It was really hard at first when I just randomly click the characters, but I decided that there must be a way to it. So I wrote down all the characters like how long they take to cross the bridge so I tried some combinations they seem to be able to work. Then I wrote down the total time taken and try to get the 30 mins... But I don't know how is that a mathematical knowledge/skill
I agree with what Ming Yi said, especially the last part, and I noticed that you must group them according to the fastest to the slowest, so the fastest go with another 2nd fastest and the slowest go with the 2nd slowest. But another factor is, there must be at least 1 person crossing back, so I decided that the fastest ones would be the one bring the light back and forth.I had to try a couple of times to get the solution in which at first I tried to randomly move them like Ming Yi.I found 2 solutions:First:1 and 3 go first.1 go back right.12 and 8 go left.3 go back right.1 and 6 go left. (From this step on, it is not 3 who goes back because from him travelling back and forth, he consumes extra 4 min (6-2) more than 1)1 go back.1 and 3 go left.Second:1 and 3 go first.1 go back right.1 and 6 go left.1 go back.12 and 8 go left.3 go back to right.3 and 1 go left.Both solutions leave a minute.But all I did was to try and minimize the time taken.
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The problem is that when any 2 people are crossing, the time taken is dependent on the person with the longest time to cross the bridge. You have to ensure that the 1 min person is left for the last while the 8 & 12 min people get over as fast as possible. I think that is chapter 9 rate, ratio and speed.