Let A is the fastest
B is the second fastest
C is the third fastest
D is the slowest
Strategy 1
1 step
A and B go across the bridge (total time: 2)
2 step
A go back (total time: 3)
3 step
C and D go across the bridge (total time: 13)
4 step
B go back (total time: 15)
5 step
A and B go across the bridge (total time: 17)
Strategy 2
1 step
A and D go across the bridge (total time =10)
2 step
A go back (total time =11)
3 step
A and C go across the bridge (total time =18)
4 step
A go back (total time =19)
5 step
A and B go across the bridge (total time =21)
My first strategy is to let A and B go across the bridge, then let A back. Then let C and D go across the bridge, then let B back. Then let the rest( A and B) go across the bridge.
My second Strategy is to let A and D go across the bridge, then A back. Then let A and C go across the bridge, then A back. Then let A and B go across the bridge.
My first strategy time = B+A+D+B+B = A+3B+D
My second strategy time = D+A+C+A+B = 2A+B+C+D
My first strategy = my second strategy
A+3B+D = 2A+B+C+D
2B-A-C = 0
2B = A+C
Therefore, the sets of speeds for which both strategies give the same crossing time has to have 2B = A+C