You are given one fuse that burns for exactly 10 minutes and two other fuses that burn for about 30 minutes each (+- 5 minutes) - these two fuses burn for identical times.
The fuses burn unevenly.
Measure an exact time of 30 minutes.
2006-09-10 17:01:46 · 4 answers · asked by debarun p 1 in Science & Mathematics ➔ Mathematics
hansolosby - you can't do that. The fuses burn unevenly.
2006-09-10 17:15:31 · update #1
The two approx 30 min fuses DO NOT burn unevenly in the same way. So any kind of measurement won't work.
2006-09-10 18:01:27 · update #2
DutchProf - you are assuming that there will be some of B2 left after burning both ends of B1 - this assumes the fuses burn fairly regularly. I never said that.
2006-09-10 18:24:03 · update #3
You can assume that the fuses burn from both ends, and that they can be extinguished.
2006-09-10 18:34:50 · update #4
You've stated that the tolerances of the two unknown fuses as exactly +/- 5 minutes, I'm now assuming that I either have 2x25min fuses or 2x35min fuses. By forming the 10min fuse in a ring, it will burn for 5min exactly. I can give you an exact burn of 30min. If I place the fuses like this:
--------------O --------------
light here^
The dashes are the fuses that are either 25min or 35min. The "O" is the 5min ring. Only one of the straight fuses is touching the ring.
By lighting the point where one fuse meets the ring, the ring will begin to burn for 5 minutes as soon as the ring goes out light the second 25 minute fuse, giving you 30min.
If they are the 2x35min fuses, set them up the same way & light the same point. As soon as the 5min fuse is out, there will be 30min left on the fuse that was lit at the same time!
In both cases 1 of the straight fuses is actually not needed, this was just the easiest way to explain what I was doing.
2006-09-11 07:58:47 · answer #1 · answered by Brendan R 4 · 0⤊ 0⤋
similar to dutchprof:
burn 10-min fuse and both ends of the 1st 30-min fuse
(10 min elapsed)
extinquish the 1st 30-min fuse
The missing portions (either ends) of the 1st 30-min fuse represent 10 mins each.
Take the 2nd fuse and remove the same portion of the remaining of 1st fuse (so you have 3 pieces, 1 which is the same as the remaining 1st fuse and 2 which are the same as the burnt portions).
Then, burn each one which equal 10 mins each.
Total 30 mins.
2006-09-10 17:54:56 · answer #2 · answered by Anonymous · 0⤊ 0⤋
I have a solution without measurement; it only assumes you can put out fuses when necessary.
Call the fuses A (10 min.), B and C (30 + x min.)
First, cut the 30 minute fuses into approximately equal parts. I call them B1, B2, C1 and C2.
Light at the same time
* one end of A
* both ends of B1
* one end of B2
As soon as B1 is finished burning, light the other end of B2.
CASE 1: Fuse A is done before fuse B2.
In that case, there are x minutes left on fuse B2. Light one end of C1. As soon as fuse A is done, start the timer. When fuse C1 is done, light C2. Stop the timer when C2 is done.
CASE 2: Fuse B2 is done before fuse A. In that case, there are -x/3 minutes ;eft on fuse A. Light one end of C1 and both ends of C2, and start the timer. As soon as fuse A is done, put out fuse C2. When C1 is done, re-light one end of C2. Stop the timer when C2 is done.
2006-09-10 17:42:10 · answer #3 · answered by dutch_prof 4 · 0⤊ 0⤋
Set the 10 min going and one of the 30 mins going. After 10 mins mark the position of the 1st 30 min fuse and start the 2nd one. When the 2nd one reaches the point the 1st one reached, you know 20 mins has passed, mark the new position of the 1st one. When the 2nd one reaches the new position of the 1st one, 30 mins has passed.
Right?
2006-09-10 17:08:55 · answer #4 · answered by Anonymous · 0⤊ 0⤋