A.10.a.
Tony and I were comparing our baseball card collections. I had twice as many cards as Tony and 3 more besides. I have 21. How many did Tony have?
A.10.b
. Our friend Mark came and wanted to share our cards but didn’t have any of his own. How many cards do Tony and I have to begin with? If we share our cards equally, how many will each person get?
A.11.a
. Tony and I were comparing our baseball card collections. I had twice as many cards as Tony and 3 more besides. Tony has 9 cards. How many cards do I have?
A.11.b.
Our friend Mark came and wanted to share our cards but didn’t have any of his own. We split our cards equally. How many cards do the three of us have altogether?
First Step: Identify the givens and the question
A.10.
Givens:
-
- (a) I had twice as many cards as Tony and 3 more besides
-
- (a) I had 21 cards
-
- (b) Mark wants to share our cards and we will all have an equal number of cards
Question:
-
- (a) How many did Tony have to begin?
-
- (b) How many cards do Tony and I have to begin with?
-
- (b) How many can we each get (Tony, Mark, and I) to have an equal number of cards?
A.11.
Givens:
-
- (a) I had twice as many cards as Tony and 3 more besides
-
- (a) Tony has 9 cards
-
- (b) Mark is sharing our cards and between the three of us, we each get 10 cards
Question:
-
- (a) How many cards do I have?
-
- (b) How many cards do the three of us have altogether?
Second Step: Problem Analysis
A.10.a. This problem resembles A.5. and A.6. It states that I have 21 cards, which is twice the amount plus 3 more than Tony. He must then have three less than 21 and then half of that difference. Subtraction and division must be used to find Tony’s total number of cards.
A.10.b.
The problem then asks how many cards do Tony and I have combined. By taking the amount I have (21) and Tony’s found amount of (9) and adding them, the sum of our cards may be found. Mark then enters the problem with no cards of his own and wants to share. There are then three groups that must divide the 30 cards up evenly.
A.11.a.
Tony has 9 cards. I have twice as many plus 3. If I have twice as many (plus 3) then the amount of cards that Tony has (9) should be multiplied by two. Then the 3 additional cards should be added to find my sum total of cards.
A.11.b.
Mark comes and wants to share the cards. We know that combined we have 30 cards (Tony’s 9, plus my 21). Even if Mark comes and we split our cards 3 ways, the sum total of cards will not change and we will still have 30 cards altogether. Multiplication can be used if it is found that each person gets 10 cards and if there are 3 groups, the total number of cards can be found.
Third Step: Identify the operation(s) needed to find the answer and then solve
A.10.
(a) Subtraction, Division
21 - 3 = 18
18 / 2 = 9 cards
(b) Addition, Division
21 + 9 = 30
30 / 3 = 10 cards each
A.11.
(a) Multiplication, Addition
9 x 2 = 18
18 + 3 = 21 cards
(b) Multiplication
10 x 3 = 30 cards altogether
Fourth Step: Identify the connection between both problems and reasoning.
The same numbers are used in both problems. The questions change and therefore require reverse operations to be used. When A.10.a. is compared with A.11.a., it can be seen that subtraction and division are used, to find Tony’s total number of cards and then multiplication and addition are used to find “my” total number of cards. The key words are
twice
, and
more besides
. After the given information is found, those key words tell which operations are necessary. In A.10.b., a third person is added to the mix, so the total number of cards to be shared among the three must be found before finding out how many can be dispersed evenly. Since the problem asks how many can each person have, it must be divided. In A.11.b. the opposite is asked. The total number of cards that each person has is given, and the question then asks how many the group has altogether. Since there are three groups of cards and the total number of cards among all three is asked for, then multiplication is the operation used.
It can also be noted that these problems parallel A.5. and A.6. but are more complex, showing how scaffolding is important for these word problems.
