Three people (A, B, and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. There is also a bicycle available and any person can cross the bridge in 1 minute with the bicycle. What is the shortest time that all men can get across the bridge? Each man travels at his own constant speed.
On first reading, I thought this was too easy. Looking for another interpretation, I came up with the answer that would be gross with another two pairs of seconds.
C rides the bike while A starts walking. A then bikes back 2 minutes after he started (3 minutes), and give the bike to B. A then walks back in 2 more minutes for a total of 5?
Unless there is a basket on the handlebars, and any two of the people are small enough to fit in it, then everyone crosses on the bike in 1 minute...
1) 1 minute. The question stated times for "persons" to cross, but asked about fastest time for "all men" to get across. (I do learn from my own puzzles ;-) ) If there were no man, there would be no answer. One man takes the bike across, and "all the men" wait for the women on the other side.
2) 1:48. "Men" implies plural. A is the only woman. B rides 80% of the way across the bridge, in 48 seconds, and leaves the bike and walks the remaining 1/5 in a minute. C walks 80% of the way across the bridge in 1:36, then hops on the bike to finish the trip in 12 more seconds.
3) 2:20. This is the answer I originally gave, thinking that the trick only required that all three reach the other end, not that they stay there. A rides to the other end (1:00), then back to B at the 1/3 way point (0:40). B rides to the other end (0.40), and finds that C has already arrived. Now, they have all REACHED the other side, but A didn't stay, and is now back near the middle of the bridge.
I THINK I have the solution right, but am very suspect of my algebra. My answer is a weird fraction, about 3 seconds more than yours. Maybe a spreadsheet to check my arithmetic...
A's speed is 1/10 (in bridges per minute), B's speed is 1/5, C's speed is 1/2, and the bicycle's speed is 1.
The fastest way I found to get everyone across is for B and C to start out on foot and A to start out with the bicycle. At a point y A will get off the bicycle and walk the rest of the way. Eventually C will get to the bicycle abondoned by A, then ride backwards to a point x, leaving the bicycle there, then turning around and walk until he reaches the end. Person B will walk until he reaches the bicycle left by C and then ride the rest of the way.
Below are the times that each will take to cross, in terms of x and y:
A: 1*y + 10*(1-y)
B: 5*x + 1*(1-x)
C: 2*y + (y-x) + 2*(1-x)
Next equate these equations: 10 - 9y = -3x + 3y + 2 = 4x + 1.
To solve set up two linear equations:
10 - 9y = -3x + 3y + 2 -> 3x - 12y = -8
10 - 9y = 4x + 1 -> 4x + 9y = 9
Then solve for x and y:
x = 12/25, y=59/75.
Given these points it will take each person 73/25 = 2.92 minutes to cross.
I agree completely with your reasoning, and with your equations. I made an error in how I expressed the distance that C rode back, which resulted in error in my answer. I believe your answer is correct. Good puzzle.
I assumed that the minimum would be with maximum use of the bicycle, so I had C ride the bicycle back until he met up with B. Then B took the bike, and left it near the end of the bridge, where A picked it up again for the last leg. But my time was a little longer than yours. I did my algebra on intervals of time rather than distances.