Three students had a twin bicycle (similar to above, but only 2 seats) and wished to go just 40 miles. The bike could complete the journey with 2 riders in one hour, but could not carry the three simultaneously.
Well, one who was a good pedestrian could walk at the rate of one mile in 10 minutes. another could walk the same in 15 minutes, and the other in twenty.
What would be the best possible time in which all three could get to the end of their journey?
* the bike can backtrack
* the bike travels at the same speed regardless of which pair of students is riding it
Not sure how many will be able to get this hard math puzzle; if you can figure it out out, feel free to enter your answer below. Will unmask in several days…
the best I can get is 2h 16m 10.95s
Let’s call A (fastest walker), B and C (slowest walker)
Strategy: A starts walking while B and C ride the bike. At a certain point B jumps off the bike and starts walking, C backtracks and picks up A. C walking and A and B riding, they finish the journey altogether.
I made a mistake in my last sentence:
B walking and A and C riding, they finish the journey altogether.
I will go out on a limb an guess one person can peddle the bike at the same speed as 2 people. So we have..
bike speed = 40mph
pedestrian 1 = 6mph
pedestrian 2 = 4mph
pedestrian 3 = 3mph
Immediately we see that the best solution is to keep pedestrian 3 on the bike at all times as he/she has the worst time. We might start off by saying we let pedestrian 1 start walking while we ride the bike all the way to the end, drop one person off, and go back and get pedestrian 1 in which we turn around and cross the finish line. If we assume no lag for acceleration, we see this time takes just under 2.48 hours.
The flaw in this line of reasoning is that the Pedestrian 2 is making no progress (being at the finish line already) for ~1.48 hours while he waits on the bike and pedestrian 1 to catch up. So what we need to do is drop Pedestrian 2 off at a point, where he can make forward progress while the bike gets pedestrian 1, and they all meet at the finish line at precisely the same time.
I’ll elaborate on this solution on my next comment…
Conceptually I solved this as the following…
Pedestrian 1= P1
Pedestrian 2 =P2
Pedestrian 3 stays on the bike.
t1= time when bike drops off P2 and heads back to pick up P1
t2= time from t1 until bike back tracks and picks up P1
t3= time from t2 until bike and P2 crosses finish line at same time
Now we can extract 3 equations from this by using the fact that speed *time = distance. We know the distance is 40 miles.
So equation 1
P1*t1 + P1*t2 + 40*t3 = 40
Verbally, this equation is P1 walks while the bike drops off P2, and continues to walk until the bike picks him up, upon they continue towards the finish line. No back tracking, just the distance traveled at different times and speeds.
Equation 2
40*t1 + P2*t2 + P2*t3 = 40
Verbally, this equation is the bike travels until it drops off P2, which P2 walks the rest of the way. Again no back tracking.
Equation 3
P1*t1 + P1*t2 + 40*t2 + P2*t2 + P2*t3 = 40
P1 travels until the bike picks him up, the bike back tracks from P2, P2 walks the rest of the distance.
Now, we have 3 equations and 3 unknowns which we can work out.
We see if we use P1 at 6mph and P2 at 4mph, we get.
t1 = 0.858921162
t2 = 0.634854772
t3 = 0.77593361
If we reverse P1 and P2, we see the reflection…
t1=0.77593361
t2=0.634854772
t3=0.858921162
The total time is the same however, ~2.26971 hours.
Need to address some assumptions.
Bicycle can backtrack with 1 rider? Yes
At the same speed as if it were 2 riders? Yes
The 15 & 20 min per mile walkers will cycle to the end point in 1 hr. while the 10 min per mile walker would have walked 6 miles.
1 of the cyclist will get off while the other returns to the walker.
(40 - 6) miles = (40miles/hour)t + (6miles/hour)t
=> t = 17/23 hour.
Therefore, total time taken = 1 + (17/23 x 2)
= 2hr 28min 41.7sec
a quick question: can the bicycle be used by just 1 rider. if so is it at same speed?
If the question that the_god_dellusion post has a negative answer then the time would be: 400 minutes or 6 hour and 40 minutes(the time that the faster pedestarian takes to make the journey).
