Multi-step
This is by far the most daunting category. Based on the four operations, considering order, there are 16 possible types of problems. However, when all of the categories mentioned above are considered, there are a much greater number of combinations. My collection of multi-step problems, (see appendix) are by no means exhaustive of all of the different combinations. After collecting the problems, solving them, and classifying them, I soon realized the challenge of grouping them. I decided to group them based on the first step. Now I will discuss some of them from each category in more detail.
Here is problem number twenty-nine:
8 students sold 272 concert tickets at $3 each. Each student sold the same number of tickets. How much did each student collect?
The first step is to divide, since this is a grouping/partition problem. I must divide the number of tickets by the number of students. That means that I am doing partitive division. 272 divided by 8 equals 34. The next step is to multiply. This is a price problem with multiplication. 34 times $3 equals $102. Each student collected 102 dollars.
Interestingly enough, this problem could also be solved a different way. The number of tickets could have been multiplied by the ticket price first, to find the total amount of money collected. Then this would be a price problem with multiplication. Then the next step would be to divide the total amount of money, $816 by the 8 students. This would be a partitive division problem since I am dividing into a certain number of groups. The solution is $816 divided by 8 yields the same answer of $102. These alternative solution methods reflect the fact that multiplication is commutative (and that division by 8 is multiplication by 1/8).
Regardless of the order, the same operations and classification for those operations was used in the two methods of solving this problem. Certainly, my classifications of these multi-step problems are not the only way they can be grouped. However, it is a very reasonable first pass. Remember, Ma urged teachers to be more flexible. So, if a different way is chosen, then that is showing flexibility, which is a good thing. For the purpose of brevity, I am not going to continue to explore multiple ways of solving each problem as I go through the multi-step domain.
Here is problem 30:
Mr. Stone bought some books for $301. Each book cost $7. If his wife carried 18 books and he carried the rest. How many books did Mr. Stone carry?
This problem falls into the price category and uses measurement division in the first step because the total number of books has to be found. To do this, the total cost $301 is divided by the cost of one book. After performing this computation, we know that Mr. Stone had 43 books in total. The next step is to subtract, but since I am analyzing the problem, just knowing to subtract is not enough. I know that Mr. Stone and his wife are carrying them some place, presumably to the same place, so the books are not being separated. So, I considered this to be a part-part-whole problem involving a missing part. Now, I can complete the calculation by subtracting 18 from 43. The difference is 25, so Mr. Stone carried 25 books.
In number 33,
Rebecca bought four boxes of doughnuts. There were 6 doughnuts in each box. She divided the doughnuts equally among eight people. How many doughnuts did each person receive?
Based on my classification by first step, this is a grouping/partitioning problem involving multiplication. Six times four equals twenty-four. The next step is a grouping/partition problem that uses partitive division. 24 doughnuts divided by 8 people. So the answer is 3. Each person received 3 doughnuts. Interestingly all of the numbers in this problem are factors of 24.
Number 37 states:
A toy car costs $5. A toy airplane costs 4 times as much as the toy car. How much more does the toy airplane cost than the toy car?
The first step for me is multiplicative comparison. The cost of the car is multiplied by four since the airplane costs four times as much as the car. Therefore, $5 times 4 equals $20. The next step is to compare the prices using subtraction to find the unknown difference. The difference is $15 since $20 minus $5 equals $15.
In problem 40,
Rani had $47. After paying for 3kg of shrimp, she had $20 left. Find the cost of 1 kg of shrimp.
The first step is subtraction. It is a separate problem with an unknown change. Rani had $47. She spent some unknown amount and now is left with $20. After performing the calculation ($47-$20), the unknown change is determined to be $27. The next step is to divide $27 by 3. This is a price problem using partitive division. The cost of 1 kg would be $9.
A final problem from my collection of multi-step problems is from the part-part-whole domain. Number forty-three states:
A farmer had 2,000 chickens and ducks. After selling some of them, he had 650 chickens and 520 ducks left. How many chickens and ducks did he sell altogether?
First I chose to perform addition to find how many chickens and ducks a farmer has left after selling some of them. 650 chickens plus 520 ducks equals 1,170 chicken and ducks. Now I am left with a separate problem to find the unknown difference. I know that it is a separate problem because the chickens and ducks are being sold and separated from the group. 2,000 take away 1,170 equals 830 chicken and ducks that were sold.
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