Sand-glass II.

[rookie1ja]

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.

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.

Pls visit New Puzzles section to see always fresh brain teasers.

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.

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.

[Guest]

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?

[rookie1ja]

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).

[Guest]

Oh yeah!! gotcha... that's a good one.

[Guest]

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!!

[Guest]

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?

[Guest]

tat aint a great one now is it..

[Guest]

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.

[Guest]

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)

[Guest]

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?

[Guest]

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.

[Guest]

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.

[Guest]

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.

[Guest]

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

[Guest]

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.

[Guest]

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.

[Guest]

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.

[Guest]

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.

[Guest]

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.

[Guest]

misread... I got it

Edited March 11, 2008 by txskydiver

[Guest]

Here's the solution...

1. First, run both the sand-glasses..

2. After 7-min, flip the 7-min sand glass and let the 11-min flow..

3. Again after 4 more minutes, flip the 7-min sand-glass again..

Thats it !! 7+4+4=15..

[Guest]

Sand-glass II. - Back to the Water and Weighing Puzzles

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.

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.

All very good I see that it works - just one problem you turn the sand glass over four times not three - twice to get the 7 & 11 going then when the 7 finishes you turn that once more and that is now three - when the 11 runs out you need to turn the 7 back over to get the four minutes that it has just used and that makes four turns.

Samuelpt you seem to be ignoring the two times to set both A11 and B7 off

[Guest]

Wouldn't it be better to simply turn both sand timers, 7 minute and 11 minute, and start the test when the 7 minute runs out, leaving 4 minutes on the larger timer. When the 11 minute timer empties, turn it again adding another 11 minutes to the four. Only two turns.

[Guest]

Is it...

1. Turn them both over so they start to spill

2. when the 7 min timer is up, turn that one back over. there will be 4 mins left in the 11 min one

3. when the 11 min timer is finished, in the 7 timer it will be 3 mins at the top and 4 at the bottom. You need 4 more mins, so you turn it back over so the 4 mins (in the 7) will spill back.

so it will be the 11 mins (from the 11 timer) and the remainder 4 mins (from the 7 timer)