Question: Students A,B and C compete in a series of tests. For coming in first on any given test, one is awarded $x$ points; for placing second, $y$ points; and for third place, $z$ points. Let $x, y$, and $z$ be natural numbers with $x > y > z$. Assume no ties on any of the tests. Student A accumulated 20 points; student B got 10 points, and student C got 9 points. Student A came in second on the Algebra test. Who finished second on the Geometry test?
My Attempt: Logic would state that:
Student A came in 1st and 2nd, cause she has the highest points
Student B came in 3rd and 1st, cause B has more points than C
Student C came in 2nd and 3rd, cause C has less points than B
Let $x=$ 1st place point allocation, $y=$ 2nd place point allocation and $z=$ 3rd place point allocation.
Thus, Student C got 9 pts so $9 = y + z$, Student B got 10 points so $10 = x + z$ and Student A got 20 points so $20= x + y$. However, solving this out we get that $y = 19/2$ which is not a natural number. Any help?
The total score is $$A+B+C=20+10+9=39=3\times13$$ from which it is reasonable to assume there are three tests (one seldom speaks of two tests as a series) and that $$x+y+z=13$$ Because $A=20$, we must have $x\ge7$. If we assume $x\ge10$ then either (a) $A>20$ or (b) both $B,C\ge10$ of which are both conditions are false, so $x\le9$. So
$$(x,y,z)\in\{(7,6,0),(8,5,0),(8,4,1),(8,3,2),(9,2,0)\}$$
The only tuple that allows $A=20,B=10,C=9$ is $(8,4,1)$, for which
$A=8+8+4,B=1+1+8,C=4+4+1$. Since A came second with 4 points in Algebra, C must have come second in Algebra and the other subject (Trigonometry perhaps?).