Division of Whole Numbers with and without Remainders for Primary 4

Division of Whole Numbers with and without Remainders

Performance Objectives

By the end of the lesson, students should be able to:

1. divide 2- or 3-digit numbers by:
   1. 1- digit number with or without remainder;
   2. multiples of 10 up to 50;
2. solve quantitative aptitude problems involving division of 1- and 2-digit numbers

What is Division?

Division is a basic mathematical operation that involves splitting a number into equal parts or groups. It helps to find out how many times one number is contained in another number.

Imagine sharing a certain number of items equally among a group of people. Division helps to find out how many items each person gets.

The symbol for division is $\div$ or $/$.

Parts of a Division Problem

In a division problem, there are specific terms used:

1. Dividend: This is the total number being divided. It is the number that is being split into equal parts.
2. Divisor: This is the number by which the dividend is divided. It tells how many equal parts the dividend is being split into, or the size of each group.
3. Quotient: This is the result of the division. It tells how many items are in each group or how many groups can be made.
4. Remainder: This is the number left over after dividing, when the dividend cannot be divided equally by the divisor.

For example, in $12 \div 3 = 4$:

1. 12 is the dividend.
2. 3 is the divisor.
3. 4 is the quotient.
4. There is no remainder.

If 13 sweets are shared among 3 children, each child gets 4 sweets, and 1 sweet is left over. Here, 13 is the dividend, 3 is the divisor, 4 is the quotient, and 1 is the remainder.

Division of 2- or 3-Digit Numbers by 1-Digit Numbers

To divide larger numbers by a single-digit number, the long division method is commonly used. This method breaks down the division into smaller, manageable steps.

Steps for Long Division

The procedure for long division involves the following steps:

1. Set up the problem: Write the dividend inside the division symbol and the divisor outside.
2. Divide the first digit (or first few digits): Look at the first digit of the dividend. If it is smaller than the divisor, take the first two digits. Divide this part of the dividend by the divisor.
3. Write the quotient: Place the result of the division (the quotient) above the digit(s) you divided.
4. Multiply: Multiply the quotient you just wrote by the divisor.
5. Subtract: Write the product under the part of the dividend you divided and subtract.
6. Bring down: Bring down the next digit from the dividend to form a new number.
7. Repeat: Repeat steps 2-6 until all digits of the dividend have been brought down and divided.
8. Identify the remainder: The number left after the last subtraction is the remainder. If the remainder is zero, the division is exact.

Worked Examples: Division by 1-Digit Numbers

Example 1: Divide $48 \div 4$ (2-digit by 1-digit, no remainder)

$$
\begin{array}{r}
12 \\
4 \overline{)48} \\
-4 \downarrow \\
\hline
08 \\
-8 \\
\hline
0
\end{array}
$$

Explanation:

1. Divide 4 (first digit of 48) by 4. The answer is 1. Write 1 above the 4.
2. Multiply 1 by 4. The answer is 4. Write 4 below the first 4.
3. Subtract 4 from 4. The answer is 0.
4. Bring down the next digit, 8, to make 08 (or just 8).
5. Divide 8 by 4. The answer is 2. Write 2 above the 8.
6. Multiply 2 by 4. The answer is 8. Write 8 below the 8.
7. Subtract 8 from 8. The answer is 0.

So, $48 \div 4 = 12$ with a remainder of 0.

Example 2: Divide $53 \div 5$ (2-digit by 1-digit, with remainder)

$$
\begin{array}{r}
10 \\
5 \overline{)53} \\
-5 \downarrow \\
\hline
03 \\
-0 \\
\hline
3
\end{array}
$$

Explanation:

1. Divide 5 (first digit of 53) by 5. The answer is 1. Write 1 above the 5.
2. Multiply 1 by 5. The answer is 5. Write 5 below the first 5.
3. Subtract 5 from 5. The answer is 0.
4. Bring down the next digit, 3, to make 03 (or just 3).
5. Divide 3 by 5. Since 3 is smaller than 5, it cannot be divided. The answer is 0. Write 0 above the 3.
6. Multiply 0 by 5. The answer is 0. Write 0 below the 3.
7. Subtract 0 from 3. The answer is 3. This is the remainder.

So, $53 \div 5 = 10$ with a remainder of 3.

