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. Each can row the boat. How many one-way trips does it take for all of them to cross the river?
What if there are 2 children and 100 adults?
What if there are 2 children and any number of adults?
What happens if there are different numbers of children? For example, 8 adults and 2 children? 8 adults and 4 children?
Write a rule for finding the number of trips needed for X adults and Y children.
Explain why the rule works in at least two different ways.

A = 8, C = 2 => T = 33

A=100, C = 2 => T = 401

A = any, C = 2 => T = 4A + 1

A = 8, C = 4 => T = 37

As long as the boat can carry one adult or one or two children, the following equation should solve the problem.

T = 4A + 2(C-2) + 1

Step-by-step explanation:

To transfer one adult, 4 one-way trips need to be made as follows:

1. Two children cross the river

2. One of them takes the boat back to the other side

3. One adult crosses the river

4. The other child takes the boat back

These four steps are repeated until all adults are crossed. At this point no children has crossed yet.

Then for every 2 children except the last 2, 2 one-way trips need to be made if the number of children is greater than 2:

1. Two children cross

2. One takes the boat back

For the last 2,only one trip needs to be made.

A = number of adults

C = number of children >= 2

T = total trips

T = 4A + 2(C-2) + 1

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !