Random Variable's Stupid Word Problem of the Day

[Random Variable]

You have two sandglasses/sand timers--one that measures 7 minutes and one that measures 11 minutes. How could you use the two sandglasses to measure 15 minutes?

 

[Syringer]

Turn both over at the same time.
When the 7 min one runs out, stop the 11 min one--that has 4 minutes remaining.
Restart the 7 minute one, let it run at the same time as the 11 min one, then stop the 7 min one when the 11 min one runs out.
Reset the 11 min one, let that run for 11 minutes, flip the 7 min one which has 4 mins left, 11 + 5 = 15.

 

[hypn0tik]

Turn both over at the same time.

When the 7 minute one runs out, there will be 4 minutes on the 11 min. one. Use a sharpie to mark the level of sand in the 11 min one. Given the symmetry of the timers, you can mark off the level on the other side as well. Let all the sand drain to one side (unnecessary, but we all know that beauty lies in symmetry).

Next, let the 11 min one run though completely, flip it over and then stop when the sand reaches the mark you made.

11+4 = 15

 

[hypn0tik]

Oops.

 

[Random Variable]

Originally posted by: Syringer
Turn both over at the same time.
When the 7 min one runs out, stop the 11 min one--that has 4 minutes remaining.
Restart the 7 minute one, let it run at the same time as the 11 min one, then stop the 7 min one when the 11 min one runs out.
Reset the 11 min one, let that run for 11 minutes, flip the 7 min one which has 4 mins left, 11 + 5 = 15.

What do you mean by "stop?"

 

[Random Variable]

Originally posted by: hypn0tik
Turn both over at the same time.

When the 7 minute one runs out, there will be 4 minutes on the 11 min. one. Use a sharpie to mark the level of sand in the 11 min one. Given the symmetry of the timers, you can mark off the level on the other side as well. Let all the sand drain to one side (unnecessary, but we all know that beauty lies in symmetry).

Next, let the 11 min one run though completely, flip it over and then stop when the sand reaches the mark you made.

11+4 = 15

What if you had no way to mark the timers?

 

[hypn0tik]

Originally posted by: Random Variable

Originally posted by: hypn0tik
Turn both over at the same time.

When the 7 minute one runs out, there will be 4 minutes on the 11 min. one. Use a sharpie to mark the level of sand in the 11 min one. Given the symmetry of the timers, you can mark off the level on the other side as well. Let all the sand drain to one side (unnecessary, but we all know that beauty lies in symmetry).

Next, let the 11 min one run though completely, flip it over and then stop when the sand reaches the mark you made.

11+4 = 15

What if you had no way to mark the timers?

You didn't specify. I win!

And by stop I guess he means turn the timer sideways.

 

[Syringer]

Originally posted by: Random Variable

Originally posted by: Syringer
Turn both over at the same time.
When the 7 min one runs out, stop the 11 min one--that has 4 minutes remaining.
Restart the 7 minute one, let it run at the same time as the 11 min one, then stop the 7 min one when the 11 min one runs out.
Reset the 11 min one, let that run for 11 minutes, flip the 7 min one which has 4 mins left, 11 + 5 = 15.

What do you mean by "stop?"

By stop I mean do your own homework

 

[Furyline]

start both at the same time, when the 7 min timer runs out there are 4 min left in the 11 min timer. Now the 15 minutes timing starts, let the rest of the 4 minutes finish, then run the 11 minute timer again to get 15 min.

 

[sygyzy]

Originally posted by: Furyline
start both at the same time, when the 7 min timer runs out there are 4 min left in the 11 min timer. Now the 15 minutes timing starts, let the rest of the 4 minutes finish, then run the 11 minute timer again to get 15 min.

Best Answer!

 

[Inspector Jihad]

Originally posted by: Random Variable

Originally posted by: hypn0tik
Turn both over at the same time.

