Maths Stages
Moving from stage 4 - 7
Intro
I created this slide to support our understanding of the Maths strategies and how to teach it to our students. You can refer to the following links for more information where needed.
Year 4/5/6 Maths Add/Sub Strategies
28 Students
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Stage 4 - 5 Advanced Counting to Early Additive Domain: Addition and Subtraction
Key Idea 1: Our number system is based on ten. slides 3-5
Key Idea 2: Basic fact knowledge can be used to add and subtract tens
Key Idea 3: Numbers can be rearranged and combined to make ten.
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Stage 4 - 5 Advanced Counting to Early Additive Domain: Addition and Subtraction
Key Idea 1: Our number system is based on ten.
Adding Tens
30 + 40 = , so 34 + 42 = .
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Key Idea 1: Our number system is based on ten.
Adding Ones and Tens
Subtracting Ones and tens
34 + 25 as 30 + 20 and 4 + 5
84 – 51 = 80 - 40 and 4 - 1
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Key Idea 1: Our number system is based on ten.
Adding Ones and Tens
Subtracting Ones and tens
34 + 25 as 30 + 20 and 4 + 5
84 – 51 = 80 - 40 and 4 - 1
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Stage 4 - 5 Advanced Counting to Early Additive Domain: Addition and Subtraction
Key Idea 2: Basic fact knowledge can be used to add and subtract tens.
Missing Ones and Tens
.
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Stage 4 - 5 Advanced Counting to Early Additive Domain: Addition and Subtraction
Key Idea 3: Numbers can be rearranged and combined to make ten.
Make Ten
5 + 2 + 5
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Key Idea 6: Addition is associative, so addends can be regrouped to solve a problem more efficiently.
Compatible Numbers
6 + 2 + 3 - 9
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Stage 4 - 5 Advanced Counting to Early Additive Domain: Addition and Subtraction
Key Idea 4: Addition and subtraction problems can be solved by partitioning one of the numbers to go up or back through ten.
Adding in Parts
8 + 6 = (8 + 2) + 4 = 10 + 4 = 14.
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Key Idea 5: Subtraction problems can be solved by going back through ten, partitioning numbers rather than counting back.
Subtracting in parts
Step one: 15 – 7 = (15 – 5) – 2
Step two: 10 – 2 = 8.
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Key Idea 7: Change unknown problems can be solved by using place value knowledge of tens and ones or by partitioning through tens.
Up and Over Ten
Missing Ones and Tens
7 + ? = 13
21 + ? = 34
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Key Idea 7: Change unknown problems can be solved by using place value knowledge of tens and ones or by partitioning through tens.
Problems lik 67 - ? = 34
37 + ? = 79
21 + ? = 34
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Key Idea 8: Subtraction can be used to solve difference problems in which two amounts are being compared.
Comparisons: Finding Difference in Data
More Comparison
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Stage 4 - 5 Advanced Counting to Early Additive Domain: Addition and Subtraction
Key Idea 9: Knowledge of doubles can be used to work out problems close to a double
Near Doubles
79 + 79 as 80 + 80 = 160 - 2 = 158
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Stage 4 - 5 Advanced Counting to Early Additive Domain: Addition and Subtraction
Key Idea 10: The equals sign represents balance.
The Balancing Act
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Stage 5 - 6 Early Additive to Advanced Additive Domain: Addition and Subtraction
Key Idea 1: Introduction to using the number line to solve change unknown problems
Jumping the Number Line
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Key Idea 2: 10 tens make one hundred and 10 hundreds make one thousand
How many ten dollar notes?
How many tens and hundreds?
19 ten-dollar notes = $190
The Bank of Mathematics has run out of $1000 notes. Alison wants to withdraw $2315 in $1, $10 and $100 dollar. How many $100 notes does she get?
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Key Idea 3: Solve addition and subtraction problems using place value
Addition and Subtraction on the numberline
Problems-like-....-+-29-81
56 + 37 = ? solved on number line as follow: 56 +30 - 86: 86 + 7 = 93
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Key Idea 4: Solve addition and subtraction problems by using rounding and compensating
When one number is near 100:
When one number is near 100:
Addition
Subtraction
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Key Idea 4: Solve addition and subtraction problems by using rounding and compensating
Problems like 73 - 19 = ?
Problems like 23 + ? = 71
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Stage 5 - 6 Early Additive to Advanced Additive Domain: Addition and Subtraction
Key Idea 4: Solve addition and subtraction problems by using rounding and compatible
Problems-like-....-+-29-81
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Stage 5 - 6 Early Additive to Advanced Additive Domain: Addition and Subtraction
Key Idea 5: Addition and subtraction are inversely related
Don’t subtract add
43 – 39 = as 39 + = 43 (reversibility)
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Key Idea 6: Solve subtraction problems with the mental strategy of equal adjustment
Equal Additions
445 – 398, the fact that 398 is very close to a tidy number, namely, 400, suggests that a useful way of solving it is by equal additions, in this case, of 2. The problem then becomes 447 – 400, whose answer is obviously 47.