If it has a possitive answer then the problem is harder:
initial time: 0
2 hours the bikers(slower pedestrians)
1 hour biker(one of the previous)
3 hours on foot (faster pedestrian)
distance biker: half the road backtrack direction(mile 20)
distance pedestrian: 18 miles (mile 18)
time passed: 3 hours
now look for the approach between the biker and the pedestrian (2 miles of distance)
the biker takes 3 min per mile ( assuming the same speed that the 2 bikers), the pedestrian get 30% of a mile in 3 minutes so :
time passed: 3 hours 3 minute
biker mile 19 backtrack, pedestrian mile 18.3
time passed: 3 hours 4 minutes
biker mile 18.67 backtrack, pedestrian mile 18.4
time passed: 3 hours 4 minutes 30 seconds
they finally meet seems forever right jajaja in mile 18.5 now change the direction an go back to the line 21.5 miles at the same speed
1 hour and 4 minutes and 30 second has passed finally they arrive:
total time 4 hours and 9 minutes
cool I think.
I was too tired yesterday to post the solution.
With A, B and C being the riders (A being the fastest and C the slowest), I think the fastest solution would be to let A walk while B and C ride the bike for 51.53527min (da=5.153527mile, db=dc=34.356846), then B would walk while C goes back to pick up A then they head towards the destination. A and C would meet at 89.626556min (da=dc=8.962656, db=36.896266), and all three would reach the destination at 136.182573min (da=db=dc=40).
Of course we could let B start walking, which would yield the same end result, but with different intermediate timings and distances.
C should be chained to the bike
If the three students can walk A=6 mph, B=4 mph, and C=3 mph, then the fastest way for them to reach their destination is as follows:
1) B and C start out riding, and A starts out walking. This takes “x” hours.
2) B dismounts and continues on foot, C goes back to meet A. This takes “y” hours.
3) A and C ride to the destination, B walks the rest of the way there. This takes “z” hours.
(The roles of A and B are reversible, but it’s important that C, the slowest walker, be the one to go back in step 2.)
There seems to be some question as to whether C can ride the twin bicycle by himself, or if he has to walk it back (Thank you, the_god_dellusion, for your insight). We can solve each case separately:
CASE #1: A twin bike can be operated by one person, and C can ride back to meet A.
The equations you would solve are:
(Distance traveled, A’s perspective): 6x + 6y + 40z = 40
(Distance traveled, B’s perspective): 40x + 4y + 4z = 40
(Distance traveled, C’s perspective): 40x – 40y + 40z = 40
The solution is x=.859, y=.635, and z=.776, for a total time of 2.27 hours.
CASE #2: A twin bike cannot be operated by one person, and C must walk back to meet A.
The equations you would solve are:
(Distance traveled, A’s perspective): 6x + 6y + 40z = 40
(Distance traveled, B’s perspective): 40x + 4y + 4z = 40
(Distance traveled, C’s perspective): 40x – 3y + 40z = 40
The solution is x=0.689, y=2.603, and z=0.506, for a total time of 3.798 hours.
(I think Case #1 is more in keeping with the spirit of the problem, but that’s only an opinion.)
I have the same question as the delusional god. If this is permitted, my answer is 2:16:10.95
BTW, ever try to ride a tandem bike at 40mph? That would not be a fun experience!!
The fast walker walks while the others ride. After 1 hour the riders arrive a their destination, while the walker has 34 miles to go. The cyclists head toward the walker, closing at 46 mph. They meet in 17/23 hours at a point 680/23 miles from the end. The walker can take the place of one of the cyclists (who has already completed the journey) and finish in another 17/23 hours. Total time is 2 & 11/23 hours, or 2 and a half if you round up to account for turning around and changing riders.
If they all must end up at their destination together, then this is a much harder problem. I’ll get back to you.
If all 3 must get there at the same time, it will take 4 & 21/44 hours.
The fast walker walks while the others ride. After 24 & 6/11 miles, the second fastest walker gets off and continues on foot while the slowest walker waits with the bike.
sorry the_god_delusion, the comment i posted the other day never made it up; yes the bike can be used by just 1 rider, and it moves at same speed.
since this is a very hard problem, will give some more time on this; next week will unmask submitted answers. We have several already, and not everyone concurs…
I’ve posted an illustration to help with my explanation at http://www.tulane.edu/~venom/images/pedestrian.jpg
I called the fastest (6mph) walker “person A”, and the 2nd fastest (4mph) walker “person B”. Person C never gets off the bike.
The total travel time is broken up into three intervals as follows :
At time zero, all three people set out, A is walking, B & C are on the bike.
At the end of time interval 1, the bicycle drops off person B who begins to walk the remaining distance, while person C turns the bike around to go fetch person A.
At the end of time interval 2, person A meets the returning bicycle, jumps on, and heads for the finish line. Person B will have covered some of the remaining distance to the finish line.
At the end of time interval 3, person B will cross the finish line at exactly the same time as the bike carrying persons A and C.
Mathematically speaking:
If the duration of the first interval is T1 hours, the bike will go a distance of 40 miles/hour * T1 hours, and person A will go a distance of 6 miles/hour * T1 hours.
