Unit 1: NUMBER SENSE
VALUE vs PLACE VALUE
We are starting out learning about place value up to the ten thousands place. We are asking that students know the VALUE vs PLACE VALUE. The value is how much the number is "worth." For example: What is the VALUE of the underlined digit in 6,458? The answer: 6,000. Place value is the location of the number. For example: What is the PLACE VALUE of the underlined digit in 6, 458? The answer: thousands place.
With a solid understanding of place value, we expect that students will be able to compare numbers up to 1,000 with <, >, or = sign. This may include comparing numbers in standard, word, or expanded form.
For example: two thousand four hundred eighty < 2,000 + 400 + 80 + 8
3,592 = 35 hundreds and 92 ones
NUMBER FORMS:
We will learn about the three types of number forms: standard form, word form, and expanded form.
For example: My number is 4,856.
NUMBER COMBINATIONS - MAKING NUMBERS - DIFFERENT WAYS TO REPRESENT THE SAME NUMBER:
Numbers can be modeled with various combinations of thousands, hundreds, tens, and ones. Students need to be able to regroup hundreds, tens, or ones to model different combinations.
We are starting out learning about place value up to the ten thousands place. We are asking that students know the VALUE vs PLACE VALUE. The value is how much the number is "worth." For example: What is the VALUE of the underlined digit in 6,458? The answer: 6,000. Place value is the location of the number. For example: What is the PLACE VALUE of the underlined digit in 6, 458? The answer: thousands place.
With a solid understanding of place value, we expect that students will be able to compare numbers up to 1,000 with <, >, or = sign. This may include comparing numbers in standard, word, or expanded form.
For example: two thousand four hundred eighty < 2,000 + 400 + 80 + 8
3,592 = 35 hundreds and 92 ones
NUMBER FORMS:
We will learn about the three types of number forms: standard form, word form, and expanded form.
For example: My number is 4,856.
- Standard form: 4,856
- Word form: four thousand eight hundred fifty six
- Expanded form: 4000 + 800 + 50 + 6.
NUMBER COMBINATIONS - MAKING NUMBERS - DIFFERENT WAYS TO REPRESENT THE SAME NUMBER:
Numbers can be modeled with various combinations of thousands, hundreds, tens, and ones. Students need to be able to regroup hundreds, tens, or ones to model different combinations.
Students should also be able to take a combination of hundreds, tens, and ones and tell us what number is modeled.
For example: 8 hundreds, 12 tens, and 6 ones = 800 + 120 + 6 = 926 Some students realize that the extra 10 tens would make an extra hundred and come up with 926 by regrouping. Others simply add the values up. |
Unit 2 ~ MULTIPLICATION and DIVISION
Five different strategies to solve multiplication equations
arrays, equal sized groups, repeated addition, an open number line, and area models
arrays, equal sized groups, repeated addition, an open number line, and area models
Arrays
An array is made up of straight rows and columns. If you have the problem 3 x 6, you can create an array to solve. There are 3 rows of 6 which when counted up is 18. When presented with an array and asked to figure out the corresponding multiplication problem, students can add up the rows to find one of the factors and the columns to find the next factor in the multiplication problem.
An array is made up of straight rows and columns. If you have the problem 3 x 6, you can create an array to solve. There are 3 rows of 6 which when counted up is 18. When presented with an array and asked to figure out the corresponding multiplication problem, students can add up the rows to find one of the factors and the columns to find the next factor in the multiplication problem.
Equal sized groups
If you have 3x4, you can make 3 equal groups of 4. Students can clearly see through counting the dots in this picture that the answer is 12.
If you have 3x4, you can make 3 equal groups of 4. Students can clearly see through counting the dots in this picture that the answer is 12.
Repeated Addition
If you have 3 x 8, you can add 8, three times. When presented with an repeated addition problem and asked to find the corresponding multiplication problem, students can look at the number being added to figure out one of the factors and they can count how many times it was added to find the next factor in the multiplication problem.
If you have 3 x 8, you can add 8, three times. When presented with an repeated addition problem and asked to find the corresponding multiplication problem, students can look at the number being added to figure out one of the factors and they can count how many times it was added to find the next factor in the multiplication problem.
