Difference between revisions of "2002 AMC 12P Problems/Problem 22"
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Either way, we find <math>c, u, w \implies c-5, u+12, w-7.</math> If there are three ways, this implies the first way <math>c, u, w</math> the second way is <math>c, u, w \implies c-5, u+12, w-7,</math> and the third way is <math>c, u, w \implies c-10, u+24, w-14.</math> It follows that our three cases' first ways are as follows: | Either way, we find <math>c, u, w \implies c-5, u+12, w-7.</math> If there are three ways, this implies the first way <math>c, u, w</math> the second way is <math>c, u, w \implies c-5, u+12, w-7,</math> and the third way is <math>c, u, w \implies c-10, u+24, w-14.</math> It follows that our three cases' first ways are as follows: | ||
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<math>11</math> correct answers, <math>0</math> unanswered questions, <math>14</math> wrong answers | <math>11</math> correct answers, <math>0</math> unanswered questions, <math>14</math> wrong answers | ||
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<math>10</math> correct answers, <math>0</math> unanswered questions, <math>15</math> wrong answers | <math>10</math> correct answers, <math>0</math> unanswered questions, <math>15</math> wrong answers | ||
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<math>10</math> correct answers, <math>1</math> unanswered questions, <math>14</math> wrong answers. | <math>10</math> correct answers, <math>1</math> unanswered questions, <math>14</math> wrong answers. | ||
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Thus, our answer is <math>6(11+10+10)+2.5(0+0+1)+0(14+15+14)=\boxed {\text{(D) }188.5}</math> | Thus, our answer is <math>6(11+10+10)+2.5(0+0+1)+0(14+15+14)=\boxed {\text{(D) }188.5}</math> | ||
Revision as of 00:29, 13 July 2024
Contents
Problem
Under the new AMC 
scoring method, 
poitns are given for each correct answer, 
points are given for each unanswered question, and no points are given for an incorrect answer. Some of the possible scores between 
and 
can be obtained in only one way, for example, the only way to obtain a score of 
is to have 
correct answers and one unanswered question. Some scores can be obtained in exactly two ways, for example, a score of 
can be obtained with 
correct answers, 
unanswered question, and 
incorrect, and also with 
correct answers and 
unanswered questions. There are (three) scores that can be obtained in exactly three ways. What is their sum?
Solution 1
Let 
denote the amount of correct answers, 
denote the amount of unanswered questions, and 
denote the amount of wrong answers.
There are two ways to realize the conversion rate of wrong answers to unanswered questions and correct answers to unanswered questions. We will start with the cheesiest method:
Sub Solution 1.1
Notice how the way to get points directly tells us the conversion rate. correct questions, unanswered question, and wrong answers is equivalent to correct answers, unanswered questions, and wrong answers. Since 
and 
, the conversion rate is 
Sub Solution 1.2
The second way to find out the conversion rate is to actually do the math. If a wrong answer is converted to an unanswered question, the total value will be more than the total value. If a correct answer is converted to an unanswered question, the total value will be 
less than the total value. Thus, we want the "more than" to be equal to the "less than".
Let 
denote the amount of correct answers converted to unanswered questions and 
denote the amount of wrong answers converted to unanswered questions. Since we want "more than" to be equal to the "less than", 
, where and are integers. Multiply by 
on both sides, 
implies 
and 
Again, the conversion rate is
Sub Solutions Combine
Either way, we find If there are three ways, this implies the first way 
the second way is 
and the third way is 
It follows that our three cases' first ways are as follows:
correct answers, unanswered questions, 
wrong answers
correct answers, unanswered questions, 
wrong answers
correct answers, unanswered questions, wrong answers.
Thus, our answer is 
See also
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 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. 
