Lesson Notes By Weeks and Term v5 - Grade 4
Whole numbers: place value and operations (Grade 4) – Week 4 focus
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Whole numbers: place value and operations (Grade 4) – Week 4 focus
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Subject: Mathematics
Class: Grade 4
Term: 1st Term
Week: 4
Theme: General lesson support
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This week, we delve deeper into whole numbers, focusing on place value up to at least four-digit numbers (thousands) and applying our understanding to perform various operations: addition, subtraction, multiplication, and division. Understanding place value is crucial because it forms the foundation for all arithmetic operations. In everyday life, we constantly use these skills: calculating the cost of groceries at a shop, figuring out how much pocket money you'll have over several weeks, sharing sweets equally amongst friends, or understanding the numbers used in sports scores. Without a firm grasp of place value and operations, these everyday calculations would be much more challenging.
Place Value: Place value is the value of a digit based on its position in a number. For example, in the number 3456: The digit 6 is in the ones place, so its value is 6 x 1 = 6 The digit 5 is in the tens place, so its value is 5 x 10 = 50 The digit 4 is in the hundreds place, so its value is 4 x 100 = 400 The digit 3 is in the thousands place, so its value is 3 x 1000 = 3000 Understanding place value allows us to break down numbers and understand their magnitude. It is essential for understanding larger numbers and performing calculations.
Example 1: Place Value What is the place value of the digit 7 in the number 2741? 2741 can be broken down as: 2000 + 700 + 40 + 1 The digit 7 is in the hundreds place.
Therefore, its place value is
7
0
0. Decomposing and Composing Numbers: Decomposing a number means breaking it down into its place value components. Composing a number means putting these components back together to form the whole number.
Example 2: Decomposing a Number Decompose the number 5283. 5283 = 5000 + 200 + 80 + 3 Example 3: Composing a Number Compose the number: 8000 + 100 + 60 + 9 8000 + 100 + 60 + 9 = 8169 Addition and Subtraction: We use column addition and subtraction to efficiently add and subtract numbers. Remember to align the digits according to their place value (ones under ones, tens under tens, etc.).
Example 4: Addition Add 2345 and 1587. ``` 2345 + 1587 3932 ``` Explanation: 5 + 7 =
1
2. Write down 2, carry-over 1 to the tens column. 4 + 8 + 1 (carry-over) =
1
3. Write down 3, carry-over 1 to the hundreds column. 3 + 5 + 1 (carry-over) =
9. Write down 9. 2 + 1 =
3. Write down
3. Example 5: Subtraction Subtract 1278 from 3456. ``` 3456 1278 2178 ``` Explanation: 6 - 8: We cannot subtract 8 from 6, so we borrow 1 from the tens place (5 becomes 4). The 6 becomes 16. 16 - 8 =
8. Write down 8. 4 - 7: We cannot subtract 7 from 4, so we borrow 1 from the hundreds place (4 becomes 3). The 4 becomes 14. 14 - 7 =
7. Write down 7. 3 - 2 =
1. Write down 1. 3 - 1 =
2. Write down
2. Multiplication: We multiply 2-digit numbers by single-digit numbers using the distributive property.
Example 6: Multiplication Multiply 23 by 4. 23 x 4 = (20 + 3) x 4 = (20 x 4) + (3 x 4) = 80 + 12 = 92 Alternatively, we can use the column method: ``` 23 x 4 92 ``` Explanation: 4 x 3 =
1
2. Write down 2, carry-over 1 to the tens place. 4 x 2 =
8. Add the carry-over 1, so 8 + 1 =
9. Write down
9. Division: Division involves splitting a number into equal groups.
Example 7: Division Divide 36 by 3. 36 ÷ 3 = ?
We can think of this as: How many groups of 3 can we make from 36?
We can use repeated subtraction: 36 - 3 = 33 33 - 3 = 30 30 - 3 = 27 27 - 3 = 24 24 - 3 = 21 21 - 3 = 18 18 - 3 = 15 15 - 3 = 12 12 - 3 = 9 9 - 3 = 6 6 - 3 = 3 3 - 3 = 0 We subtracted 3 twelve times.
Therefore, 36 ÷ 3 =
1
2. Alternatively, we can use short division: ``` 12 3 | 36 ``` Explanation: How many 3s are there in 3? There is
1. Write down 1 above the
3. How many 3s are there in 6? There are
2. Write down 2 above the
6. Therefore, 36 ÷ 3 =
1
2. If there's a remainder: Example 8: Division with remainder Divide 23 by 5. 23 ÷ 5 = ? 5 x 4 = 20 (closest we can get to 23 without going over) 23 - 20 = 3 (remainder) Therefore, 23 ÷ 5 = 4 remainder
3. Guided Practice (With Solutions)
Question 1: What is the place value of the digit 8 in the number 5821?
Solution: 5821 = 5000 + 800 + 20 +
1. The digit 8 is in the hundreds place.
Therefore, its place value is
8
0
0. Question 2: Add 4567 and
2
8
9
4. Solution: ``` 4567 + 2894 7461 ``` Explanation: 7+4 = 11, write down 1, carry over 1 6+9+1 = 16, write down 6, carry over 1 5+8+1 = 14, write down 4, carry over 1 4+2+1 = 7, write down 7 Question 3: Subtract 1654 from
3
2
1
0. Solution: ``` 3210 1654 1556 ``` Explanation: 0 - 4, borrow from the tens (1 becomes 0), 10-4 = 6 0 - 5, borrow from the hundreds (2 becomes 1), 10-5 = 5 1 - 6, borrow from the thousands (3 becomes 2), 11-6 = 5 2 - 1 = 1 Question 4: Multiply 18 by
5. Solution: 18 x 5 = (10 + 8) x 5 = (10 x 5) + (8 x 5) = 50 + 40 = 90 Alternative: ``` 18 x 5 90 ``` Question 5: Divide 47 by
6. Solution: 47 ÷ 6 = 7 remainder 5, because 6 x 7 = 42 and 47 - 42 =
5. Independent Practice (Questions Only) What is the place value of the digit 3 in the number 9135? Decompose the number
6
0
7
2. Compose the number: 2000 + 500 + 30 +
8. Add 5829 and
3
1
7
5. Subtract 2345 from
6
7
8
9. Multiply 27 by
6. Divide 53 by
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