Open number line
The video below shows examples of how to create an open number line to solve multiplication problems. |
Area Model
This is very similar to an array, except the rows and columns are connected. Your rows and columns equal the two factors. You count out all the squares to find the product or area.
This is very similar to an array, except the rows and columns are connected. Your rows and columns equal the two factors. You count out all the squares to find the product or area.
Four different strategies to solve division equations
sharing equal groups, partitioning or splitting into groups, repeated subtraction, and inverse operation
sharing equal groups, partitioning or splitting into groups, repeated subtraction, and inverse operation
Sharing Groups:
This strategy is used when trying to figure out how many objects are in each group. If the equation is 12/3, we have to equally share the 12 stars into 3 groups and count the number of items in each group.
This strategy is used when trying to figure out how many objects are in each group. If the equation is 12/3, we have to equally share the 12 stars into 3 groups and count the number of items in each group.
Partitioning or Splitting into Equal Groups:
This strategy is used when trying to figure out how many groups of objects are present. If the equation is 12/4, we would draw 12 objects. We would then circle groups of 4 objects and figure out how many groups we have circled.
This strategy is used when trying to figure out how many groups of objects are present. If the equation is 12/4, we would draw 12 objects. We would then circle groups of 4 objects and figure out how many groups we have circled.
Repeated subtraction
If the equation was 20/4, we would start with 20 and continue to subtract 4 from the answer until there is nothing left to subtract or 0. We would then count how many times the number 4 was subtracted to find the quotient.
This strategy is often difficult for students. If they miscalculate only one subtraction problem through the process, the answer will be incorrect.
If the equation was 20/4, we would start with 20 and continue to subtract 4 from the answer until there is nothing left to subtract or 0. We would then count how many times the number 4 was subtracted to find the quotient.
This strategy is often difficult for students. If they miscalculate only one subtraction problem through the process, the answer will be incorrect.
Unit 3 ~ MULTIPLICATION and DIVISION STRATEGIES
Now that your child has the basic understanding of how to multiply and divide, we will now be working on when to use each operation in word problems and how to become fluent with our multiplication and division facts.
MULTIPLICATION AND DIVISION WORD PROBLEMS
Students will need to be able to read a word problem and analyze it to decide whether to multiply or divide. Some students may need to draw pictures to visualize what is happening in the story.
MULTIPLICATION
Do we have repeated groups or rows of an item and we need to FIND THE TOTAL? Are we repeatedly adding?
MULTIPLICATION AND DIVISION WORD PROBLEMS
Students will need to be able to read a word problem and analyze it to decide whether to multiply or divide. Some students may need to draw pictures to visualize what is happening in the story.
MULTIPLICATION
Do we have repeated groups or rows of an item and we need to FIND THE TOTAL? Are we repeatedly adding?
OR
DIVISION
DIVISION
- Do we HAVE THE TOTAL and we are splitting or sharing? Are we repeatedly subtracting?
MULTIPLICATION AND DIVISION FLUENCY STRATEGIES
What is multiplication and division fact fluency? What does that look like in third grade?
“According to CCSSM, fluency is “skill in carrying out procedures flexibly, accurately, efficiently and appropriately” (CCSSO 2010, p. 6). Thus, far from just being a measure of speed, fluency with multiplication facts involves flexibly and accurately using an appropriate strategy to find the answer efficiently.”
What is multiplication and division fact fluency? What does that look like in third grade?
“According to CCSSM, fluency is “skill in carrying out procedures flexibly, accurately, efficiently and appropriately” (CCSSO 2010, p. 6). Thus, far from just being a measure of speed, fluency with multiplication facts involves flexibly and accurately using an appropriate strategy to find the answer efficiently.”
SKIP COUNTING STRATEGY:
Third grade students should be able to skip count very easily by 2s, 5s, and 10s. Some can even skip count fluently by 3s. When 2, 5, or 10 is one of the factors in a multiplication problem, skip counting can be an efficient strategy to help find the product.
Example: 5 x 8 = 40 (5, 10, 15, 20, 25, 30, 35, 40)
6 x 10 = 60 (10, 20, 30, 40, 50, 60)
DOUBLING STRATEGY:
Knowing that 2s double to make 4s, 3s double to make 6s, and 4s double to make 8s can help us when trying to figure out larger multiplication facts!
