This problem is around a mother with 2 kids who stroll by a gumball machine and both her children need a gumball of the same shading. The problem had three sections to it. The first being what is the most extreme sum the mother would need to spend all together for her children to have the same shading gumballs. The second part inquired as to whether there were more hues or more children? What's more, the last part inquired as to whether now the children needed diverse hues?
For this problem I began with the 2 children and 2 colors then proceeded onward to 2 children and 3 colors and continued changing the numbers while making table. In any case, this table was not quite the same as the past ones since there is a third variable in this problem. When I had different circumstances, I found the example in my table and found an equation. Once I understood why the equation worked, I went onto the following part where there are just 20 gumballs in the machine and it lets you know what number of number of every shading there are. This was somewhat different on the grounds that the past circumstance was accepting there was a 50/50 shot of getting either shading however with this new one there was not a 50/50 chance furthermore now the children need distinctive hued gumballs.
Our solution for the first part was the equation c(k-1)+1=p. “c” being the amount of colors of gumballs, “k” being the amount of kids, and “p” being the pennies used. For the next part where the kids want one gumball each of different color, we got 9 since there are 8 yellow gumballs, 7 red, and 5 black, if by chance, you kept getting the same color gumballs, by the time you got to the 9th gumball, there is no way it could be the same color as the 8 before since there are no more than 8 gumballs of the same color. We used the same process for when the twins each wanted 2 gumballs of the same color but different from each other. If you were to get the first 8 gumballs the same color, the 9th gumball would be a different color, then if the next gumball happens to be the third color you have not gotten yet, the next gumball (11th) has to be the same color as any gumball you’ve already gotten. Therefore the maximum the mom would spend would be 11 pennies.
This problem was hard to grasp because we did not know what to do if we did not know how many gumballs were in the gumball machine in total. If there were 400 gumballs in the machine, and half are one color and half the other, even though it’s nearly impossible, it could happen that you get 200 gumballs of the same color, then it would take you 201 pennies in order to get different colored marbles. Then we realized that it didn’t matter since we were trying to get them the same color marbles.
I am proud of myself for completing this problem, because it is definitely one of the ones in which I struggled the most out of all of these. I relied heavily on my partners when finding a solution, but am proud of us for figuring it out on our own without asking for help from teachers.