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Topic: HOURGLASSES (Read 11760 times) |
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singh181
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Q: You have two hourglasses: a 7 minute one and an 11 minute one. Using just these hourglasses, accurately time 15 minutes.
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Start 7 min and 11 min hourglasses together. When 7 min Hourglass is over we have 4 min remaining in 11 min Hourglass.
Now, again start 7 min hourglass, now 4 min which was remaining in 11 min hourglass gets over.
Now, we have 3 min left in 7 min Hourglass. again start 11 min hourglass. when 3 min left in 7 min Hourglass gets over we have 8 min left in 11 min hourglass.
Now, start 8 min left in 11 min hourglass. when 8 min gets over then start 7 min hourglass. when 7 min gets over, we have successfully measured 15 mins.
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towr
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Re: HOURGLASSES
« Reply #1 on: Jun 5th, 2007, 8:42am » |
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You can do it much faster. 2 x 11 - 7=15
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| « Last Edit: Jun 5th, 2007, 8:43am by towr » |
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RedCrest
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Re: HOURGLASSES
« Reply #2 on: Jun 6th, 2007, 9:57pm » |
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How about this?
| hidden: | | Set both hourglasses running, but when the 7-minute one runs out, flip the 11-minute one back over immediately, and it will measure out 4 minutes for you (because 11-7 = 4). After the four minutes in the 11-minute hourglass runs out, flip the 11-minute one back over and it will measure out 11 minutes for you. 4 + 11 = 15, so you have your 15 minutes exactly. |
~Red
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Grimbal
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Re: HOURGLASSES
« Reply #3 on: Jun 7th, 2007, 12:38am » |
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on Jun 6th, 2007, 9:57pm, RedCrest wrote:
You should NOT flip the 11-minute hourglass to measure the remaining 4 minutes. This comes to determine the start at 7 minutes with one hourglass, and determine the end at 22 minutes (2x11) with the other hourglass.
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Grimbal
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Re: HOURGLASSES
« Reply #4 on: Jun 7th, 2007, 1:50am » |
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Damn! I found a better way:
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time 0: start both hourglasses.
time 7: 7 runs out, flip 7
time 11: 11 runs out, flip both. 7 now has 4 minutes.
time 15: 7 runs out.
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sitai
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Re: HOURGLASSES
« Reply #5 on: Jun 7th, 2007, 4:23pm » |
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i guess you got it... you took the scenic route though
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srn437
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Re: HOURGLASSES
« Reply #6 on: Aug 26th, 2007, 9:12pm » |
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I don't know how or why to hide stuff but the answer is as stated at top except when the 7 minute one has 3 left turn it upside down.
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maha
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Re: HOURGLASSES
« Reply #7 on: Jul 23rd, 2008, 9:37am » |
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allow 7 min and 11 min hourglass simultaneusly to start...when 7 min hourglass is over, there ll be 4 mins remainig in 11 hourglass.invert tat to count 4 mins .again invert it to count 11 mins..tat totals to 15 mins... 
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| « Last Edit: Jul 23rd, 2008, 9:40am by maha » |
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rmsgrey
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Re: HOURGLASSES
« Reply #8 on: Jul 24th, 2008, 7:23am » |
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on Jul 23rd, 2008, 9:37am, maha wrote:| allow 7 min and 11 min hourglass simultaneusly to start...when 7 min hourglass is over, there ll be 4 mins remainig in 11 hourglass.invert tat to count 4 mins .again invert it to count 11 mins..tat totals to 15 mins... |
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If you invert the 11 minute hourglass after it's run for 7 minutes, it will take another 7 minutes for the 7 minutes worth of sand to run back out...
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Grimbal
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Re: HOURGLASSES
« Reply #9 on: Jul 25th, 2008, 8:38am » |
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on Jun 7th, 2007, 12:38am, Grimbal wrote:| You should NOT flip the 11-minute hourglass to measure the remaining 4 minutes. ... |
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pasta12
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Re: HOURGLASSES
« Reply #10 on: Jun 18th, 2012, 8:32am » |
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How about this. Flip the 11 minute hourglass. When it runs out, immediately flip the 11 minute hourglass again. When the 11 minute hourglass has the same amount of sand remaining as the 7 minute hourglass has, 4 minutes has passed. 11 minutes and 4 minutes makes 15.
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| « Last Edit: Jun 18th, 2012, 8:33am by pasta12 » |
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rmsgrey
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Re: HOURGLASSES
« Reply #11 on: Jun 19th, 2012, 5:12am » |
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on Jun 18th, 2012, 8:32am, pasta12 wrote:| How about this. Flip the 11 minute hourglass. When it runs out, immediately flip the 11 minute hourglass again. When the 11 minute hourglass has the same amount of sand remaining as the 7 minute hourglass has, 4 minutes has passed. 11 minutes and 4 minutes makes 15. |
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Or you could flip the 11 minute glass, let it run out then flip it again. When the sand in the top and the sand in the bottom are in a 7:4 ratio, 4 minutes have passed.
