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# Sand-glass II.

23 replies to this topic

### #11 M3nTaL*Wr3Ck

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Posted 10 August 2007 - 07:01 PM

The Correct answer is most definitely to start the 7-min hourglass and the 11-min hourglass at the same time. When the 7-min hourglass runs out "Start" the test. Upon completion of the 11-min hourglass, simply flip it over. When that time runs out the test is "Finished".

The other proposed answer requires four turns of the hourglass. Initially turning over two, the 7-min and the 11-min simultaneously, followed by two more flips of the 7-min hourglass. Does not work within the constraints of the problem.

No thanks is necessary, the answer was there all along.
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### #12 geekmanager

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Posted 01 October 2007 - 08:32 PM

From the wording of the original puzzle ("he turned sand-glasses only three times"), I think it could either way. That is to say that it could be argued that turning two glasses at one time constitutes one turn or that each is counted as a separate turn. This would force one of the two answers depending on what you decide a "turn" means. Each correct for it's particular constraints.

However, I like pranavojha's solution much better. It is a much more elegant solution. Though it does have the pesky problem of an awkward seven minute delay before the start of the test.
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### #13 twittbrod

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Posted 16 October 2007 - 06:20 PM

From the wording of the original puzzle ("he turned sand-glasses only three times"), I think it could either way. That is to say that it could be argued that turning two glasses at one time constitutes one turn or that each is counted as a separate turn. This would force one of the two answers depending on what you decide a "turn" means. Each correct for it's particular constraints.

However, I like pranavojha's solution much better. It is a much more elegant solution. Though it does have the pesky problem of an awkward seven minute delay before the start of the test.

A 7 minute delay before starting is more elegant? If you say so!

It just depends how big your hands are as to whether you can start the timing with one turn or two. Obviously, two glasses needed to be turned. If we do it in one motion, it's one turn. If it takes two, it's two turns.

Anyway, enjoy that 7 minute delay before you can even start. Meanwhile, I'll be downing my first pint, waiting for you to finish.
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### #14 Flash10

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Posted 03 January 2008 - 09:01 PM

Start both glasses at the same time like you said. When the seven minute glass runs out tell the class to start so the 11 minute glass will run for four more minutes. When the 11 minute glass runs out simply flip it.(11-7)+11=15
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### #15 Riddari

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Posted 11 January 2008 - 01:47 AM

Both answers seem fine to me. I interpreted a flip as a point in time when you had to make a change to the timers. So, it would not matter to me whether you were flipping one or two of the timers, as long as the flips occurred at the same time. Also, the seven minute delay for pranavojha's solution does not have to be wasted time. The teacher could simply flip the timers, then hand out the papers and give the testing instructions. As long as he timed this properly, it could flow right into the beginning of the test.
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### #16 samuelpt

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Posted 30 January 2008 - 06:24 PM

I think he only needs to turn the sandglasses two times during the experiment (one more in the beginning):
0:00 - Turn both sandglasses A11 and B7.
7:00 - sandglass B7 finishes, turn it around. sandglass A11 still has another 4 minutes to go.
11:00 - sanglass A11 finishes. sandglass B7 has been running for 4 minutes. Turn B7 around, to measure 4 minutes. We can ignore A11.
15:00 - Sandglass B7 runs out of sand.
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### #17 I_like_chocolate

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Posted 01 February 2008 - 05:28 PM

I did this another way, using fractions, but I still got 15. If the teacher turned over the 7-min. sand glass, then there would be 8 minutes left. Saying the teacher marked his sand glasses at 1/4 intervals, could'nt he have then turned the sand glass to fall until 3/4 of the way, at 5 and 1/4 minutes, then turned over the 11-min. sand glass to fall until 1/4 of the way, which is 2 and 3/4 min.? It still adds up to 15 minutes, because 7+ 5.25 +2.75 =15, and he makes 3 turns total.
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### #18 Kazure

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Posted 14 February 2008 - 10:48 PM

I did this another way, using fractions, but I still got 15. If the teacher turned over the 7-min. sand glass, then there would be 8 minutes left. Saying the teacher marked his sand glasses at 1/4 intervals, could'nt he have then turned the sand glass to fall until 3/4 of the way, at 5 and 1/4 minutes, then turned over the 11-min. sand glass to fall until 1/4 of the way, which is 2 and 3/4 min.? It still adds up to 15 minutes, because 7+ 5.25 +2.75 =15, and he makes 3 turns total.

Well, I think it's assumed that height of glass is not in the same ratio as volume of the glass.
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### #19 DrJim

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Posted 14 February 2008 - 11:10 PM

Start both glasses. After 7 minute glass is finished turn it over as 11 minute glass continues for 4 more minutes. When 11 minute glass is done, turn over 7 minute glass which will have 4 minutes in bottom giving you 11 + 4 = 15 minutes.
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### #20 txskydiver

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Posted 11 March 2008 - 08:42 PM