Problem 11 · 1993 AJHSME
Hard
Arithmetic & Operations
mediancumulative-count
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Answer: C — 70.
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Hint 1 of 2
You don't need the actual scores — the median is just the middle student when everyone lines up shortest-score to tallest. With 81 students lined up, who's in the exact middle? (Hint: 81 is odd, so there's a single middle person.)
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Hint 2 of 2
Don't read the tallest bar — that's the most common score, not the median. Instead walk along the bars from the low end, adding heights, and stop the moment your running count reaches the middle position.
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Approach: find the middle student's position, then walk the bars
- With 81 students, the middle one is the (81+1)/2 = 41st from the bottom. The median is whichever interval that 41st student falls in.
- Add bar heights from the lowest interval up: 1, then 1+2=3, +4=7, +5=12, +6=18, +10=28, +14=42. The total first reaches 41 inside the bar of height 14 — the interval labeled 70.
- Trap to dodge: the median is a position, not a height. The tallest bar (16, at 75) is the most frequent score, and it's tempting — but the 41st student lands one interval earlier. Counting up to the middle position is what separates median from mode.
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