Example: 4 x 6 = ? I know that 2 x 6 = 12, so 4 x 6 = 12 + 12 = 24! Just double 12!
6 x 8 = ? I know that 3 x 8 = 24, so 6 x 8 = 24 + 24 = 48! Just double 24!
LANDMARK or FRIENDLY NUMBERS STRATEGY:
If you know a multiplication fact that is near the one you are solving, you can use that problem to solve the new one by adding an extra group or taking an extra group away!
Example: 6 x 6 = ? I know 5 x 6 = 30, so I can add one more group of 6. 30 + 6 = 36
4 x 7 = ? I know 5 x 7 = 35, so I can take away one group of 7. 35 - 7 = 28
9 x 7 = ? I know 10 x 7 = 70, so I can take one away one group of 7. 90 - 7 = 63
PARTIAL PRODUCTS/DISTRIBUTIVE STRATEGY:
Sometimes you can take a more difficult multiplication fact and break it into 2 easier problems and add the products. This works especially well when breaking the factors of 7 or 12.
Example: 12 x 6 = (10x6) + (2x6) = 60 + 12 = 72
7 x 8 = (5x8) +(2x8) = 40 + 16 = 56
Third grade students should be able to skip count very easily by 2s, 5s, and 10s. Some can even skip count fluently by 3s. When 2, 5, or 10 is one of the factors in a multiplication problem, skip counting can be an efficient strategy to help find the product.
Example: 5 x 8 = 40 (5, 10, 15, 20, 25, 30, 35, 40)
6 x 10 = 60 (10, 20, 30, 40, 50, 60)
DOUBLING STRATEGY:
Knowing that 2s double to make 4s, 3s double to make 6s, and 4s double to make 8s can help us when trying to figure out larger multiplication facts!
Example: 4 x 6 = ? I know that 2 x 6 = 12, so 4 x 6 = 12 + 12 = 24! Just double 12!
6 x 8 = ? I know that 3 x 8 = 24, so 6 x 8 = 24 + 24 = 48! Just double 24!
LANDMARK or FRIENDLY NUMBERS STRATEGY:
If you know a multiplication fact that is near the one you are solving, you can use that problem to solve the new one by adding an extra group or taking an extra group away!
Example: 6 x 6 = ? I know 5 x 6 = 30, so I can add one more group of 6. 30 + 6 = 36
4 x 7 = ? I know 5 x 7 = 35, so I can take away one group of 7. 35 - 7 = 28
9 x 7 = ? I know 10 x 7 = 70, so I can take one away one group of 7. 90 - 7 = 63
PARTIAL PRODUCTS/DISTRIBUTIVE STRATEGY:
Sometimes you can take a more difficult multiplication fact and break it into 2 easier problems and add the products. This works especially well when breaking the factors of 7 or 12.
Example: 12 x 6 = (10x6) + (2x6) = 60 + 12 = 72
7 x 8 = (5x8) +(2x8) = 40 + 16 = 56
Unit 4 ~ ADDITION and SUBTRACTION
EQUATIONS:
Students need to be able to solve an equation with a variable in place of the unknown. Students can solve by thinking of basic facts they know. Or they can think that a part + part = total, and a total - part = part.
Examples:
8 = n + 5 n = 3
11 - n = 4 n = 7
9 = n - 12 n = 21*
*Students often will want to say 3 because they want to read the equation BACKWARDS. They should not! Alwasy read left to right, just like in regular reading.
n + 4 = 20 n = 16
Students need to be able to solve an equation with a variable in place of the unknown. Students can solve by thinking of basic facts they know. Or they can think that a part + part = total, and a total - part = part.
Examples:
8 = n + 5 n = 3
11 - n = 4 n = 7
9 = n - 12 n = 21*
*Students often will want to say 3 because they want to read the equation BACKWARDS. They should not! Alwasy read left to right, just like in regular reading.
n + 4 = 20 n = 16
ROUNDING TO TENS AND HUNDREDS
We want students to be able to visualize the numbers they are rounding on a number line. We will teach students a few tricks to help them remember how to round but stress the importance of understanding WHAT they are doing when rounding. For example: Round 47 to the nearest ten (you are between 40 and 50 of a number line but closer to 50.) Answer: 50. Round 845 to the nearest hundred (you are between 800 and 900 on a number line but closer to 800.) Answer: 800.
