Problem 24 · 2012 Math Kangaroo
Stretch
Logic & Word Problems
careful-countingsum-constraint
12 children were at a birthday party. The children were 6, 7, 8, 9, and 10 years old (every one of these ages was present). Four of them were 6 years old. There were more 8-year-olds than any other age group. What is the average age of the children?
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Answer: B — 7·5
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Hint 1 of 3
All five ages 6, 7, 8, 9 and 10 are present, so each of those groups has at least one child.
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Hint 2 of 3
The 8-year-olds must outnumber every other group, including the four 6-year-olds, so there are at least five 8-year-olds.
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Hint 3 of 3
Once the counts are pinned down, just add all twelve ages and divide by 12.
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Approach: pin down the counts from the clues, then average
- Four children are 6, so the other eight share the ages 7, 8, 9 and 10, with at least one child at each age.
- The 8-year-olds are the biggest group, so they must beat the four 6-year-olds: at least five children are 8.
- Five 8s plus one each of 7, 9 and 10 already uses 5 + 3 = 8 children, so the only fit is exactly one 7, five 8s, one 9 and one 10.
- The ages are four 6s, one 7, five 8s, one 9, one 10; their total is 24 + 7 + 40 + 9 + 10 = 90, and 90 / 12 = 7.5, choice B.
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