Problem 17 · 2003 AMC 8
Hard
Logic & Word Problems
casework
The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?
| Child | Eye Color | Hair Color |
|---|---|---|
| Benjamin | Blue | Black |
| Jim | Brown | Blond |
| Nadeen | Brown | Black |
| Austin | Blue | Blond |
| Tevyn | Blue | Black |
| Sue | Blue | Blond |
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Answer: E — Austin and Sue.
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Hint 1 of 2
A sibling of Jim must MATCH Jim on eyes or hair — cross off anyone who shares neither of his traits.
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Hint 2 of 2
Look for three children who share a single trait among themselves; if they make one valid family, the leftovers must be the other.
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Approach: shortlist by Jim's traits, then split the six into two valid families
- Jim has brown eyes and blond hair. A sibling must match at least one, so anyone with blue eyes and black hair is out. That leaves only Nadeen (brown eyes), Austin (blond), and Sue (blond) as possible siblings — and Jim has exactly 2 siblings, so two of these three are his.
- Now use the "other family" to break the tie. Benjamin, Nadeen, and Tevyn all have black hair — they form a complete, valid family of three on their own. That uses up Nadeen.
- The remaining three — Jim, Austin, Sue — are all blond, a valid family too. So Jim's siblings are Austin and Sue.
- You'll see this again: when a group must split into valid sub-groups, finding one forced sub-group (the three black-haired kids) automatically determines the other — you rarely have to test every option.
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