Children, Humor

Riddle Me This One

A zebra walks into a bar – no wait, that’s the beginning of another type of story entirely.

Imagine this if you can.

Nine children invited to a birthday party.

Nine (different) stuffed safari animals hidden about the yard.

Nine children loosed upon the camouflaged beasts.

In a perfect equation, each child would find one animal and go home with said animal.

In reality, there are any number of permutations:

  • Overzealous  child with 20/20 (or x-ray) vision finds more than one (or all) animal(s)
  • Each child finds one animal, but it is not the one he or she wants
  • An all-out brawl, not unlike a lion mauling a zebra on the savannah, ensues over who gets which animal

Now imagine you’re taking one of those standardized tests that are all the rage these days and solve for Z (which is for zebra, by the way).  What is the solution?  What say you?

Well?

Well?

 

Standard

8 thoughts on “Riddle Me This One

  1. Little Mighty says:

    Maybe you could tell whoever finds an animal that the animal has to be put into a large covered box. When all the animals are in the box, then each child could be blindfolded and then can reach into the box and grab an animal. So that the birthday child doesn’t get disappointed, let that child have first choice. Hopefully, each child will be delighted with his prize. If not, you could suggest they could trade with another child, but only if he wants to trade.

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    • Jennifer Butler Basile says:

      Ooh, making a blind grab in a box of caged wild animals! Isn’t it against one’s better judgment to stick an appendage into an animal burrow, hole, cave, etc? Ha!

      That just may work. Aside from the implied lesson that wild animals should be removed from their habitat and caged 😉

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  2. Sid Dunnebacke says:

    Ooh, yes. Those are dangerous waters to tread. As my daughters are the algebra experts in the family, I’ll have to pose the question to them. 🙂

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