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Topic: 3 easy ones (Read 1706 times) |
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Speaker
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Re: 3 easy ones
« Reply #25 on: Oct 31st, 2003, 12:34am » |
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on Oct 30th, 2003, 11:03pm, Darcal wrote:Hi, please help with the following, 52c in a p (WJ's) |
| I think that this original riddle is mis-worded. How about 52c in a p (W/OJ's) or 54c in a p (WJ's) There are also a bunch of other riddles like this someplace. You might find some similar ones if you search for your riddle.
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They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety. <Ben Franklin>
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Darcal
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Re: 3 easy ones
« Reply #26 on: Oct 31st, 2003, 1:53am » |
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Maybe this would help, a previous riddle read, 66 b in the b. What is the answer?
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Darcal
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Re: 3 easy ones
« Reply #27 on: Oct 31st, 2003, 2:47am » |
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66 b in the b = 66 books in the bible 52c in a p (WJ's) = 52 cards in a pack (With Jokers) So the next one is: 39b of the ot
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« Last Edit: Oct 31st, 2003, 2:57am by Darcal » |
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ThudnBlunder
wu::riddles Moderator Uberpuzzler
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Re: 3 easy ones
« Reply #28 on: Oct 31st, 2003, 3:43am » |
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Quote:So the next one is: 39b of the ot |
| :39 Books of the Old Testament
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
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Re: 3 easy ones
« Reply #29 on: Oct 31st, 2003, 9:50am » |
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on Oct 31st, 2003, 2:47am, Darcal wrote:52c in a p (WJ's) = 52 cards in a pack (With Jokers) |
| This is what Speaker was refering to as being misstated, since there are 54 cards in a deck with jokers. Presumably, though what was meant is "52 cards in a pack (Without Jokers). However, if they are going to throw in the "s" for Jokers (with a non-existant apostrophe, no less), they should have also included the o for without.
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
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rmsgrey
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Re: 3 easy ones
« Reply #30 on: Nov 1st, 2003, 7:55pm » |
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on Feb 1st, 2003, 1:49am, udippel wrote:1.) You drive in a race car and overtake the second fastest. Which position will you be in? |
| If you overtake the second fastest, you must be driving the fastest - which puts you in first (assuming the fastest car always wins). Quote:2.) You drive in a race car and overtake the last car. Which position do you reach? |
| I started the race with the other ten cars ahead of me. I overtook the first car, then the second. The third, fourth and fifth followed swiftly. I overtook the sixth on a sharp corner before overtaking the seventh on the next straight. The ninth strayed wide to cut off the eighth, letting me ovetake them both. Finally, I overtook the last car just before the finish line to take first place! Alternatively, if I overtake the last car mentioned (the second fastest), I win again! Thirdly, if the "last" describes the position of the car overtaken after the overtaking, I end up in second last place. Surely any of the interpretations which makes the situation described possible is more correct than insisting that the question is not valid... Quote:3.) Susan's father has five daughters: Nani Neni Noni Nuni What's the name of the fifth? |
| OK, this one got me at first glance, and I didn't think of the various quibbles either, but I spotted my mistake inside the three second time limit...
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