1990 AMC 8 Problem 9

Attempt Problem 9 of the 1990 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1990 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

9.

The grading scale shown is used at Jones Junior High. The fifteen scores in Mr. Freeman’s class were: 89,89, 72,72, 54,54, 97,97, 77,77, 92,92, 85,85, 74,74, 75,75, 63,63, 84,84, 78,78, 71,71, 80,80, 90.90.
A 93–10093\text{--}100
B 85–9285\text{--}92
C 75–8475\text{--}84
D 70–7470\text{--}74
F 0–690\text{--}69
In Mr. Freeman’s class, what percent of the students received a grade of C?
20%20\%
25%25\%
30%30\%
3313%33\dfrac{1}{3}\%
40%40\%
Answer: D
Concepts:data and graph interpretationpercentage
Small hint:
Count how many scores fall in the range 75–8475\text{--}84
Big hint:
Divide that count by the 1515 students to get the fraction, then convert to a percent
Solution:
A grade of C corresponds to scores from 7575 to 8484. The qualifying scores are 77,77, 75,75, 84,84, 78,78, 8080, which is 55 students.
So the percent is 515=13=3313%.\dfrac{5}{15} = \dfrac13 = 33\dfrac13\%.
Thus, the correct answer is D .
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