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#1 User is offline   rookie1ja Icon

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Posted 30 March 2007 - 05:42 PM

Hourglass II. - Back to the Water and Weighing Puzzles
A teacher of mathematics used an unconventional method to measure a 15-minute time limit for a test. He just used 7 and 11-minutehourglasses. During the whole time he turned sandglasses only 3 times (turning both hourglasses at once count as one flip).
Explain how the teacher measured 15 minutes.



Spoiler for Solution:
Sand-Glass II. - solution
When the test began, the teacher turned both 7min and 11min sand-glasses. After the 7min one spilt its last grain, he turned it upside down (the 11min one is still to spill sand for another 4 minutes). When the 11min sand-glass was spilt, he turned the 7min one upside down for the last time. And that’s it.



Spoiler for old wording:
A teacher of mathematics used an unconventional method to measure time for a test lasting 15 minutes. He used just a sand-glass, which spills in 7 minutes and a second sand-glass, which spills in 11 minutes. During the whole time he turned sand-glasses only 3 times. Explain how the teacher measured 15 minutes.

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#2 User is offline   pranavojha Icon

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Posted 07 May 2007 - 06:18 AM

I don't get it? If you start both sand glass together, the 7min will run out after seven minutes and we turn it over. After four minutes (total 7+4=11), the second glass also runs out. So the 7min glass has sand for only three more minutes (7-4=3). Turning it (7min glass) over again therefore calculates to 11+3=14mins and not 15.
Did I do something wrong?
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#3 User is offline   rookie1ja Icon

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Posted 07 May 2007 - 09:24 AM

Quote

I don't get it? If you start both sand glass together, the 7min will run out after seven minutes and we turn it over. After four minutes (total 7+4=11), the second glass also runs out. So the 7min glass has sand for only three more minutes (7-4=3). Turning it (7min glass) over again therefore calculates to 11+3=14mins and not 15.
Did I do something wrong?

In your last step, the 7min glass has sand for 3 more minutes after 11 minutes - but if you turn it upside down, there is still sand to spill for another 4 minutes (which were already measured from 7th to 11th minute).
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#4 User is offline   pranavojha Icon

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Posted 15 May 2007 - 05:11 AM

Oh yeah!! gotcha... that's a good one.
Don't be so concerned about your rights that you forget your manners.
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#5 User is offline   pranavojha Icon

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Posted 17 May 2007 - 07:49 AM

Another method which involves only two turning of the sand glass is given below.

1. Start both the Sand glasses.
2. Once the 7-minute glass spills over, start the test.
3. By the time 11-minutes glass spills over, we would have finished 4-minutes.
4. Once the 11-minutes spills over, turn it over again. That's all!!
Don't be so concerned about your rights that you forget your manners.
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#6 User is offline   Bdw725 Icon

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Posted 22 June 2007 - 04:56 AM

I have a question... If the teacher turns both of the hour glasses when the test starts, that is 2 turns. Then when the 7 minute one runs out he turn that over, that is 3 turns. Then your solution says to turn that again once the 11 minute one runs out, wouldn't that be four turns of an hour glass? The puzzle said to do it in 3, so how is this correct?

I think pranavojha's answer is the correct one? Am i wrong or is he the one that solved the puzzle?
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#7 User is offline   ggsarah Icon

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Posted 22 June 2007 - 02:28 PM

tat aint a great one now is it..
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#8 User is offline   pranavojha Icon

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Posted 22 June 2007 - 03:02 PM

Well, when you start timing, then it is obvious that the sand in the two glasses have to start pouring. So instead of saying it as "Startin the sand-glass", for which on must assume that there is a switch that starts the sand pouring in, hence the turning counting will start only after the first two turns.
Don't be so concerned about your rights that you forget your manners.
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#9 User is offline   fantomexloa Icon

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Posted 05 July 2007 - 01:44 AM

Quote

I have a question... If the teacher turns both of the hour glasses when the test starts, that is 2 turns. Then when the 7 minute one runs out he turn that over, that is 3 turns. Then your solution says to turn that again once the 11 minute one runs out, wouldn't that be four turns of an hour glass? The puzzle said to do it in 3, so how is this correct?

I think pranavojha's answer is the correct one? Am i wrong or is he the one that solved the puzzle?


they did not turn the 11 min glass anymore after it ran out of time...

starting when they turned the 7min sandglass and 11min sandglass(that is 2 turns)..when the 7min sandglass ran out of sand,we now wait for the 11min sandglass to ran out of sand as well and if it does, then we will be turning over the 7min sandglass(thus the third and last turn)
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#10 User is offline   ninjamentat Icon

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Posted 19 July 2007 - 06:34 PM

This is how it goes.

1st turn: you turn both sand glasses over to start them both at the same time.

2nd turn: The 7-minute glass runs out. Immediately turn it over.

3rd turn: The 11-minute glass runs out. Immediately turn the 7-minute glass over again.

When the 7-minute glass runs out, 15 minutes have elapsed.

Easy huh?
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