CEMC Fryer, 2016

Algebra Difficulty 2.0 Junior Prove it Canada

Three schools each sent four students to a competition. The scores earned by nine of the students are given in the table below. The scores of the remaining three students are represented by $x,y$ and $z$ . The total score for any school is determined by adding the scores of the four students competing from the school.

	Student 1	Student 2	Student 3	Student 4
School A	12	8	10	6
School B	17	5	7	$x$
School C	9	15	$y$	$z$

What is the total score for School A?
The total scores for Schools A and B are the same. What is the value of $x$ , the score for Student 4 at School B?
The total scores for Schools A and C are the same. If the score for Student 4 at School C is twice that of Student 3 at School C, determine these two scores.

Solution

The total score for School A is found by adding the scores of the four students competing from the school.

Therefore, the total score for School A is $12+8+10+6=36$ .
Since the total score for School A is 36, then the total score for School B is also 36.

The scores of the four students competing from School B are $17,5,7$ , and $x$ , and so $17+5+7+x=36$ or $29+x=36$ and so $x=7$ .
The score for Student 4 at School C, $z$ , is twice that of Student 3 at School C, $y$ .

Thus, $z=2y$ .

The scores of the four students competing from School C are $9,15,y$ , and $z$ .

Since the total score for School C is also 36, and $z=2y$ , then $9+15+y+2y=36$ or $24+3y=36$ or $3y=12$ , and so $y=4$ .

Therefore, the score for Student 3 at School C is 4, and the score for Student 4 at School C is $2(4)=8$ .

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