We want students to be able to visualize the numbers they are rounding on a number line. We will teach students a few tricks to help them remember how to round but stress the importance of understanding WHAT they are doing when rounding. For example: Round 47 to the nearest ten (you are between 40 and 50 of a number line but closer to 50.) Answer: 50. Round 845 to the nearest hundred (you are between 800 and 900 on a number line but closer to 800.) Answer: 800.
Unit 5 ~ AREA and PERIMETER
Perimeter ~ The distance around an object.
Area ~ The space inside an object.
There are several ways to find the PERIMETER of this object:
- Add up all side lengths: 8 + 5 + 8 + 5 = 26 cm
- Double same or different sides and add: 16 + 10 = 26 cm or 13 + 13 = 26 cm
- Multiplying: (2 x 8) + (2 x 5) = 16 + 10 = 26 cm
If students know the length and width of a rectangle, they can multiply them together to figure out the AREA. Using the example above, 8 x 5 = 40 sq cm.
Students can add up the outside edges when dealing with square units to figure out the PERIMETER. In this example, 4 + 4 + 3 + 3 = 14 units or 8 + 6 = 14 units.
Students can add up the inside squares to figure out the AREA or they can find a length and width of the outside edges and multiply them together. 4 x 3 = 12 sq units.
Students can add up the inside squares to figure out the AREA or they can find a length and width of the outside edges and multiply them together. 4 x 3 = 12 sq units.
Different rectangles and squares can be created when given a specific PERIMETER.
To do this, you must figure out 2 of the same lengths and 2 of the same widths that add up to be the perimeter.
Examples that may work with a perimeter of 14 would be:
3 + 4 + 3 + 4 = 14 units (example given)
1 + 6 + 1 + 6 = 14 units
2 + 5 + 2 + 5 = 14 units
Students should know that multiple rectangles/squares with different lengths and widths
can have the same PERIMETER.
To do this, you must figure out 2 of the same lengths and 2 of the same widths that add up to be the perimeter.
Examples that may work with a perimeter of 14 would be:
3 + 4 + 3 + 4 = 14 units (example given)
1 + 6 + 1 + 6 = 14 units
2 + 5 + 2 + 5 = 14 units
Students should know that multiple rectangles/squares with different lengths and widths
can have the same PERIMETER.
Much like figuring out the perimeter, different rectangles/squares can be created when given a specific AREA.
This is done by figuring out a length and width that would multiply together to become the AREA.
Examples that would work with an AREA of 18 sq. cm:
can have the same AREA.
This is done by figuring out a length and width that would multiply together to become the AREA.
Examples that would work with an AREA of 18 sq. cm:
- 3 x 6 = 18 sq. cm (example)
- 1 x 18 = 18 sq. cm
- 2 x 9 = 19 sq. cm
can have the same AREA.
Finding Unknown Sides When Given at Least One Side Length and the PERIMETER or the AREA
PERIMETER = 20 in
- Side lengths in a rectangle or square will match. Therefore, we know that 4 + 4 = 8.
- To find the other side lengths, we must subtract: 20 - 8 = 12.
- We now know that the missing side lengths when added together must equal 12.
- Half of 12 equal 6. Therefore, the missing side lengths are 4 , 6, and 6 or P = 4 + 4 + 6 + 6 = 20 in
AREA = 24 in
- To find the AREA of a rectangle or square, we must know that length (l) x (w) = area (A)
- Therefore, 4 x ____ = 24.
- The missing side length is 6 because 4 x 6 = 24.
AREA and PERIMETER word problems
Students need to visualize and draw pictures to help them solve word problems. Sometimes the word problems specifically asks them for the AREA or PERIMETER and sometimes students have to be able to visualize if the word problem is asking for the outside edge (PERIMETER) or inside (AREA.) Below are some examples of area and perimeter word problems.
-A gardener digs a flower bed that is 8 meters long and 3 meters wide. What is the area of the flower bed?
- Answer: 8 meters x 3 meters= 24 sq meters
- Lisa’s bedroom is 6 meters long and 4 meters wide. How much carpet will Lisa need to cover the floor of her bedroom?
