Thread: Riddle Me This, SRP

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?

Originally Posted by xman

Something you neglected to tell us is that we should know if the odd ball out is heavier or lighter than the others.

X

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.
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.

Originally Posted by sidneykidney

Whats the maximum number of people that could have gone round on the ferris wheel in that time?

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?

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?

Originally Posted by sidneykidney

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.

Well I don't like to guess when I can just compute the correct answer, but that would necessitate unambiguously defined problem

Originally Posted by gugi

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

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.

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?

Originally Posted by sidneykidney

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!

You must be glad I'm not the operator...

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.