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 2003-09-17, 10:44 #1 koal   Nov 2002 Vienna, Austria 41 Posts Steam Boat Rendezvous Two steam boats start to cross a lake from different locations, let's call them A-Junction and B-Ville. One travels from A-Junction to B-Ville, the other from B-Ville to A-Junction. They start at the same time, weather is bright, no wind and there is nothing (no torch, no Zippo (c), no submarine, …) that could disturb their journey. As they travel at different speeds, the boats meet at a point 7 miles away from A-Junction. They both reach their destination and - after waiting one hour - they leave and their steam engines take them back where they came from, but now meeting at a point 4 miles away from B-Ville. What's the distance between A-Junction and B-Ville? 1 mile = 1760 yards = 5280 feet = 63360 inches = 160934.4 mm, but a mile ist still a mile
 2003-09-17, 12:33 #2 wblipp     "William" May 2003 New Haven 236110 Posts Is there a strong current in this lake? On the first leg the boat that leaves from A travels 7 miles before they meet. On the second leg that same boat leaves from B and travels 4 miles before they meet. If the boats each had constant speed with no other effects, the two legs would be mirror images and the distances would be identical, so something else is going on.
 2003-09-17, 13:42 #3 wblipp     "William" May 2003 New Haven 3·787 Posts Nevermind. They each wait 1 hour - they don't start the return trips at the same time.
 2003-09-17, 14:09 #4 koal   Nov 2002 Vienna, Austria 4110 Posts You got it!
 2003-09-19, 00:23 #5 Maybeso     Aug 2002 Portland, OR USA 2×137 Posts Bump I've PM'd a try at the solution. Perhaps a diagram will help, and bump the topic too. Code: |<------- 7 ------->|------------------|<-- 4 -->| A-Junction P1 P2 B-Ville
 2003-09-19, 12:44 #6 dweick   Apr 2003 2×5 Posts Here is my attempt Two steamships, Abel and Baker Abel starts from the Port of A-Junction Baker starts from the Port of B-Ville Distance between the two ports is D Speed of Abel is Sa, Speed of Baker is Sb Travel time for Abel is (1 + 2D/Sa) (A spends an hour in port at B-Ville before returning) Travel time for Baker is (1 + 2D/Sb) (B spends an hour in port at A-Junction before returning) After Abel has traveled 7 miles, 7/Sa hours, it meets Baker After Backer has traveled (D-7 miles), (D-7)/Sb hours, it meets Abel Since they meet we know that (1) 7/Sa = (D-7)/Sb After Abel has traveled D+4 miles, 1 + (D+4)/Sa hours, it meets Baker a second time After Baker has traveled 2D - 4 miles, 1 + (2D-4)/Sb hours, it meets Abel a second time Since they meet this means that 1 + (D+4)/Sa = 1 + (2D-4)/Sb or (2) (D+4)/Sa = (2D-4)/Sb We solve (1) for Sa (3) Sa = 7*Sb/(D-7) Substituting (3) into (2) gives us (4) (D+4)*(D-7)/7*Sb = (2D-4)/Sb Multiply both sides by Sb gives us (5) (D+4)*(D-7)/7 = (2D-4) Or (6) D^2 - 17D = 0, D - 17 = 0 Therefore D = 17 miles Both equations (2) and (3) now tell us that Sa = .7Sb
 2003-10-03, 10:18 #7 koal   Nov 2002 Vienna, Austria 41 Posts Here's the solution. Most people are NOT able to solve it by use of equations. Congratulations to those who succeeded. But there exists a tricky way to get through the puzzle, without any equations. Watch it: First meeting 7 miles offshore (no matter, which shore we use, let's take A-Junction) As we can see, the sum of the distances is exactly the lake's width. 1) A|--------B1-------->7<---------------B2---------------|B Both steam boats continue their journey until they reach the other side. Again, the combined distances are equal to the lake's width. 1) A|--------B1-------->7<---------------B2---------------|B 2) A|<-------B2---------7----------------B1-------------->|B Now each boat waits a time (I said one hour), but the time has no impact on the solution, as long as both ships wait the same time. The boats start anew and now they meet 4 miles away from the other shore 1) A|--------B1-------->7<---------------B2---------------|B 2) A|<-------B2---------7----------------B1-------------->|B 3) A|------------------B2------------------>4<-----B1-----|B Again the distances' sum is the lake's width. So the boats crossed the lake 3 times "together" until now. That means, both boats have travelled the triple distance as they had, when they met first. At the first meeting point, boat B1 travelled 7 miles, so it must have travelled 21 miles until the second meeting point. Because it comes from the other side, we must subtract 4 miles. The calculation one hast to perform is 7*3-4 which gives the lake's width of 17 miles. 1) B1: 07mi, B2: 10mi 2) B1: 10mi, B2: 07mi 3) B1: 04mi, B2: 13mi --------------------- T) B1: 21mi, B2: 30mi
 2003-10-09, 06:59 #8 dweick   Apr 2003 1010 Posts Very good, I missed the "obvious" simplification. For example, I stated that when the boats first met they had traveled 7 miles and D-7 miles respectively. I missed that that added up to a total of D miles. When the boats met again I stated that they had traveled D+4 and 2D-4 respectively. That adds to a total of 3D miles or 3X the distance they had traveled when they first met. What I didn't see was that the time in port was irrelevant since they both spent the same amount of time there so I could simply say that 3*7 = D-4 or 3*(D-7) = 2D-4, both reduce to the same answer of D=17. Of course there are written down mathematical equations and then there are verbal story equations, I'm not sure that you solved it without any equations but you certainly solved it more simply than I did!

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