There was this thread about logic puzzles a while ago here at Sajha. Needless to say, it was quite popular. The thread was also a confluence of 'great' minds of sajha, who displayed there abilities to approach problems from different view points, not necessarily the correct point of view, however.

One way to keep this thread moving is to come up with puzzles each time a puzzle is solved. And just to avoid any stress to Google Inc., I encourage people to phrase the puzzles as accurately as they can in their own words.
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Here's the one that I would like to start with.
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There are two threads, with DIFFERENT lengths and DIFFERENT diameters. The threads are NOT uniform, each thread has different diameter at different points along the thread. But if you light up the thread at any one end, it takes exactly one hour for EACH thread to finish buring from one end to the other end. You can light up any end first-- it does not matter. Now, how do I measure 45 minuties by using those threads?

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This is not a riddle, so the aim is not trick anyone in small details.

light up thread 1 , when thread 1 is half done light up thread 2 , and make sure you mark the half of thread 1 , now when the other half of thread 1 is half done you have reached 45 minutes point countdown, now countdown your 45 minutes on thread 2( the remaining of thread 2 will last for 45 minutes.) or you have just reached 45 minutes from the time you lit up thread1.

Hyper thread what bull s-hit answer is that?

Here is how it works ;)


1. Light one thread at both ends and the other at one end at the same time. Since it takes 1 hour to completely burn one thread if you light it at one end, it should take exactly 1/2 hour for it to burn completely if you light it at both ends at the same time.
2. The other end of the second thread should be lighted at the same time when the first thread completely burn out. The complete burn out of the second thread should give you 15 mins
3. 30 min + 15 min = 45min

ya, fold the damn thread 4 times in equal lenght. so it has 4 parts..so it takes 15 mins to burn each fold ( 1/4th of the total thread)

I think I got this one.

haude_ko_bhai I got solution from your solution. you assumed that the second thread would burn in 15 mins, I dont know how you got that one coz it should take 1 hour for both.

here is how you do it. Take two ends of thread one and one end of thread 2 and light them up. the moment thread one ends burning indicates the half hour mark. The second end of thread two should be burnt at the instance the thread one ends burning. Which means that half an hour of thread two is burnt. Thus half an hour of thread 2 should burn at instance thread one stops burning. Thus we get fourty five minutes.

Thread one burns for 30 mins. Thread tow burns of 30 mins. The remaining of thread 2 equals 30 mins. When we burn both ends of thread 2 at that moment, thread 2 should burn for 15 mins. 30+15=45 mins. Man it took almost half an hour for that. Good one though Guest4

I got another one for you. There is a 200 mile distance in which runners are allowed to carry 2 gallons of water in 2 hands, thats it. The 1 gallons of water is enough for the runner to run 25 miles. He can go a certain miles and store gallons as many times as he likes provided he has enough water to get back to the start. He can come back and get maximum of 2 gallons. So he runs a certain distance and stores gallons of water, comes back to start get more water. What is the lowest number of trips he has to make back to make the 200 mile distance?

here's one:-)

"Sente, Violet, Scarlett and Kurt have spent the day hiking in the hills to the south of the city. On their way home is a rickety old bridge that's only strong enough for two of them to be on it at the same time.

The light is fading fast and so they need to use a flashlight every time they make a crossing. The bridge is too long and narrow to throw the flashlight across, so someone will always have to carry it back and forth. What's more, Sente was injured earlier in the day and will take 10 minutes to cross the bridge. Scarlett can cross in 5 minutes, Kurt in 2 minutes and Violet in just 1 minute. A storm is brewing and they can't waste any time. Kurt thinks for a moment and claims they can all cross the bridge in just 17 minutes. How?"

