Difference between revisions of "2013 AMC 12A Problems/Problem 6"
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<math>18</math>, which is <math>B</math> | <math>18</math>, which is <math>B</math> | ||
| − | == | + | ==Solution 2== |
Since the problem doesn't specify the number of 3-point shots she attempted, it can be assumed that number doesn't matter, so let it be <math>0</math>. Then, she must have attempted <math>30</math> 2-point shots. So, her score must be: | Since the problem doesn't specify the number of 3-point shots she attempted, it can be assumed that number doesn't matter, so let it be <math>0</math>. Then, she must have attempted <math>30</math> 2-point shots. So, her score must be: | ||
<math>\frac{3}{10}*30*2</math>,which is <math>B</math>. | <math>\frac{3}{10}*30*2</math>,which is <math>B</math>. | ||
| − | + | ==Solution 3== | |
| − | == | ||
(similar to Solution 1, however a slightly more obvious way) | (similar to Solution 1, however a slightly more obvious way) | ||
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Because the total number of shots is <math>30</math>, <math>x + y = 30</math> | Because the total number of shots is <math>30</math>, <math>x + y = 30</math> | ||
| − | However, Shenille was only successful on <math>20%</math> of the 3-pt shots, and <math>30%</math> of the 2-pt shots, so | + | However, Shenille was only successful on <math>20\%</math> of the 3-pt shots, and <math>30\%</math> of the 2-pt shots, so |
<math>0.2x + 0.3y</math> = #number of successful shots | <math>0.2x + 0.3y</math> = #number of successful shots | ||
| Line 51: | Line 50: | ||
this evaluates to <math>0.6 (x + y)</math> = her score | this evaluates to <math>0.6 (x + y)</math> = her score | ||
| − | <math>x + y</math> is already determined to be 30, so her score is <math>0.6 (30) = | + | <math>x + y</math> is already determined to be 30, so her score is <math>0.6 (30) = \boxed{\textbf{(B)}\ 18}</math> |
| − | |||
~amuppalla | ~amuppalla | ||
Latest revision as of 11:39, 10 March 2024
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
Let the number of 3-point shots attempted be 
. Since she attempted 30 shots, the number of 2-point shots attempted must be 
.
Since she was successful on , or 
, of her 3-pointers, and , or 
, of her 2-pointers, then her score must be
, which is 
Solution 2
Since the problem doesn't specify the number of 3-point shots she attempted, it can be assumed that number doesn't matter, so let it be 
. Then, she must have attempted 2-point shots. So, her score must be:
,which is .
Solution 3
(similar to Solution 1, however a slightly more obvious way)
Say that
x = # of 2-pt shots
y = # of 3-pt shots
Because the total number of shots is , 
However, Shenille was only successful on of the 3-pt shots, and of the 2-pt shots, so
= #number of successful shots
For each successful shot, there is an associated number of points with it.
Therefore, 
= her score
this evaluates to 
= her score
is already determined to be 30, so her score is 
~amuppalla
Additional note
It is also reasonably easy to find all possibilities for the number of two-point and three-point shots she made. Just note that both numbers of successful throws have to be integers. For " of her two-point shots" to be an integer we need the number of two-point shots to be divisible by 10. This only leaves four possibilities for the number of two-point shots: 0, 10, 20, or 30. Each of them also works for the three-point shots, and as shown above, for each of them the total number of points scored is the same.
Video Solution
https://youtu.be/CCjcMVtkVaQ ~sugar_rush
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. 
