ==See Also== |
Revision as of 16:48, 22 December 2019
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
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.
-- I-_-I
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.