-Answer: (Students would need to be able to visualize that they are figuring out the area.) 6 meters x 4 meters= 24 sq meters
-The perimeter of a rectangular sticker is 12 centimeters. The sticker is 4 centimeters tall. How wide is it?
-Answer: 4+4= 8.....12- 8= 4 .......half of 4 is 2. The missing side length is 2 cm.
Students need to visualize and draw pictures to help them solve word problems. Sometimes the word problems specifically asks them for the AREA or PERIMETER and sometimes students have to be able to visualize if the word problem is asking for the outside edge (PERIMETER) or inside (AREA.) Below are some examples of area and perimeter word problems.
-A gardener digs a flower bed that is 8 meters long and 3 meters wide. What is the area of the flower bed?
- Answer: 8 meters x 3 meters= 24 sq meters
- Lisa’s bedroom is 6 meters long and 4 meters wide. How much carpet will Lisa need to cover the floor of her bedroom?
-Answer: (Students would need to be able to visualize that they are figuring out the area.) 6 meters x 4 meters= 24 sq meters
-The perimeter of a rectangular sticker is 12 centimeters. The sticker is 4 centimeters tall. How wide is it?
-Answer: 4+4= 8.....12- 8= 4 .......half of 4 is 2. The missing side length is 2 cm.
Unit 6 ~ FRACTIONS
That Quiz: on this website students can take mini quizzes to review their fraction skills. When you pull up the website click on the identify button under the Fractions title. After that it will take you to the quiz. On the left side of the screen you can modify the quiz to your student's ability; change the length, the level, and the types of questions asked. Select "All" under the Fractions section.
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Math Skill Builder: This website has a short summary with the different strategies to compare fractions.
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COMPARING FRACTIONS
Compare fractions by drawing a picture of each fraction, showing the fractions on a number line, or by comparing the fractions to a benchmark such as 1/2.
Compare fractions by drawing a picture of each fraction, showing the fractions on a number line, or by comparing the fractions to a benchmark such as 1/2.
"To compare fractions, you must be looking at the same size WHOLE!"
Unit 7 ~ TIME, MONEY, PROBLEM SOLVING
Students in third grade must be able to read time
to the nearest minute on an analog clock.
Steps for reading a clock ---------------->
to the nearest minute on an analog clock.
Steps for reading a clock ---------------->
ELAPSED TIME:
There are 3 types of elapsed time problems third graders should be able to complete:
The types are listed below with examples and the method third grade uses to solve the problem.
Method: We utilize the "mountain" and "hills" to determine hours and 5 minute intervals.
There are 3 types of elapsed time problems third graders should be able to complete:
The types are listed below with examples and the method third grade uses to solve the problem.
Method: We utilize the "mountain" and "hills" to determine hours and 5 minute intervals.
On Line Review Games
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Unit 8 ~ GEOMETRY
Students in third grade must understand that shapes (rhombuses, rectangles, and others) may share attributes (having four sides), and that the shared attributes can define a larger category (quadrilaterals).
BASICS: point, line, line segment, ray, types of lines, and types of angles.
BASICS: point, line, line segment, ray, types of lines, and types of angles.
- Students should be able to draw and identify lines, line segments, rays and points.
- Students should be able to draw lines and line segments at specified lengths (ie. 5cm or 3 1/2 inches).
POLYGONS: closed figures made of line segments.
Students should be able to identify the following polygons (based on the number of line segments and angles):
Students should be able to identify the following polygons (based on the number of line segments and angles):
- Triangle (3)
- Quadrilateral (4)
- Pentagon (5)
- Hexagon (6)
- Octagon (8)
QUADRILATERALS: A quadrilateral is a polygon that has exactly four sides. (This also means that a quadrilateral has exactly four vertices, and exactly four angles.)
- Students should be able to recognize the similarities and differences between the following quadrilaterals:
- Parallelogram
- Rectangle
- Rhombus
- Square
- Trapezoid
- Kite (image is included but is not part of the standard)
SOLID FIGURES: Solid figures are three-dimensional figures that have length, width and height.
- Students should be able to identify the following solid figures and name the number of faces, edges, and vertices:
- Cube
- Cone
- Rectangular Prism
- Sphere
- Square Pyramid
- Cylinder