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Alice, Betty, and Carol took the same series of examinations. For each examination there was one mark (grade) of $x$, one mark of $y$, and one mark of $z$; where $x, y, z$ are distinct positive integers. After all the examinations, Alice had a total score of $20$, Betty a total score of $10$, and carol a total score of $9$. If Betty placed first in Algebra, who placed second in Geometry?

What I analyzed(tell me if I'm wrong): There is 3 girls, and they take 3 exams each. If Algebra and Geometry are two of the exams, then should there should be a third subject as well?

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  • $\begingroup$ To answer the question asked in the title, Betty did. Because your problem statement says that Betty placed first in algebra. $\endgroup$ Sep 24, 2018 at 20:30
  • $\begingroup$ Sorry about that. In the quest of consolidating the title I completely missed that point of the question. $\endgroup$ Sep 24, 2018 at 20:39

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It doesn't explicitly state that there are three examinations, but the total score on one exam is $x+y+z$, so the total score over $n$ exams is $n(x+y+z)$. The total score over all the exams is given as 39 (20+10+9), so $n$ divides $39$. This means $n$ is either $3$ or $13$. If $n=13$, then $x+y+z=3$, but there aren't three distinct positive integers that add to $3$, so $n$ has to be $3$.

At first glance, since there are two more subjects, there is no way to tell which of those remaining scores correspond to geometry. However, if everyone scored the the same on both geometry and the third subject, then t doesn't matter which is geometry. So you need $x,y,z$ such that $x+y+z=13$, and $20$, $10$, and $9$ can be made by taking twice one of the numbers plus one of the other numbers.

Hint: since Alice averaged over $6$, at least one of $x,y,z$ must be greater than $6$.

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This is impossible; Based off Acccumulation's Answer, because there are 3 exams, and Betty placed first in one of them, all of the scores have to be less than 10. Therefore, because Alice has a total score of 20, the only possible largest numbers are 9, 8, and 7. Because they are all distinct and positive, the only solution is 7, 2, and 1; But, this does not create a working solution for Alice. Therefore, this is impossible.

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