Difference between revisions of "2002 AMC 12P Problems/Problem 22"
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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> |
== Solution 2 == | == Solution 2 == | ||
| − | This solution is formulated in | + | This solution is formulated in formal language rather than cheesy methods, though the idea is the same. Additionally, we do not need to take into account the amount of wrong answers, which is easier to deal with. |
| − | Let <math>c</math> denote the amount of correct answers, <math>u</math> denote the amount of unanswered questions | + | Let <math>c</math> denote the amount of correct answers, <math>u</math> denote the amount of unanswered questions. Then <math>c+u \leq 25.</math> We want to find the LCM of <math>2.5</math> and <math>6</math> so we can convert between different ways, similar to solution <math>1.</math> The LCM is equal to <math>30</math>, since <math>6 \cdot \textbf{5}=30</math> and <math>2.5 \cdot \textbf{12}=30.</math> |
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| + | The first way will take the form <math>(c,u).</math> The second way will take the form <math>(c+5, u-12).</math> The third way will take the form <math>(c+10, u-24).</math> In particular, since we cannot have a negative amount of unanswered questions, <math>u \leq 24.</math> Combined with <math>c+u \leq 25</math>, our three cases are <math>(0,24), (1,24),</math> and <math>(0,25).</math> | ||
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| + | Thus, our answer is <math>6(0+1+0) + 2.5(24+24+25)=\boxed {\text{(D) }188.5}.</math> | ||
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| + | ===Note=== | ||
| + | The conversion rate can either be <math>(c+5, u-12).</math> or <math>(c-5, u+12).</math> Both solutions use a different conversion rate to illustrate this point. The "first way" is defined as the one with the most correct in the first solution, while the second solution defines the "first way" as the one with the most unanswered in the second solution. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=P|num-b=21|num-a=23}} | {{AMC12 box|year=2002|ab=P|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 05:02, 14 July 2024
Contents
Problem
Under the new AMC 
scoring method, 
points 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. The conversion rate is what differentiates between two ways to get the same score.
There are two ways to realize the conversion rate of wrong answers to unanswered questions and correct answers to unanswered questions.
Sub Solution 1.1 (Cheese)
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 
Since 
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 
Solution 2
This solution is formulated in formal language rather than cheesy methods, though the idea is the same. Additionally, we do not need to take into account the amount of wrong answers, which is easier to deal with.
Let denote the amount of correct answers, denote the amount of unanswered questions. Then 
We want to find the LCM of and so we can convert between different ways, similar to solution 
The LCM is equal to 
, since 
and 
The first way will take the form 
The second way will take the form 
The third way will take the form 
In particular, since we cannot have a negative amount of unanswered questions, 
Combined with 
, our three cases are 
and 
Thus, our answer is 
Note
The conversion rate can either be or 
Both solutions use a different conversion rate to illustrate this point. The "first way" is defined as the one with the most correct in the first solution, while the second solution defines the "first way" as the one with the most unanswered in the second solution.
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. 
