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# Igniter Cords

28 replies to this topic

### #11 otasyn

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Posted 15 August 2007 - 04:09 PM

I believe this solution is also possible .. Please comment ...

1. Ignite the first cord at the two ends and somewhere approximately in the middle so that the fire from the middle propagates towards the two ends. It should take 15 minutes for this first cord to burn out completely (There would be two pieces of the first igniter cord since it was ignited at the middle as well - EVERY LAST BIT should burn out)

2. Ignite at the second cord at the two ends as soon as the first cord "completely" burns out. This would take another 30 minutes.

This would not be an accurate measurement. If the cords burned at constant speed throughout the whole cord, and you started at exactly the center, this would be fine, but this is not the case in the rules.

Even though you're trying to start one at the approx. middle, you might get differing times. One might burn up in 12 and the other 18. Without thinking too hard on it, the average would probably be 15 minutes, but even then, we cannot measure something on averages.
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### #12 questionable

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Posted 27 August 2007 - 08:48 PM

I do not see how this one can be solved. The given answer is incorrect; It assumes that there is 30 minutes left to burn on the second cord. There may be 50 or only 2. Please correct my logic if necessary.
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### #13 Martini

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Posted 28 August 2007 - 05:03 AM

It assumes that there is 30 minutes left to burn on the second cord. There may be 50 or only 2. Please correct my logic if necessary.

There is only 30 minutes to burn on the second cord for the following reason:

If three ends of two cords are all lit at the same time, the one lit at both ends will be finished burning in 30 minutes. At this point the second cord will be finished burning in 30 minutes if left alone. However, if instead of leaving it alone, the other end of that cord is lit at the very moment the other cord finished burning, it will now finish burning in 15 minutes.
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### #14 zyanbrain

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Posted 11 September 2007 - 06:37 PM

easy
45=30+15=60/2+30/2;

first ignite both ends of first one cord, but only one end of second cord,
when the first one burn out, it is 30 mins,
ignite the other end of second one, left 30,
when all burn out, it is 30+15=45.
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### #15 Taidaishar

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Posted 02 November 2007 - 07:54 PM

I do not see how this one can be solved. The given answer is incorrect; It assumes that there is 30 minutes left to burn on the second cord. There may be 50 or only 2. Please correct my logic if necessary.

While I knew the answer was correct, I was still having trouble getting straight in my head the non-constant rate of burning. However, if a rope takes 1 full hour to burn, and you've been burning it for 30 minutes, no matter what the rate of burning, then the rest of the rope will have to take another 30 minutes to burn. Therefore, if you've set both ends of a 1-hour rope on fire, it will burn in 30 minutes meaning that the rest of the OTHER rope that hasn't burned yet is 30 minutes worth of rope... then, you light the other end of that making it 15 minutes worth of rope.
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### #16 al_rawi

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Posted 22 November 2007 - 11:00 AM

fold the first cord at the middle (you can make sure this is the case by making sure the opposite ends meet when folded), and light it up (such that both ends start burning at the same time)

While it is burning (and it should take 30 minutes) fold the other cord twice (so that you have it in quarters).. once the first cord finishes burning, light the other cord (such that all four bits start burning).. this should take 15 minutes, and you have a total of 45!

ok, so technically, you might argue that the parts might burn eachother, but you can always space them out such that they don't!
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### #17 aishi_khurana

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Posted 10 December 2007 - 09:19 AM

don't u need 2 mention dat composition of both the cords is xactly same...?
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Posted 12 December 2007 - 07:20 AM

don't u need 2 mention dat composition of both the cords is xactly same...?

NO. the beauty of this problem is know you dont give a crap what they are made off. they could be even tapered so that one end is really skinny. The only thing you know is that the full length takes 1 hour. When you fold it in half and light both ends simultaneously the flames will not meet in the middle but they will definitely meet after exactly 30 mins. The other rope (which may be very different) has also been burning for exactly 30 mins. This leaves 30 mins left on the second rope. Not half the length of the rope but half the rope in terms of time. That rope is immediately folded and lit. The two flames meet 15 minutes later at an arbitrary location. The total elapsed time is exactly 30+15=45 minutes
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### #19 marxercise

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Posted 24 January 2008 - 12:25 AM

I believe this solution is also possible <!-- s:o --><!-- s:o --> .. Please comment ... <!-- s:roll: --><!-- s:roll: -->

1. Ignite the first cord at the two ends and somewhere approximately in the middle so that the fire from the middle propagates towards the two ends. It should take 15 minutes for this first cord to burn out completely (There would be two pieces of the first igniter cord since it was ignited at the middle as well - EVERY LAST BIT should burn out)

There's no telling the time it takes each piece to burn. One half lit at each end may burn up in 5 minutes, while the other half takes 25 (even though they are of equal length). All we know is 60 minutes in total.

In that case:
----------10 min----------|----------50 min----------

Or in another, 22 minutes and 8 minutes:
----------44 min----------|----------16 min----------
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### #20 eric m

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Posted 22 February 2008 - 12:24 AM

A little off topic, but I just had an interview yesterday for a software engineering job, and one of the technical questions was this exact problem. The worst part about it was that I read this riddle before on this site, but I couldn't think of it for the life of me when it mattered! Ugh. I got it after a hint though, but still, would have been nice to nail it.

So it turns out reading riddles all day at work isn't a total waste of time afterall
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