Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum

Monthly Update sign up
Mailing List signup
Search
Follow The Annenberg Learner on LinkedIn Follow The Annenberg Learner on Facebook Follow Annenberg Learner on Twitter
MENU
Learning Math Home
Number Session 10, Grades 3-5: Solutions
 
Session 10 Session 10 6-8 Part A Part B Part C Homework
 
Glossary
number Site Map
Session 10 Materials:
Notes
Solutions
 

A B C 

Video

Solutions for Session 10, Part C

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


Problem C1

a. 

Solution:

1. 

The greatest possible score is 8 + 8 + 8, or 24.

2. 

The lowest possible score is 5 + 5 + 5, or 15.

3. 

There are 10 different score combinations possible, but since a score of 21 can be gotten two different ways, there are only nine different scores possible:

8 + 8 + 8 = 24
8 + 8 + 7 = 23
8 + 8 + 5 = 21
8 + 7 + 7 = 22
8 + 7 + 5 = 20
8 + 5 + 5 = 18
7 + 7 + 7 = 21
7 + 7 + 5 = 19
7 + 5 + 5 = 17
5 + 5 + 5 = 15

4. 

John and Mary Beth both got 21, because that's the only score that can be gotten two different ways.

b. 

The number and operations content of this problem involves the ability to use the commutative and associative laws to find a sum with three or more addends. Students will also develop flexibility with numbers.

c. 

Students need to use logical thinking and to be able to recognize that the order in which they add three whole numbers doesn't affect the sum. They also need to be able to compare sums of numbers in an organized way.

d. 

If students are struggling, ask them to start by writing out the sums. It is important to have an organized system -- for example, add all the sums that start with the 8s (first with three 8s, then two 8s, then one 8) and then repeat for the 7s and 5s.

e. 

To get students to think beyond this problem, you can turn it around. Make up several scores that players said they got, and have students decide if these scores are possible.

<< back to Problem C1


 

Problem C2

a. 

Solution:

1. 

Five-sixths is greater. You know this because five parts out of six is more than five parts out of eight -- the pieces are larger.

2. 

Three-fourths is greater. Two pies are each missing one piece. The pie with four pieces has less missing, so it is bigger.

3. 

Two-thirds is greater. If you change the 2/3 to 6/9, then you have two pies each missing three pieces. The pie with nine pieces has less missing, so it is bigger.

4. 

The order is 2/5, 1/2, 5/8, 2/3, 3/4, 5/6, 9/10, and 5/3. People will explain their reasoning in different ways, but here's one example:

 

Two-fifths is the only fraction less than 1/2, so it is first. (Here you used 1/2 as the benchmark.)

 

Five-thirds is the only fraction greater than 1, so it is last. (Here you used 1 as the benchmark.)

 

The others are ordered according to the procedures explained above.

b. 

The number and operations content of this problem involves understanding the concept of fractions (the roles of the numerator and denominator; understanding that the larger the denominator is, the smaller the piece is; etc.), how to compare and order fractions, equivalent fractions, and benchmarks.

c. 

Students will probably want to convert to decimals or get common denominators for all the fractions. The procedures discussed above will help students learn to reason about fractions so that they can avoid tedious computation. Students will also need to use logical thinking.

d. 

If students are struggling, you can ask the following questions to try to elicit the following answers:

 

Why not just get a common denominator for all the fractions and then order them? (Because this is much more time-consuming and does not help to give us an intuitive feeling about the magnitude of fractions.)

 

Why not convert all the fractions to decimals and then order the decimals? (Because this is much more time-consuming and loses the fractions -- and thus our understanding of the fractions -- entirely.)

Essentially, students need to think about the benchmarks. They should be able to look at a fraction and know if that fraction is between 0 and 1/2, between 1/2 and 1, or greater than 1, and which of those benchmarks it's closer to. In addition, if the denominators are the same, they should know to compare numerators, and vice versa.

e. 

There are several ways to help students think beyond this problem. The simplest is to change the fractions. You could also increase the number of fractions, use larger numbers, or include some decimals in the list. You could also have students put the numbers on a number line.

<< back to Problem C2


 

Problem C3

a. 

Solution:

1. 

The smallest number of blocks you could remove is 4 -- 3 red blocks and 1 green block. To have 3 red blocks for every 4 green blocks, you must have a multiple of 3 + 4, or 7 (e.g., 7, 14, 21) blocks remaining in the box. This could be 3 red and 4 green, 6 red and 8 green, 9 red and 12 green, and so forth. However, since there are only nine of each color in the box, the most you could have is 6 red and 8 green blocks.

2. 

The smallest number of blocks you could remove is 3 (green blocks). To have 3 red blocks for every 2 green blocks, you must have a multiple of 3 + 2, or 5 (e.g., 5, 10, 15) blocks remaining in the box. This could be 3 red and 2 green, 6 red and 4 green, 9 red and 6 green, and so forth. However, since there are only nine of each color in the box, the most you could have is 9 red and 6 green blocks.

b. 

The number and operations content of this problem is ratio and proportional reasoning, patterns, and developing mathematical language (such as understanding what "for every" means).

c. 

Students need to use logical thinking. They also need to understand the inverse relationship between the number of parts and the size of the parts. Students need to be able to find equivalent fractions and compare fractions. They need to understand that when they're doing part-part relationships, they must identify the whole. For example, in the block problem, to have 3 red for every 4 green blocks, they need to be aware that the total will be 7 or a multiple of 7.

d. 

If students are struggling, have them use actual blocks. They can make two piles and then notice the relationships that are evident.

e. 

To help students think beyond the problem, ask them to solve a more complicated version of it, using nine red blocks, nine blue blocks, and nine green blocks. Tell them that there must be 3 red blocks for every 7 non-red blocks and 3 blue blocks for every 7 non-blue blocks. (There are two possible answers: 6 red, 6 blue, and 8 green; and 3 red, 3 blue, and 4 green.)

<< back to Problem C3


 

Problem C4

a. 

Solution:

1. 

The sum of 3 + 4 + 5 = 12. Yes, because 12 = 4 • 3.

2. 

The sum of 7 + 8 + 9 = 24. Yes, because 24 = 8 • 3.

3. 

The sum of any three consecutive integers is three times the middle number.

b. 

The number and operations content of this problem is number theory, addition of whole numbers, fluent computation, the idea of average, and patterns.

c. 

Students need to be able to use logical thinking. They need to be able to divide or multiply by 3, add three numbers, and understand what is meant by "three consecutive integers."

d. 

If students are struggling, encourage them to make an organized list in which they list the options one by one. They can also use manipulatives, such as Unifix cubes, to help make the patterns more evident. Once they have the numbers represented by stacks, they can take the top cube off the tallest stack and put it on the smallest to see that they are all the same height.

e. 

To help students think beyond this problem, ask them to think about five consecutive numbers. (In this case, the sum will be five times the middle number.)

<< back to Problem C4


 

Learning Math Home | Number Home | Glossary | Map | ©

Session 10, Grades 3-5 | Notes | Solutions | Video

© Annenberg Foundation 2014. All rights reserved. Legal Policy