If the duration of the second interval is T2 hours, the bike will go backwards for a distance of 40 miles/hour * T2 hours, person A will go forward a distance of 6 miles/hour * T2 hours, and person B will go forward a distance of 4 miles/hour * T2 hours.
So, when the returning bike meets person A:
(6 * T1) + (6 * T2) = (40 * T1) - (40 * T2)
which can be simplified to read: 34 * T1 = 46 * T2
and simplified even further: T1 = 23/17 * T2.
This equivalence will always be true, irrespective of how far the bike travels before turning around.
With this in mind, I had the bike go all the way to the finish line during the first interval, and drop off person B. I know that isn’t the fastest way, but here’s why I did it…
For this scenario I know that T1 is exactly 1 hour, and I can calculate (using my simplified equation above) that T2 = 0.73913 hours.
If the bike turns around again at this point, and continues all the way back to the finish line, then the remaining distance is equal to the total of 40 miles, minus the distance from the start to the point where person A meets the bike, which we already set to be (6 * T1) + (6 * T2). Using this info, we can calculate T3 as the time it would take for the bike to cover the remaining distance at a rate of 40 miles/hour: (40 -[(6 * T1) + (6 * T2)])/40
In this scenario, T3 = T2, but we’ll change that soon enough.
So when person B gets dropped off, he starts walking, and walks until the bike crosses the finish line carrying person A, which is a time span equal to T2 + T3. We know that person B goes 4 miles/hour, so he’ll cover a distance of 4 miles/hour * (T2 + T3) hours.
Now, if you follow my illustration, and if my logic is sound (big ifs) regarding the most efficient scenario, then the total time will be the shortest when (40*T1)+(4*(T2+T3)) = 40.
At this point, my mathematical sophistication ends, and I just input these equations into a spreadsheet, starting with a T1 of 1 hour, and then I incrementally decreased the value of T1 until the preceding equivalence was approximately true. Here’s how it breaks down:
T1 lasts 0.85892 hours, and person A walks 5.15352 miles, while the bike travels 34.3568 miles.
T 2 lasts 0.634854 hours, by which time person A will have walked an additional 3.809123 miles, the bike will have gone backwards a distance of 25.39416 miles, and they will meet up at the 8.962643 mile marker. Meanwhile, person B will have walked a further 2.539416 miles.
T 3 lasts 0.775934 hours, during which time the bike will have gone 31.03736 miles to reach the finish line, and person B will have covered the last 3.103736 miles to finish at the exact same time.
So if I’m correct, the total time to get everyone across the 40 mile finish line is (very, very close to) 2.269708 hours.
I should say that I chose to have had the 6mph walker start on the ground because it seemed most efficient to have the fastest person on the ground for the longest time. I just tested this out, and you can reach the same final answer with the 4mph walker starting out, and the 6mph walker being dropped off. T1 just gets shorter so that the fastest person is still on the ground for the longest time, but at the opposite end of the path.
2h 18m 15s is what I’m getting.
Okay, if all three must arrive at the same time AND the bike goes just as fast with one rider as with two, then the shortest time is 2 & 65/241 hours (about 2 hours and 16 minutes).
The fast walker walks while the others ride. After 34 & 86/241 miles, the second fastest walker gets off and continues on foot while the slowest walker rides back to pick up the first guy.
Let t = time, x = the distance traveled by the first two cyclists, and y = the distance backtracked. Then the time for the fastest walker is t = (x - y)/6 + (40 - x + y)/40,the time for the medium walker is t = x/40 + (40 - x)/4, and the time for the slowest walker (who stays on the bike throughout) is t = (40 + y + y)/40. Solve for t algebraically.
I get 3 hours 41 9/11 minutes. Fastest walker starts on foot. Bike rides forward and drops off the second fastest walker, returning for the fastest walker and arriving at the finish at the same time as the walker.
A starts walking while B & C ride most of the 40 miles; then B gets off and walks the rest of the way while C on the bike goes back to get A. I have A as the 4mph walker and B as the 6mph walker; I think the bike can turn back at the 31mi mark and do the whole thing in 2h 16min (2.27h).
i got approx 136 mins, ie 2 hrs 16 mins. do we need to be a lot more precise?
It was a lot harder than I initially thought, cool/>>
Yes, this is a really hard problem.
The best official answer I have for this is just under 2.3 hours. It seems many of you (MFox, MichaelC, Bilbao, Mashplum, Hex, Shawn,the_god_delusion, kitchen, MrMan) are coming to this answer, and several very nice detailed explanations given above.