A father of triplets plans to surprise their mom by having all four of them meet her downtown for dinner. Unfortunately, when he goes to start his eleven-year-old van, he discovers that it has finally given up the ghost. Undeterred, he looks around the garage and discovers a tandem bicycle in fine working order. Alas, it can only safely transport him and one of the three boys at a time.
He realizes that if he leaves Kevin alone with Andrew while ferrying Justin, they will get into mischief with super glue and silly string. He also knows he cannot leave Andrew alone with Justin or they will make prank phone calls to India on the cell phone. Giving it a moment's thought he determines that he can get all four of them to dinner without any monkeyshines. How does he do it?
[This puzzle is for entertainment purposes only; any similarities to real triplets, naughty or obedient, is purely coincidental.]
I get it, using some logic if you take Andrew along first then Keving and Justin won't get into any mischief. Then you take Justin or Kevin it does not matter wich. Then you come back and take the one you decided to leave behind when you had the option of taking either Kevin or Justin.
ReplyDelete-Rory
I agree with Rory.
ReplyDeletehi person
ReplyDeleteYou were so close Rory!!! You just missed one step. Here is how it works:
ReplyDeleteFirst you bring Andrew (me) to the restauraunt. Kevin and Justin wont get into any trouble.
(By the way, this Brain teaser makes me sound like the trouble-maker! Well, I can't say this Brain Teaser isn't accurate!) Then you go back to the house and pick up eather Kevin or Justin, it does not matter which one. You drop them off at the restaurant. But here is the catch! You bring Andrew back to the house! Then you bring the one brother you left behind. Finally, you take Andrew (me!) to the restaurant (again!)
Dad must have gotten one heck of a workout!
Keep checking in!
-Andrew
Thats confusing
ReplyDeleteThat's a surprise
ReplyDeleteKP