Riddles!
05-16-2013, 06:36 AM
Post: #261
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RE: Riddles!
Yes. Each light ball weigh the same and each heavy ball weigh the same.
No. I hear there's a secret Mobi hidden here somewhere… |
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05-16-2013, 07:50 AM
(This post was last modified: 05-16-2013 07:59 AM by TheGreatErenan.)
Post: #262
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RE: Riddles!
I think this is one among several solutions:
Label the balls B1, R1, W1, and B2, R2, and W2. Weigh B1R1 vs B2W1. If the two sides are equal: We know that B1 != B2, so R1 != W1. Furthermore, a light and a heavy must be on each side, so B1 = W1 and B2 = R1. So if we weigh B1 vs B2, we instantly learn the weight relationship between B1 and B2. Since B1 = W1, then B2 = W2. Since B2 = R1, then B1 = R2. Thus, we know all balls' weight relationships. In other words, if B1 < B2, then W1 < W2 and R1 > R2. Or if B1 > B2, then W1 > W2 and R1 < R2. If B1R1 < B2W1: This could not happen if B1 > B2. So we know B1 < B2. Observation 1: Either R1 = W1 or R1 < W1, since R1 > W1 and B1 < B2 implies B1R1 = B2W1. Now, weigh R2 vs W1. If R2 < W1, then R1 > R2 and W1 > W2. Therefore B1 R2 W2 are light, and B2 R1 W1 are heavy. If R2 > W1, then R1 < R2 and W1 < W2. Therefore B1 R1 W1 are light, and B2 R2 W2 are heavy. If R2 = W1, then R1 = W1 is false, and from observation 1, we know that R1 < W1. Therefore, B1 R1 W2 are light, and B2 R2 W1 are heavy. If B1R1 > B2W1: This could not happen if B1 < B2. So we know B1 > B2. Observation 2: Either R1 = W1 or R1 > W1, since R1 < W1 and B1 > B2 implies B1R1 = B2W1. Now, weigh R2 vs W1. If R2 < W1, then R1 > R2 and W1 > W2. Therefore B2 R2 W2 are light, and B1 R1 W1 are heavy. If R2 > W1, then R1 < R2 and W1 < W2. Therefore B2 R1 W1 are light, and B1 R2 W2 are heavy. If R2 = W1, then R1 = W1 is false, and from observation 2, we know that R1 > W1. Therefore, B2 R2 W1 are light, and B1 R1 W2 are heavy. |
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05-16-2013, 06:33 PM
(This post was last modified: 05-16-2013 06:51 PM by awpertunity.)
Post: #263
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RE: Riddles!
Here are some easier ones...
What's the missing number below? 4 6---2 9---?---1 19--10--7--6 (The dashes don't mean anything, the spaces just didn't show properly) You have 2 sandglass timers that run for 5 hours and 7 hours. How can you use them to time 16 hours? |
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05-16-2013, 07:12 PM
Post: #264
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RE: Riddles!
let both sandglass timers run. after 5 hours the 5-hour glass would have run out and the 7-hour glass will have 2 hours left. then flip the 5 hour glass and let it run until the remaining 2hours on the 7 hour glass gets finished. ( checkpoint: 7 hours total)
so now theres 3 hours left on the 5 hour glass. flip the 7-hourglass and let it run for the remaining 3 hours of the 5 hour glass. (checkpoint: 10 hours) with 4 hours left on the 7 hour glass. flip the 5 hour glass and let it run until the remaining 4 hours on the 7 hour sandglass finishes. (checkpoint: 14) flip the 7 hour glass and let it run until the 1 remaining hour on the 5 hourglass finishes. (checkpoint: 15) then flip the 7 hour glass again to get the last hour making it 16 hours total. |
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05-16-2013, 10:01 PM
Post: #265
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RE: Riddles!
(05-16-2013 06:33 PM)awpertunity Wrote: Here are some easier ones... The answer is: 3 The Gruff Billies Clan:
Yeah. We ate your grass, bub. Whatcha gonna do? |
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05-18-2013, 08:57 AM
(This post was last modified: 05-18-2013 09:00 AM by TheGoldenGriffin.)
Post: #266
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RE: Riddles!
You have a 13 inch tube, 12 inch tube, and 7 inch tube. You need to cut 10 inch tube, a 6 inch tube exactly from those parts with out a ruler.
I hear there's a secret Mobi hidden here somewhere… |
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05-18-2013, 09:16 AM
Post: #267
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RE: Riddles!
(05-18-2013 08:57 AM)TheGoldenGriffin Wrote: You have a 13 inch tube, 12 inch tube, and 7 inch tube. You need to cut 10 inch tube, a 6 inch tube exactly from those parts with out a ruler. Lay the 13 inch and 12 inch tubes together. Mark the 13 inch tube where the 12 inch tube ends. We'll go back here later. Lay the bottom ends of the 13 inch tube (opposite, unmarked end) and the 7 inch tube together. Cut the 13 inch tube where the top end of the 7 inch tube is. You'll get the 6 inch tube here. (13 - 7 = 6) Go back to the cut off but marked 7 inch remnant from the 13 inch tube. Cut where it is marked. This piece should be a 1 inch tube. Use it to make a 1 inch mark on the 12 inch tube, then make another. Cut the 12 inch tube at the second mark. You now have a 10 inch tube. (12 - 2x1 = 10) You don't know me? Let me introduce myself. I am Anonymous. May the wits ever favor you. |
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05-18-2013, 09:25 AM
(This post was last modified: 05-18-2013 09:26 AM by TheGoldenGriffin.)
Post: #268
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RE: Riddles!
Well thats was easier than expected, lol, try to get 10 3 3 1 1 and 6 inch pipes
I hear there's a secret Mobi hidden here somewhere… |
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05-22-2013, 07:54 AM
(This post was last modified: 05-22-2013 07:58 AM by TheGoldenGriffin.)
Post: #269
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RE: Riddles!
You friend has 12 sugar cubes he lets you take as much as you need so you divide it into 2 equal groups then one group goes bad so you throw them away then you cut the half of the remaining in half then take away 6 pieces then add the bad ones back to that because it turns out they weren't really bad and you eat 2 pieces. How many sugar cubes do you have?
I hear there's a secret Mobi hidden here somewhere… |
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05-23-2013, 02:08 AM
Post: #270
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RE: Riddles!
4, because the cut pieces aren't cubes.
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