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, 2018

Algebra Difficulty 1.0 Junior Prove it Canada

Mr. Singh gives his students a test each week.

Aneesh’s scores on the first six tests were 17, 13, 20, 12, 18, and 10. What was the average (mean) of his test scores?
Jon scored 17 and 12 on his first two tests. After the third test, his average (mean) score was 14. What was his score on the third test?
After the first six tests, Dina had an average (mean) test score of 14. On each of the next $n$ tests, Dina’s score was 20 out of 20. After all of these tests, her average (mean) test score was 18. Determine the value of $n$ .

Solution

The average of Aneesh’s first six test scores was $\dfrac{17+13+20+12+18+10}{6}=\dfrac{90}{6}=15$ .
After Jon’s third test, his average score was 14, and so the sum of the scores on his first three tests was $14\times3=42$ .

The sum of his scores on his first two tests was $17+12=29$ , and so the score on his third test was $42-29=13$ .
(We may check that the average of $17,12$ and 13 is $\dfrac{17+12+13}{3}=14$ .)
Dina wrote six tests followed by $n$ more tests, for a total of $n+6$ tests.

After Dina’s first $6$ tests, her average score was 14, and so the sum of the scores on her first $6$ tests was $14\times6=84$ .

Dina scored 20 on each of her next $n$ tests, and so the sum of the scores on her next $n$ tests was $20n$ .

Therefore, the sum of the scores on these $n+6$ tests was $84+20n$ .

After Dina’s $n+6$ tests, her average score was 18, and so the sum of the scores on her $n+6$ tests was $18(n+6)$ .
Thus, $84+20n=18(n+6)$ or $84+20n=18n+108$ or $2n=24$ , and so $n=12$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.