Something you neglected to tell us is that we should know if the odd ball out is heavier or lighter than the others.
X
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Something you neglected to tell us is that we should know if the odd ball out is heavier or lighter than the others.
X
X- The wording was deliberate I think. But tbh you dont need to know whether the odd one is lighter or heavier. Just that the rest are all identical in weight and that the odd one is of a different weight.
My puzzle is shamelessly copied from a source, but I wont disclose where until the answer comes.
At a local fairground, there are 10 two-seater cars attached to the ferris wheel. The ferris wheel turns so that one car rotates through the exit every minute. The wheel begins operating at 10 in the morning and closes 30 minutes later. Whats the maximum number of people that could have gone round on the ferris wheel in that time? :hmmm:
Nope, you do not have this information ahead of time and in some cases you would find it at the end, in some cases you won't.Quote:
Originally Posted by xman
If you knew the problem is pretty trivial. In fact in that case you can find the odd one among 27 balls with three measurements.
What is the definition of closing - when it's completely empty or when it stops accepting new riders. In the later case do partial rounds count?Quote:
Originally Posted by sidneykidney
Also does it mean that each car rotates trough the exit every minute or there is a car passing through the exit every minute - that's factor of 10 in speed you know :)
The ride closes at 10.30. At 10.30 the ride operator leaves the ride and goes home. Is that clearer?
Any more than that and you'll be asking me for the answer. Take a guess gugi. If you get it wrong you can always guess again.
42 people
Assuming there is a couple in the seat at 10, 60? or is it 58?
Well I don't like to guess when I can just compute the correct answer, but that would necessitate unambiguously defined problem :)Quote:
Originally Posted by sidneykidney
So presuming each car passes through the exit once a minute, and not that 'a car' passes through the exit once a minute, the first car would've passed 30 times during the time of operation and the other 9 cars would've passed 29 times each. Which means the maximum number of people who could've had a full ride is 2*(30+9*29)=582.Quote:
Originally Posted by gugi
Lee is correct. The thing which initially caught me out is that you have to have everybody off by 10.30. When the ride closes and the operator goes home he wont leave people sitting on the ride! Can you show your working Lee? Or did you google the answer? :p
You must be glad I'm not the operator...Quote:
Originally Posted by sidneykidney
And this means the speed of the wheel is the slow one where each car passes by the exit once every ten minutes... I guess it's the old folks roundabout.