When the 7 minute one runs out, there will be 4 minutes on the 11 min. one. Use a sharpie to mark the level of sand in the 11 min one. Given the symmetry of the timers, you can mark off the level on the other side as well. Let all the sand drain to one side (unnecessary, but we all know that beauty lies in symmetry).

Next, let the 11 min one run though completely, flip it over and then stop when the sand reaches the mark you made.

11+4 = 15

What if you had no way to mark the timers?

lay it on its side

 

[DaveSimmons]

Originally posted by: Random Variable
You have two sandglasses/sand timers--one that measures 7 minutes and one that measures 11 minutes. How could you use the two sandglasses to measure 15 minutes?

Both of the above are correct given the vague wording of the problem (turning sideways or marking). The third variation is just to say "start!" (or begin whatever it is you want to measure) after the 7 minute empties.

Edit: I've run into this one (the fluid version) this year in replaying KOTOR 1 and in Xenosaga. It's right up there in "common RPG puzzles" with the "give me food and I live, give me water and I die" and "what walks on four legs at dawn, two at noon, three at dusk" riddles.

 

[GMI]

Start both hourglass, when the 7 minute finish, flip it over (to restart it persay)
When the 11 minute finish, 4 minute has pass in the 7 minute hour glass, so flip the 7 minute hour glass over again, once that one runs out, 15 minutes have passed

edit:
dammit, too slow

 

[Furyline]

Originally posted by: GMI
Start both hourglass, when the 7 minute finish, flip it over (to restart it persay)
When the 11 minute finish, 4 minute has pass in the 7 minute hour glass, so flip the 7 minute hour glass over again, once that one runs out, 15 minutes have passed

edit:
dammit, too slow

if its any consolation this is probably the best answer in that it takes you exactly 15 min to do it

 

[Random Variable]

Originally posted by: Furyline

Originally posted by: GMI
Start both hourglass, when the 7 minute finish, flip it over (to restart it persay)
When the 11 minute finish, 4 minute has pass in the 7 minute hour glass, so flip the 7 minute hour glass over again, once that one runs out, 15 minutes have passed

edit:
dammit, too slow

if its any consolation this is probably the best answer in that it takes you exactly 15 min to do it

Yeah, I don't know what he's sad about.

 

[Turin39789]

Originally posted by: Random Variable

Originally posted by: Furyline

Originally posted by: GMI
Start both hourglass, when the 7 minute finish, flip it over (to restart it persay)
When the 11 minute finish, 4 minute has pass in the 7 minute hour glass, so flip the 7 minute hour glass over again, once that one runs out, 15 minutes have passed

edit:
dammit, too slow

if its any consolation this is probably the best answer in that it takes you exactly 15 min to do it

Yeah, I don't know what he's sad about.

meh, its impossible, both scenarios require you to flip an hourglass, which burns up an undetermined period of time, and you'll be off by .5-1.5 seconds

 

[Random Variable]

Originally posted by: Syringer

Originally posted by: Random Variable

Originally posted by: Syringer
Turn both over at the same time.
When the 7 min one runs out, stop the 11 min one--that has 4 minutes remaining.
Restart the 7 minute one, let it run at the same time as the 11 min one, then stop the 7 min one when the 11 min one runs out.
Reset the 11 min one, let that run for 11 minutes, flip the 7 min one which has 4 mins left, 11 + 5 = 15.

What do you mean by "stop?"

By stop I mean do your own homework

Actually, the homework I've been working on involves determining the nature of the fixed points of a system of non-linear differential equations by using Jacbobian analysis.

 

[mobobuff]

Flip both.
Stop 11(a) when 7(b) runs out. Giving you 4(a) and 7(b).
Flip both.
Stop 7(b) when 4(a) runs out. Giving you 11(a) and 3(b).
Flip both.
Stop 11(a) when 3(b) runs out. Giving you 8(a) and 7(b).
Flip 8(a).
Flip 7(b) immediately after 8(a) runs out.

That's how I think the problem was supposed to be answered. "Stop" means to lay the timer on its side.