Difference between revisions of "2013 AMC 10A Problems/Problem 9"
(→Solution 2) |
m (→Solution 1) |
||
(6 intermediate revisions by 5 users not shown) | |||
Line 2: | Line 2: | ||
In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on <math>20\%</math> of her three-point shots and <math>30\%</math> of her two-point shots. Shenille attempted <math>30</math> shots. How many points did she score? | In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on <math>20\%</math> of her three-point shots and <math>30\%</math> of her two-point shots. Shenille attempted <math>30</math> shots. How many points did she score? | ||
− | |||
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </math> | <math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Let the number of attempted three-point shots | + | 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> | ||
Line 14: | Line 13: | ||
==Solution 2 (cheap)== | ==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. 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. |
− | |||
− | - | ||
==See Also== | ==See Also== |
Latest revision as of 11:59, 26 July 2021
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.
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.