Marc, himself, his blogs, and you reading them.
March 31, 2006
Found myself at an incredible difficult fund-raising-quiz last weekend. Which translates to "Paying your share to feel stupid all night." Go figure!
Anyways, among the load of those silly "Why can't I remember stupid facts" questions there was also the more interesting logic and riddle section. Here is one from that set:
Two ferry boats (A and B) operate on a river. Both boats have different (but constant) speeds. At a given time they start off together at opposite sides of the river and meet 155m from the side where B started. When they reach the other side both need to wait the reglementary 12 minutes before (each at their own time now) can start crossing over again. During this second crossing they meet at 85m from the side where B took it's second start.
Obvious question: How wide is the river?
# Posted by mpo at 07:28 AM | TrackBacktime to change guts...
Posted by: -marc= at March 31, 2006 09:31 AMHmmm, to me it looks like a problem with three unknows and just two equations. Let's call t1 the time when they meet the first time, Va the speed of A, Vb the speed of B, w the width of the river.
Then we have:
t1 = 155/Vb = (w - 155)/Va
Then boat B takes t2 = (w - 155)/Vb to reach the other side, while A takes t3 = 155/Va to reach its side.
They both wait 12 minutes, then B takes t4 = 85/Vb to reach the new meeting point and A takes t5 = (w - 85)/Va to reach it.
Obviously
t1 + t2 + 12 + t4 = t1 + t3 + 12 + t5
which simplifies to:
t2 + t4 = t3 + t5
or:
(w - 155)/Vb + 85/Vb = 155/Va + (w - 85)/Va
This gives us our second eq, but we still have three unknows: w, Va and Vb.
What gives?
Posted by: Ugo Cei at March 31, 2006 02:06 PMThis was exactly as far as I got, then I threw the notes... :(
Posted by: Patrick Ahles at March 31, 2006 02:09 PMWrite your equations in term of Va/Vb, eliminate Va/Vb between the two equations and you have the solution for w.
Yet I'm asking myself if there's some hidden symmetry I'm overlooking because I can't imagine you're supposed to solve equations within the time frame of a quiz?
@Pascal:
This was part of a pen and paper section in the quiz. Teams where 4 persons each and per 2 series of 10 questions that required direct answers we received another 10 questions on paper to solve in parallel...
So there was time for equations, but still: yes, as you've guessed, there is more 'symmetry' to be found.
Symmetry is quite a geometrical term... So we exploited that by drawing the situation in a distance-time diagram (where their mutual trajectories become lines with an inclination relative to their unknown speed).
In hindsight of that drawing there is this set of Socratic questions to be asked:
At meeting point one:
- How much distance did they cover together?
- How much each ?
At meeting point two:
- How much distance did they cover together? (counting from the first start)
- How much each ?
- In which way does the 12 minutes affect the measurements? What would change if they wait another time?
- What does that tell you about the relation between both meeting *times* (counted in traveled time, not real time)?
- And given the speeds were constant during travel, what can we say about the relations between all distances?
Gutt feeling: 155 + 85 = 240m
Posted by: Tom Klaasen at March 31, 2006 09:06 AMSome pictures might defend this position.