Problem 19 · 1996 AJHSME
Hard
Fractions, Decimals & Percents
weighted-percent
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Answer: C — 32%.
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Hint 1 of 2
Big trap: you canNOT just average 22% and 40% to get 31% — the two schools have different sizes, so a percent at the bigger school counts for more. Turn the percents into actual HEADCOUNTS first.
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Hint 2 of 2
Count the real tennis fans at each school (a percent OF its own total), add them, then divide by everyone combined. This is a weighted average: the larger school pulls the result toward its percent.
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Approach: count heads, then divide by everyone
- Convert to actual people: East has 22% of 2000 = 440 tennis fans, West has 40% of 2500 = 1000. Together that's 1440 tennis fans.
- Divide by all students, 2000 + 2500 = 4500: 1440 ⁄ 4500 = 32%.
- Notice 32% is NOT the plain average of 22% and 40% (which is 31%). It leans toward 40% because West is the bigger school — that's a weighted average at work.
- Why this transfers: to combine percentages from groups of different sizes, never average the percents — turn each into a count, total the counts, and divide. Equal-size groups are the only time the shortcut average is correct.
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