1) Violet and Kurt cross the bridge which will take them 2 minutes because Kurt takes 2 minutes.
2) Violet returns back with the flashlight which will take 1 minute.
3) Sente and Scarlette then cross the bidge with the flashlight which will take them 10 minutes since Sente takes 10 minutes.
4) They will hand the flashlight now to Kurt who will return to pick up Violet at the other end taking 2 minutes.
5) Kurt and Violet then cross the bridge taking them 2 minutes again

Thats 17 minutes..!!

hetterika..tauko khayo..no more puzzle..

more???

haude_ko_bhai I know what I was talking about , if you didnt get , you simply didnt

Haude ko bhai,
You got it!

Hyperthread,
You assumed that the thread are uniform, ie at half hour, the thread would finish burning half--which is not necessarily true.

Others...depends how boring the day will be tomorrow.

im not sure if someone's posted it but here's an easy solution to the first one:

Fold the first longer thread into half and fold the second shorter thread into 4 equal parts (its easy).
Line them up end to end and light up either one of the threads at one of the folded ends so that every part of the thread burns
Once they're both finished burning its exactly 45 minutes.

.you have
10 stacks of coins
an electronic scale

problem:
of the 10 stacks of coins, 9 stacks are good coins
one stack is all counterfeit coins
the countefeit coins are identical to the real ones, except they weigh one tenth of a gram less than the real coins
you are allowed to weigh any combination of coins

tara
ONLY ONE WEIGHING ALLOWED

figure out which stack is counterfiet

and remember, one weighing allowed in total

For the sake of clarity, I am assuming that each real coins weigh 10g and the fake coins weigh 9g (it does not really matter what they weigh though)

Take 1 coin from stack 1; 2 from stack 2; 3 from stack 3; 4 from stack 4; 5 from stack 5; 6 from stack 6; 7 from stack 7; 8 from stack 8; and, 9 from stack 9. This makes a total of 45 coins.

This combination of coins give different weight depending upon which stack the coin is in. For example, if the fake stack is stack 1, the total weight will be 449g.
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Try the same problem, now, using a BEAM BALANCE. Try to find the fake stack by weighing no more than 2 times.

Ye dai..ani ki..ma sanga pani auta question cha...
ani ki.. tyo Kachuwa ra Kharayo ko..
Baba le malai achel english sikaunu huncha....
Ma aja english ma lekchu la..

In the old story of Rabit( R) and Tortoise (T), I say that Rabit will never meet Tortoise if Rabit starts the race after Tortoise. It doesnt matter how fast Rabit runs. He just can't meet Tortoise.
Here is my argument:

Lets say T starts some seconds earlier than R with his very slow speed and R starts to follow T with his extremely fast speed.

After certain time "t", R reaches the place where T was. But in that time, T will cover some extra distance. Again R will reach that new position after some small time "t1" but again T will have travelled some additional distance in that period. Again R will run to cover the extra distance but T will cover some more distance in that time.. it keeps on going..

R will always be busy to cover the extra distance that T has already covered.

Hence R can never meet T becase by the time R reaches the point where T was, T will have covered some additional distance.

Ani ki ...dai..
Kharayo le ni ..Kachuwa lai kahile pani bhetauna sakdaina..
Mero school ko kitab ma galti lekheko cha..

Dai..chadai bhannu na...
Kharayo le Kasari Kachuwa lai jitcha...
Maile Bholi Miss lai bhannu cha..

kapuri kha, you've totally missed the concept of speed.
incorporate that and youll find that the book was right.

timetraveller dai, speed ko concept bujhai dinu na..

Nature should be the same from whichever perspective you observe it.
Truth should be truth.

Please point out any flaw if there is in my argumentation.

Otherwise I dont see any way the Rabit winning the match.

For the puzzle regarding the thread,

take the first thread and burn it on both ends and only one end of the second thread at the same time. This way when the first thread burns out it will be 30 minutes. Then after as soon as the first thread burns out then light the other end of the second thread as well (which would have been half burnt by now). Then the remaining part of the thread should burn out in 15 minutes. Hence you get 30 + 15 = 45 minutes.
Thank you.