Question: Supposed you have 2 hourglasses, 1 runs for 7 mins and the other runs for 11 mins. How can you time a duration of 15 mins with these 2 hourglasses only?
PS: The time starts ticking the moment you turn over either or both hourglasses.
Updated: Problem solved by Beng Keong @ 2:39 PM, January 28, 2007 in the comment section of this post. Credits must also be given to Naka who solved the problem via sms.
13 comments:
how can we time 15mins with these 2 hourglasses? simple... just flip them over while starting a stopwatch or countdown timer loh... didnt say have to time 15mins using the hourglasses huh? ;D
wah lau thought u would have provided the answers/solutions... sounds like nus student hor? haha!
Very funny naka.
@monte: Of course it has to be only using the 2 hourglasses only lar..
Run both hourglasses, when the 7min glass ends, 4mins left on 11min glass. Start timing now, once 11min glass end flip to get your other 11mins( 4+11 = 15) :D
Sorry BK, the timing starts the moment you turn over either or both the hourglasses, i.e. you cant define the starting point whenever you like.
Nice try though.
Run both glasses,when 7min end flip again, when 11min glass ends flip 7mins again. (7+4+4 =15) :D hope i got it right this time
BK, that will only end up as (7 + 4 + 3) = 14 min right??
Haha, yea BK is right!
Start: 7 full, 11 full
At 7mins: 7 empty, 11 left 4 mins
Flip 7: 7 full, 11 left 4 mins
At 11mins: 7 left 3 mins, 11 empty
Flip 7: 7 left 4mins, 11 empty
At 15mins: 7 empty, 11 empty.
(where 7 is the 7 mins hourglass and 11 is the 11 mins hourglass.)
Good job! Heh.
oh okok, yes, i seem to get it now. interesting question.
Wee! Honourable mention on jh blog! :D
i feel so stupid seeing you all...
anyway, the method not accurate, cos when u flip the hourglass a few nanoseconds are gone.
nvm, i'm just crapping...
For a duration of 15 mins, nanoseconds are negligible. In fact, even 1 second is negligible cos the percentage error of (1/(15*60))*100 = 0.1% is so insignificant and the approximate is a good one. So, the method is good enough for the engineer.
just asking, is this supposed to be a difficult question?
Erm.. not really.. heh.
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