A B C

Solutions for Session 10, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5

Problem C1

a.

Solution: There are seven different ways:

0 + 6, 1 + 5, 2 + 4, 3 + 3, 4 + 2, 5 + 1, 6 + 0

b.

The number and operations content in this problem is addition of whole numbers. There is also an exposure to equations and the symbols involved, such as the equality symbol.

c.

Students need to be able to recognize the numerals for the numbers 0 through 6 and to be able to add whole numbers with sums to 6. They also need to understand the equality and addition symbols. And they need to use logical reasoning and have the ability to make an organized list.

d.

For students who are having difficulty, give them six counters and ask them to arrange the counters in the two squares, starting with all six in the right square. Ask them to record this number fact: 0 + 6 = 6. Then ask them to move one counter to the left square and record a new number fact: 1 + 5 = 6. Have them continue until all the counters are in the left square.

e.

Choosing a sum other than 6 can extend this problem. A nice extension activity would be to ask students to make a table with all the ways to get each sum from 1 to 10:

Number

Sums

Number of Ways

 1 0 + 1; 1 + 0 2 2 0 + 2; 1 + 1; 2 + 0 3 ... 10 0 + 10; 1 + 9; 2 + 8; 3 + 7; 4 + 6; 5 + 5; 10 + 0 11

Another extension would be to allow more than two addends. Ask, "How many different ways can you add counting numbers to get 6?" (Notice that 0 is not allowed in this problem.) Solution: There are 31 different ways to get 6 by adding counting numbers. (Remember that 2 + 1 + 2 + 1 is different from 1 + 2 + 2 + 1, etc., so there are six different ways to write these four numbers.)

Problem C2

 a. Solution: Rectangle B has the most shaded because it has the greatest number of parts. As the number of parts increases, the size of the parts decreases. Rectangle B is only missing one part out of eight parts. The others have fewer parts. Rectangle D has the least shaded because it is divided into the fewest number of parts, so the missing part is greater than any of the others. b. The number and operations content is the basic concept of fractions and the inverse relationship between the number of parts and the size of the parts. c. Students need to be able to make sense of visual representations, extract information from a visual representation, and make observations about what they see. d. Students who are having difficulty with this problem need to have concrete materials to represent the fractions. They need to be able to compare the size of 1/3 to the size of 1/4 and so on. e. This problem can be extended by asking students to order all five of the rectangles from least shaded to most shaded. You could also include more fractions and not use the pictures.

Problem C3

a.

Solution:

I.

 1 You can trade 2 blue rhombuses for 2 sets of 2, or 4, green triangles. 2 There are 3 sets of 2 green triangles, so you can trade for 3 blue rhombuses. 3 Yes. Eight green triangles is 4 sets of 2, so you can trade for 4 blue rhombuses.

II.

 1 There are 3 trapezoids, so there should be 3 sets of 3 triangles. 2 There are 2 sets of 3 triangles, so there should be 2 trapezoids. 3 No. There should be 3 triangles for each trapezoid. Three sets of 3 triangles is 9 triangles, and 4 sets of 3 triangles is 12 triangles. That means that 10 triangles is not a fair trade for 4 trapezoids.

b.

The number and operations content of this problem is the concept of fractions and proportional reasoning. It also involves basic exposure to equations.

c.

Students must be able to add numbers to 9, count by twos and threes, and count backward by twos or threes. They need to understand the equality symbol. Students will also be asked to make observations based on visual representations and to use logical thinking.

d.

If students are having trouble, give them pattern blocks to use. Have them cover each blue rhombus with 2 green triangles and each red trapezoid with 3 green triangles.

e.

One extension of this problem is to use more complex relationships. For example, if 1 hexagon can be traded for 2 red trapezoids, how many green triangles can it be traded for? How many blue rhombuses?

Problem C4

a.

Solution:

 1 Even. Every counter has a partner. 2 Even. The two counters without partners can be partners, so every counter has a partner. 3 Odd. The counter without a partner still has no partner.

b.

The number and operations content of this problem is number theory (odd and even numbers) and addition.

c.

Students must understand odd and even numbers and be able to add one-digit numbers. Logical thinking and working with word problems are also present.

d.

If students are having difficulty, give them counters. Ask them to choose two odd numbers and make as many partners as they can for each number. They will see that one is left over in each number. Ask them what happens when the two numbers are added. Do the same with the other problems.

e.

One extension of this problem is to ask students to think about numbers that are multiples of 3. Describe these numbers as numbers you say when you count by threes. What happens when you add two numbers that are multiples of 3?

 Problem C5 Answers will vary, but here is one example: Many students at this level work with pattern blocks. A lesson could be structured around exploring pattern blocks and the relationships that exist among different shapes. Once students gain a familiarity with these, this knowledge can be further extended into writing up those relationships as simple equations. For example, for Problem C3, students could begin by using the triangular pattern blocks to "make" a rhombus before they work on the written equations. This way, students can use manipulatives to reason about and justify their answers. By writing these simple equations, students are not only learning about the equality symbol, but also gaining some early groundwork on fractions, multiples, and addition (for example, one half of a rhombus is a triangle, and three triangles make one trapezoid).