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The Flintstones are Fred & Wilma, with daughter Pebbles and pet Dino. The Rubbles are Barney & Betty, with son Bamm-Bamm and pet Hoppy.
The eight are having a brontoburger picnic in the backyard to celebrate Dino and Hoppy’s return. (They just got back from the vet after having been neutered. Ouch.) To help reduce the trauma of the event, the group randomly divided up into four teams to play a series of games. Each team was comprised of ONE adult, and either a child or a pet. (And no, teammates weren’t necessarily related to one another.)
Based on the following clues, can you determine the members of each team?
1. Barney’s teammate was not a child.
2. Either Wilma OR Betty’s teammate lived with her, but not both.
3. Betty’s teammate and Fred’s teammate were both males (or had been before the neutering – ouch).
HERE is the solution.
Comments (13)
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You have yesterdays answer on the link
posted by oose85 on 11-13-2008 at 8:33 am
I really didn’t know hoppy was a she. I am not used to the english but the spanish names. Better luck next time
posted by luis Alberto on 11-13-2008 at 8:56 am
Hoppy isn’t a she. It says that both the animals were neutered.
posted by flintstone on 11-13-2008 at 9:05 am
In fact, I added that whole part of the story (the pets being neutered) to make it clear that both pets were male, ’cause I had a feeling some might question whether or not Hoppy was male.
I hate to admit that I did it, but I did. :)
posted by Sandy on 11-13-2008 at 9:28 am
You can figure it out without knowing the gender of Hoppy. If Betty is paired with someone she lives with none of the other solutions could work. Because if Betty gets paired with Hoppy Fred has to have Dino or Bam Bam. Fred can’t be paired with Dino because Barney is paired with a pet and then Fred would have Bam Bam which leaves Wilma with Pebbles. By the second statement this can’t happen.
If Betty gets paired with Bam Bam, Fred will either get Dino or Hoppy and Barney will get the other leaving Wilma once again with Pebbles.
posted by Jenna on 11-13-2008 at 9:35 am
Couldn’t there be two answers to this riddle? If you swap Dino and Bamm-Bamm wouldn’t it still work as Barney-Hoppy, Fred-Dino, Wilman-Pebbles, and Betty-BammBamm. BammBamm and Dino are both males and the only criteria for Betty was that her partner didn’t live with her (since Wilma’s did). Am I missing something?
posted by Mike James on 11-13-2008 at 9:39 am
Mike, Bam Bam lives with Betty so it doesn’t work.
I also meant to add that if you think Hoppy is a girl it just makes the solution easier to come to because Hoppy is taken out of the third statement making Betty either Bam Bam or Dino.
posted by Jenna on 11-13-2008 at 9:43 am
Duh. Can’t believe I actually got this one right the first time around then tried to talk myself out of it. I’m a dummy this morning. Thanks Jenna!
posted by Mike James on 11-13-2008 at 10:03 am
I must admit, i stopped working to figure this one out. haha, it was tricky but a great way to break a mental sweat. The second clue threw me off because I am really tired.haha
posted by Ben W. on 11-13-2008 at 10:24 am
Couldn’t another solution be:
Fred + Barney
Wilam + Pebbles
Betty + Dino
Bam-bam + Hoppy
posted by Adam on 11-13-2008 at 10:57 am
Adam,
This couldn’t be a solution because the rules state that each of the 4 teams has 1 adult. By putting Fred and Barney together you have two adults on one team.
Recaptcha: banking SS (is this a statement of the times or what)
posted by wheel on 11-13-2008 at 11:46 am
I love it when the logic incorporates something more than “X goes with 1″ or “Y doesn’t go with Z”. Your puzzles definitely pass the Turing Test. Thanks for the great puzzle!
posted by Bryan on 11-13-2008 at 5:17 pm
Yay! Got this one – these are my favorite types of puzzles (always use paper or excel to work them out in a grid).
:)
posted by Dawn on 11-15-2008 at 7:59 pm