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) = 18</math>, <math>\textbf{(B)}</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) = 18</math>, <math>\textbf{(B)}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2013|ab=A|num-b=8|num-a=10}} | ||
| + | {{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}} | ||
Revision as of 20:59, 7 February 2013
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 | |
