Assuming that Bo and Jo cannot set up secret stashes from each other (as otherwise the problem is fairly trivial), but assuming that they can set up a shared stash, one strategy they could use is to:
Set up a cache anywhere between 18 ∕ 25 mile and 13 ∕ 16 mile from A, where they will first deliver all the bars to. The cache will be far enough from B to prevent either of them stealing bars and absconding on the 10:00 train. Once they've brought all bars to the cache, they can take the bars to B and catch the 11:00 train.
Proof:
It takes Bo 30 minutes to transport a bar 1 mile, and Jo 40 minutes. It takes them 20 minutes to walk 1 mile unladen. A round trip takes Bo 50 minutes, and Jo 60 minutes. It thus takes Bo 50 + 30 = 80 minutes to transport 2 bars from A to B, and Jo 60 + 40 = 100 minutes, so neither of them is stealing onto the 9:00 train.
If they take bars straight from A to B, Jo will arrive with the 4th bar (his 2nd) after 100 minutes, at 9:40. At this point Bo has completed two round trips and is back at A - even if he goes back to B unladen he won't arrive until 20 minutes later, i.e. 10:00, by which point Jo is on the departing train with 4 bars of gold. So clearly they must leave bars at an intermediate location.
Let the position of the intermediate cache be x miles from A. At maximum efficiency, the 4th bar will be delivered by Jo after 100x minutes. Assume Bo has hidden in wait the whole time, and will immediately start stealing the bars as soon as Jo sets back off for A.
Bo has to transport 4 bars, which takes him (50 × 3) + 30 = 180 minutes per mile. As he has to transport them (1 − x) miles, this takes 180 − 180x minutes.
The total elapsed time is thus 180 − 80x minutes, which we thus require to be more than 115 to prevent Bo getting onto the 10:00 train. 180 − 80x > 115, so 80x < 65 and x < 13 ∕ 16.
Note if Jo wants to steal the bars, he also has to start his theft as soon as he drops off his 2nd bar, which is the same time Bo starts his attempt - but as Jo is slower than Bo, if Bo can't steal the bars, neither can Jo.
Now assured of co-operation, we just need to sort out the logistics. Bo, being faster, will deliver 4 bars to the cache, and Jo 3.
Bo completes his task after ((50 × 3) + 30)x = 180x minutes.
Jo completes his task after ((60 × 2) + 40)x = 160x minutes.
If we have Bo then take 4 bars from the cache to B, it will take him a further 180(1 − x) minutes, which means he will only finish after a total elapsed 180 minutes - i.e. at 11:00, too late to load the train. Jo must thus use his headstart to help - Jo will pick up 4 bars from the cache, and Bo 3.
Transporting 3 bars takes Bo (50 × 2) + 30 = 130 minutes per mile, so transporting 3 bars for (1 − x) miles takes him 130 − 130x minutes for a total of 130 + 50x minutes.
Transporting 4 bars takes Jo (60 × 3) + 40 = 220 minutes per mile, so transporting 4 bars for (1 − x) miles takes him 220 − 220x minutes for a total of 220 − 60x minutes.
Bo will finish first if 130 + 50x < 220 − 60x i.e. if 110x < 90. But since x < 13 ∕ 16, 110x < 1430 ∕ 16 = 89.375, so this is always true.
When Bo finishes, Jo will be (220 − 60x) − (130 + 50x) = 90 − 110x minutes away from B. We could have Jo carry the last bar the remaining distance, but it will be quicker to have Bo carry it. In fact, if x < 3 ∕ 4, we must have Bo pick up the bar, since 220 − 60x > 220 - 45 = 175 in this case.
Bo will thus rendezvous with Jo at a combined approach speed of 4.5mph, which will take (90 − 110x) × 1.5 ∕ 4.5 = 30 − (110 ∕ 3)x minutes. Total time elapsed is now 160 + (40 ∕ 3)x minutes.
The remaining distance takes 60 − (220 ∕ 3)x minutes at 1.5mph, so at 2mph it will take Bo (60 − (220 ∕ 3)x) × 1.5 ∕ 2 = 45 − (165 ∕ 3)x minutes. Jo is faster and walks ahead to B. Total time elapsed is 205 − (125 ∕ 3)x minutes.
We thus require 205 − (125 ∕ 3)x < 175, in order for Bo to arrive with the final bar in time to load the train. Thus (125 ∕ 3)x > 30, and so x > 90 ∕ 125 = 18 ∕ 25.
Thus, so as long as the cache is between 18 ∕ 25 and 13 ∕ 16 miles from A, Bo and Jo can get all the bars to the train in time. The process is quickest with the cache at just shy of 13 ∕ 16 mile (since exactly 13 ∕ 16 mile lets Bo steal the bars), taking them just over 171 minutes.
Addendum:
It's possible to optimise this further. My above solution has Bo directly pick up a bar from Jo, but this is inefficient, since Jo is now walking towards B unladen. To make best use of their time, when walking towards B they should always be carrying a bar, and never carrying a bar away from B. This is possible by dropping bars partway along to be picked up later.
The optimal solution has both Bo and Jo finishing at the train station at the exact same time. Otherwise, the one who finishes first could be taking some work off his colleague's hands.
Let b be the distance Bo carries bars for, and j be the distance Jo carries bars for. b + j = 7 miles.
The time taken by Bo is thus (50 × (b − 1)) + 30 = 50b − 20.
The time taken by Jo is thus (60 × (j − 1)) + 40 = 60j − 20.
Thus 50b − 20 = 60j − 20, therefore 5b = 6j. b = 7 × 6 ∕ 11 = 42 ∕ 11, j = 7 × 5 ∕ 11 = 35 ∕ 11.
This can work out as Bo and Jo each carrying 3 bars the whole way, with the 7th bar being carried by Bo 9 ∕ 11 of the way, and by Jo 2 ∕ 11 of the way.
As it turns out, 2 ∕ 11 < 1 - (13 ∕ 16); our cache is always at least 2 ∕ 11 miles from B. So Jo can transport one bar for 2 ∕ 11 miles, then drop it for Bo to take the rest of the way. It thus takes him 2 ∕ 11 × 60 = 120 ∕ 11 minutes to drop off that one bar, and ((60 × 2) + 40)(1 − x) = 160(1 − x) to drop off the other 3 at the station. His total time elapsed is 160x + 160(1 − x) + 120 ∕ 11 = 160 + 120 ∕ 11 = 170 and 10 ∕ 11 minutes.
Bo transports his 3 bars in 130(1 − x) as before. He goes back (1 − x) − 2 ∕ 11 = 9 ∕ 11 − x miles to pick up the bar Jo dropped, and bring it to the station. This takes 50 × (9 ∕ 11 − x) = 450 ∕ 11 − 50x minutes. His total time elapsed is 180x + 130(1 − x) + 450 ∕ 11 − 50x = 130 + 450 ∕ 11 = 170 and 10 ∕ 11 minutes, the same as Jo.
So, Bo and Jo both arrive just before 10:51, giving enough time to load the bars and get a coffee and a ham sandwich. This is actually independent of where they place the cache too, as long as it's within the limits derived earlier. The cache just stops either one absconding onto the 10:00 train.