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sanhas
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Re: HOURGLASSES
« Reply #12 on: Sep 11th, 2012, 4:20am » |
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Wow this one is pretty hard 
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Forest
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Re: HOURGLASSES
« Reply #13 on: Dec 1st, 2012, 10:28pm » |
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How about this;
Start both hourglasses at the same time.
When the 7min hourglass runs out, your 15 minute period starts. (There is 4 min remaining on the 11 min hourglass)
When the 11 min hourglass runs out, flip it
When it runs out again, 15 minutes has passed.
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Grimbal
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Re: HOURGLASSES
« Reply #14 on: Dec 3rd, 2012, 1:21am » |
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It works. But you are loosing 7 minutes for nothing.
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marlonmark
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Re: HOURGLASSES
« Reply #15 on: Jan 2nd, 2013, 2:26am » |
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A:
Start 7 min and 11 min hourglasses together. When 7 min Hourglass is over we have 4 min remaining in 11 min Hourglass.
Now, again start 7 min hourglass, now 4 min which was remaining in 11 min hourglass gets over.
Now, we have 3 min left in 7 min Hourglass. again start 11 min hourglass. when 3 min left in 7 min Hourglass gets over we have 8 min left in 11 min hourglass.
Now, start 8 min left in 11 min hourglass. when 8 min gets over then start 7 min hourglass. when 7 min gets over, we have successfully measured 15 mins.
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towr
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Re: HOURGLASSES
« Reply #16 on: Jan 2nd, 2013, 10:13am » |
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on Jan 2nd, 2013, 2:26am, marlonmark wrote:
A:
Start 7 min and 11 min hourglasses together. When 7 min Hourglass is over we have 4 min remaining in 11 min Hourglass.
Now, again start 7 min hourglass, now 4 min which was remaining in 11 min hourglass gets over.
Now, we have 3 min left in 7 min Hourglass. |
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And 4 if you turn it over. 11+4=15
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rmsgrey
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Re: HOURGLASSES
« Reply #17 on: Jan 3rd, 2013, 5:07am » |
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Note that the "best" solution here relies on the property of hourglasses that the rate of flow of sand is independent of the amount of sand remaining (true except possibly for the very last bit of sand that requires a jolt to nudge out of its initial, stable, configuration)
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jbcrowley
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Re: HOURGLASSES
« Reply #18 on: Jan 8th, 2013, 6:43am » |
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on Dec 1st, 2012, 10:28pm, Forest wrote:How about this;
Start both hourglasses at the same time.
When the 7min hourglass runs out, your 15 minute period starts. (There is 4 min remaining on the 11 min hourglass)
When the 11 min hourglass runs out, flip it
When it runs out again, 15 minutes has passed.
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This is correct and the most efficient way to do this.
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rmsgrey
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Re: HOURGLASSES
« Reply #19 on: Jan 10th, 2013, 6:13am » |
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on Jan 8th, 2013, 6:43am, jbcrowley wrote:
This is correct and the most efficient way to do this. |
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It requires 22 minutes to time 15. It's possible, by invoking the "uniform flow rate" property of hourglasses, to time 15 minutes in just 15 minutes, though with 4 turns of the glasses rather than just 3
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jbcrowley
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Re: HOURGLASSES
« Reply #20 on: Jan 10th, 2013, 10:41am » |
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on Jan 10th, 2013, 6:13am, rmsgrey wrote:
It requires 22 minutes to time 15. It's possible, by invoking the "uniform flow rate" property of hourglasses, to time 15 minutes in just 15 minutes, though with 4 turns of the glasses rather than just 3 |
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Heh... you are correct sir! My daughter (12) actually figured this out using the method (I believe) you are referring to. I should have known better to come on a forum like this with ravenous ambiguity-eating piranhas lurking about and make an ambiguous statement!  What I meant to say was that this method is the most efficient in that it takes the least amount of flips. 
*hides and waits to see if that was unambiguous enough*
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whizen
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Re: HOURGLASSES
« Reply #21 on: May 29th, 2013, 5:40pm » |
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on Jun 19th, 2012, 5:12am, rmsgrey wrote:
Or you could flip the 11 minute glass, let it run out then flip it again. When the sand in the top and the sand in the bottom are in a 7:4 ratio, 4 minutes have passed. |
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Or you can take out your cell phone and start a 15 minute timer at the moment you flip the hourglasses, cause its cool to see the sand flowing and you can calibrate your hourglasses at the same time. This method takes just 2 flips and exactly 15 minutes.
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