The math problem:
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?
Repeat the problem using counters. I want to know the number of crossings it takes to shift the 8 adults and 2 children.
What if one of the adults was sick and didn’t end up going camping? How would that change the number of crossing?
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?
Today our teacher gave us a math problem to solve. We had to work in groups to solve it out. Here is how my group solved this problem:
Firstly, we got counters to experiment with the different possibilities. After trying for a while, we eventually came up with the solution. The solution is letting both children first get across the lake, then bringing one back to the island. Next, one adult goes over and the other child goes back to the island. The whole “cycle” begins again afterwards until everyone is across the lake.
Secondly, we started again to count how many crossings it would take for all of them to get back to their destination. We counted 33 crossings in total with eight adults and two children.
After that, we tried step two again with seven adults and two children. Using the same way to cross the river with seven adults and two children, we found out that that would take up 29 crossings, back and forth. After some thinking, we figured out the pattern in the problem. The pattern turned out to be 4n + 1 = number of crossings.
Lastly, to make it even more clear for us, we started to try this five times, all with different number of adults but keeping two children. The first time, we did it with three adults. 3 x 4 is 12 and 12 + 1 is 13 so the number of crossings for three adults and two children would be thirteen. Then, we tried it with 6. 6 x 4 + 1 = 17. We also tried it with 9, 10 and 5 using the same pattern.
This problem was helpful because it helped me with me algebra skills and it was fun and engaging for me to do. I hope this helped you a little too!