Five neighborhood children (two of whom- Henry and Kirby- are boys, and three of whom-Iris, Josephine, and Louise-are girls) are counting their marbles. Each child is a different age (7 through 11 years) and has at least one but no more than 5 of each of aggies, alleys, immies, mibs, and steelies. No two children have the same number of the same type of marble. From the information provided determine each child's age and the number of each type of marble in his or her collection.
1. The five children are Josephine, the 10 year old, the child who owns 2 alleys, the girl who has 3 immies, and the child who has 4 mibs.
2. Each child owns exactly 15 marbles.
3. The boy who has five steelies is older than at least one other child.
4. The number of steelies in one child's collection is exactly half his or her age.
5. The 11 year old child has one aggie.
6. The only types of marble of which Iris has an even number are immies and mibs.
7. One of the boys has 5 aggies.
8. Kirby doesn't own 2 of any type of marble.
9. The child with 3 immies has fewer than 3 mibs and fewer than 3 steelies.
10. One child has 3 aggies, 5 immies, and 1 steelie.
11. Each of at least 2 children has exactly twice as many alleys as immies.
(In reply to
re: Answer + Solution (explanation) by Erik O.)
Thanks. Did you get hung up on the idea of one boy have 5 aggies and the OTHER boy having 5 steelies too? Like I said, that messed me up for a while.
And Catherine, I would just like to say I really enjoyed this puzzle. Not to put others down, but there are many other similar types of logic problems (you know what I mean, the "find all the characteristics of each person" kind) that either have not enough information or redundant information. There's nothing really wrong with redundant info... it's just cleaner and nicer without it :)
Well done!
|
Posted by nikki
on 2005-04-22 22:10:13 |