AMC 10 · 2010 · #8
Grade 6 logicPick an answer.
Name the number of schools with a letter so the total count and Andrea's median rank become simple expressions. The two teammate placements act as boundaries: they squeeze the median rank from above and below (Extreme Principle). Because the total must be odd and the schools count is a whole number, only one choice survives (Eliminate Possibilities).
Name the number of schools
The total is three times the number of schools.
One letter for the thing you want lets every other quantity be written in terms of it.
6.EE.A.2Introduce A VariableLocate the median rank
An untied median forces an odd total.
A single middle value only exists when the list has an odd number of entries.
A single middle value only exists when the list has an odd number of entries.
▸ Why?
The median is whatever stands in the middle of the ordered list, so it needs one entry to stand there.
▸ Why?
An even count leaves two entries sharing the middle, so a lone middle forces the count to be odd.
Use Beth to bound n above
Being her team's best sets an upper bound.
Beating a known rank puts a ceiling on where the median can sit.
6.EE.B.8Extreme PrincipleUse Carla to bound n below
The lowest teammate sets a lower bound.
A place number that actually happened proves at least that many people showed up.
6.EE.B.8Introduce A VariableKeep only the value that fits
Only 23 survives both bounds.
When a whole-number range holds just one value with the right parity, that value is forced.
6.EE.B.5Eliminate PossibilitiesConfirm the count
Recomputing the median confirms 23, choice (B).
Plugging the winner back in should make every clue in the story come out true.
6.SP.A.3Eliminate PossibilitiesName the unknown, turn each clue into a squeeze from above and below, then keep only the whole number with the right parity.
- Name the number of schools
- Locate the median rank
- Use Beth to bound n above
- Use Carla to bound n below
- Keep only the value that fits
- Confirm the count