- On the weekend, friends of mine went camping. They were camping on a small island in the middle of the lake. There were eight adults and two children in all.
- When they went to cross the lake and return home their boat was missing. They searched and searched but all they could find was an old canoe. It wasn’t as big as the boat and they were worried it wouldn’t carry them all. So, they tested it and found the boat could carry either:
- • One adult
- • One or two children
- At first they thought some of them would be stranded forever, but finally, they figured out how to get them all safely across the lake.
- Can you figure it out?
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- Repeat the problem using counters. I want to know the number of crossings it takes to shift the 8 adults and 2 children.
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- What if one of the adults was sick and didn’t end up going camping? How would that change the number of crossing?
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- Let’s adjust the number of adults. How does the number of crossings change when the number of adults changes? Record all your data. Can you find an algebraic pattern?
That whole chunk of writing there is the math problem. Anyway our unit for math has been on patterns and algebraic thinking. If you read the problem we have to first get 8 adults and 2 children on the safe island without making the boat sink. And if you read the whole thing (if you did not then read it) only 1 adult or 2 children could get on the boat. And they can’t swim. So this is what we had to do. We had to take the 2 children to the safe island and leave 1 on the island. Then the other one would go back to the adults. Then 1 adult would go on the boat to the safe island and the 1 kid would come back. This would keep on repeating. The total number of trips was 33. Then we would try 7 adults and 2 children. This time it was 29 trips. We finally got how this works. If you have 100 adults and 2 children you would get a total of 401 trips because you have to x4+1. So if you multiply 4×100 you would get 400 and if you add 1 you get 401.
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