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, 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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