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Key idea 7: Choosing wisely
Choosing Wisely
73 – 29 = ? Prompt the students to look carefully at the numbers before deciding how they might solve this. The following are possible strategies:
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Key idea 8: Using the standard written form to solve addition and subtraction problems
A standard written form
Large Numbers Roll Over
Mental of written ?
"Which is the better way to solve these problems – mentally or using the standard written forms? 997 + 1 234 4 546 – 2 788.”
Discuss why 997 + 1 234 is easy to solve mentally, using strategies such as, 1 000 + 1234 = 2 234, 2 234 – 3 = 2 231.
Doing 4 546 – 2 788 mentally will be beyond most students, so the standard written form is needed for this one.
Examples: A mixture of addition and subtraction problems (Material Master 5–13).
Addition
Subtraction
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Stage 6 - 7 Advanced Additive to Advanced Multiplicative Domain: Addition and Subtraction
Key Idea 1: To add or subtract fractions, they must be renamed to have a common denominator
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Stage 6 - 7 Advanced Additive to Advanced Multiplicative Domain: Addition and Subtraction
Key Idea 2: Decimal fractions arise out of division
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Key Idea 3: The “ten for one” and “one for ten” canons apply when adding and subtracting with decimal fractions (one-decimal place-fractions)
Adding decimal fractions
Subtracting with tenths
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Stage 6 - 7 Advanced Additive to Advanced Multiplicative Domain: Addition and Subtraction
Key Idea 4: Subtraction can produce negative numbers
Dollars and Bills
Subtracting positive and negative numbers
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Stage 6 - 7 Advanced Additive to Advanced Multiplicative Domain: Addition and Subtraction
Dropping and RIsing Temperatures
Bucket Balance
The aim here is for students to become comfortable moving up and down a vertical number line (thermometer) to reflect the operation of adding an integer. Adding a positive integer is a rise and adding a negative integer is a drop. For example, if the temperature at Mt Ruapehu was ⁻2 degrees in the morning and rose by 5 degrees it would be recorded as ⁻2 + ⁺5 = ⁺3. If the temperature had dropped by 5 degrees, it would be recorded as ⁻2 + ⁻5 = ⁻7.
The key idea here is that subtracting a negative integer has the same effect as adding a positive integer. For example, ⁺5 – ⁻7 = ⁺12. Rather than memorising a rule such as “two negatives make a positive” (which will mislead them), students can see from this learning experience why this happens.
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Stage 6 - 7 Multiplication
Use standard place value to solve multiplication problems 3 × 44 = as 3 × 40 + 3 × 4
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Stage 6 - 7 Multiplication
Use tidy numbers to solve multiplication problems (distributive property) 4 × 26 = as 4 × 25 + 4 × 1
Use proportional adjustment like doubling and halving, thirding and trebling, to solve multiplication problems 4×6 = so 2× =24, 8 × 3 = 24
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Stage 6 - 7 Multiplication
Cut and Paste
Use standard place value to solve division problems, including written forms,
Paper Power
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Stage 6 - 7 Multiplication
Use standard place value with tidy numbers to solve division problems 96 ÷ 4 = from 100 ÷ 4 = 25
Little bites at big multiplication and division
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Stage 6 - 7 Multiplication
The royal cooking lesson
Simplify division problems by changing both numbers (halving, thirding etc.) 52 ÷ 4 = as 26 ÷ 2 =
PART 1
The royal cooking lesson
Simplify division problems by changing both numbers (halving, thirding etc.) 52 ÷ 4 = as 26 ÷ 2 =
PART 2
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Stage 6 - 7 Multiplication
Proportional Packets
Use proportional adjustment to solve division problems 24 ÷ 4 = 6 so 24 ÷ 8 = , 24 ÷ 2 =
PART 1
Proportional Packets
Use proportional adjustment to solve division problems 24 ÷ 4 = 6 so 24 ÷ 8 = , 24 ÷ 2 =
PART 2
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Stage 6 - 7 Multiplication
Cross Products
Use place value units to solve multiplication and division problems, including written multiplication algorithms, 10 × 20 = 200 so 14 × 23 =
Remainders
Solve division problems that involve remainders expressing the remainders as whole numbers, fractions or decimals depending on the context, e.g. 38 ÷ 4 = 9 r2 or 9.5 or 9½
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Stage 6 - 7 Multiplication
Nines and Threes
Use divisibility rules for 2, 3, 4, 5, 6, 8, 9
Sherpa
Anticipate what happens to a number when it is multiplied or divided by ten, one hundred, one thousand, and so on
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