This problem comes from the book, "Fostering Algebraic Thinking" by Mark Driscoll.
Crossing the River
- Eight adults and two children need to cross a river. A small boat is available that can hold one adult or one or two children. (i.e., three possibilities: 1 adult in the boat, 1 child in the boat, or 2 children in the boat). Everyone can row the boat. How many one-way trips does it take for all of them to cross the river? Can you describe how to work it out for 2 children and any number of adults? How does your rule work out for 100 adults?
- What happens to the rule if there are different numbers of children? For example: 8 adults and 3 children? 8 adults and 4 children? Write a rule for finding the number of trips needed for A adults and C children.
- One group of adults and children took 27 trips. How many adults and children were in the group? Is there more than one solution?
Relevant guiding questions:
- What steps am I doing over and over?
- When I do the same thing with different numbers, what still holds true? What changes?
- Now that I have an equation, how do the numbers (parameters) in the equation relate to the problem context?
Let's have fun discussing this problem! Be sure to explain your process so that we may have the chance to understand your reasoning - if you have a picture, please post it on the web and then post the link here for us to look at.
Do we also need someone to row the boat back across the river, or are we using math magic?
Great Question - from the context of the problem it is hard to tell. But, realistically, boats cannot magically go back so yes, you have to think about the going back piece, too, we just aren't counting it in the total.
Does this help?
The following assumes that someone has to be in the boat at all times on both the crossing and return trips. Note: This is an horribly small boat and I can not think of anything this small outside of an amusement park ride!
2 kids go over and one gets off while the other goes back to allow an adult to go over. Then the kid takes the boat back. in this pattern every adult crossing will total 4 trips (2kids, 1kid back, 1adult, 1kid back). With this pattern, no matter how many adults you have, you can multiply by 4 to see how many trips to get them over then you can simply add one more trip to the end to get the 2 kids over.
For 8 Adults 2 Kids a total of 33 trips
For 100 Adults 2 Kids a total of 401 trips
In general given X Adults and 2 kids a total of 4X + 1 trips
I will withhold my thoughts on the other two bulleted questions to read what others think about the first part.