Example 3: Divide $369 \div 3$ (3-digit by 1-digit, no remainder)

$$
\begin{array}{r}
123 \\
3 \overline{)369} \\
-3 \downarrow \downarrow \\
\hline
06 \\
-6 \downarrow \\
\hline
09 \\
-9 \\
\hline
0
\end{array}
$$

Explanation:

1. Divide 3 by 3. Write 1 above the 3. Multiply $1 \times 3 = 3$. Subtract $3 – 3 = 0$.
2. Bring down 6. Divide 6 by 3. Write 2 above the 6. Multiply $2 \times 3 = 6$. Subtract $6 – 6 = 0$.
3. Bring down 9. Divide 9 by 3. Write 3 above the 9. Multiply $3 \times 3 = 9$. Subtract $9 – 9 = 0$.

So, $369 \div 3 = 123$ with a remainder of 0.

Example 4: Divide $475 \div 6$ (3-digit by 1-digit, with remainder)

$$
\begin{array}{r}
79 \\
6 \overline{)475} \\
-42 \downarrow \\
\hline
55 \\
-54 \\
\hline
1
\end{array}
$$

Explanation:

1. 4 (first digit of 475) is smaller than 6, so take the first two digits, 47.
2. Divide 47 by 6. The largest multiple of 6 less than 47 is 42 ($6 \times 7$). Write 7 above the 7.
3. Multiply $7 \times 6 = 42$. Write 42 below 47.
4. Subtract $47 – 42 = 5$.
5. Bring down the next digit, 5, to make 55.
6. Divide 55 by 6. The largest multiple of 6 less than 55 is 54 ($6 \times 9$). Write 9 above the 5.
7. Multiply $9 \times 6 = 54$. Write 54 below 55.
8. Subtract $55 – 54 = 1$. This is the remainder.

So, $475 \div 6 = 79$ with a remainder of 1.

Division of 2- or 3-Digit Numbers by Multiples of 10 up to 50

Multiples of 10 up to 50 are 10, 20, 30, 40, and 50. Dividing by these numbers also uses the long division method, but sometimes there are shortcuts, especially for dividing by 10.

Dividing by 10

When dividing a whole number by 10:

1. If the number ends in 0, remove the 0 to get the quotient. The remainder is 0.
2. If the number does not end in 0, the last digit of the number is the remainder, and the remaining digits form the quotient.

Example 5: Divide $70 \div 10$ (2-digit by 10, no remainder)

Using the shortcut: 70 ends in 0. Remove the 0 to get 7.
So, $70 \div 10 = 7$ with a remainder of 0.

Example 6: Divide $73 \div 10$ (2-digit by 10, with remainder)

Using the shortcut: The last digit of 73 is 3. This is the remainder. The remaining digit is 7, which is the quotient.
So, $73 \div 10 = 7$ with a remainder of 3.

Dividing by 20, 30, 40, or 50

For other multiples of 10, the long division method is the most reliable way to find the quotient and remainder.

Example 7: Divide $120 \div 20$ (3-digit by 20, no remainder)

$$
\begin{array}{r}
6 \\
20 \overline{)120} \\
-120 \\
\hline
0
\end{array}
$$

Explanation:

1. 1 (first digit of 120) is smaller than 20. 12 (first two digits) is also smaller than 20. So, take all three digits, 120.
2. Divide 120 by 20. Think: how many 20s are in 120? $20 \times 6 = 120$. Write 6 above the 0.
3. Multiply $6 \times 20 = 120$. Write 120 below 120.
4. Subtract $120 – 120 = 0$.

So, $120 \div 20 = 6$ with a remainder of 0.

Example 8: Divide $135 \div 20$ (3-digit by 20, with remainder)

$$
\begin{array}{r}
6 \\
20 \overline{)135} \\
-120 \\
\hline
15
\end{array}
$$

Explanation:

1. Take all three digits, 135, because 1 and 13 are smaller than 20.
2. Divide 135 by 20. The largest multiple of 20 less than 135 is 120 ($20 \times 6$). Write 6 above the 5.
3. Multiply $6 \times 20 = 120$. Write 120 below 135.
4. Subtract $135 – 120 = 15$. This is the remainder.

So, $135 \div 20 = 6$ with a remainder of 15.

Example 9: Divide $250 \div 50$ (3-digit by 50, no remainder)

$$
\begin{array}{r}
5 \\
50 \overline{)250} \\
-250 \\
\hline
0
\end{array}
$$

Explanation:

1. Take all three digits, 250.
2. Divide 250 by 50. Think: how many 50s are in 250? $50 \times 5 = 250$. Write 5 above the 0.
3. Multiply $5 \times 50 = 250$. Write 250 below 250.
4. Subtract $250 – 250 = 0$.

So, $250 \div 50 = 5$ with a remainder of 0.

Example 10: Divide $278 \div 50$ (3-digit by 50, with remainder)

$$
\begin{array}{r}
5 \\
50 \overline{)278} \\
-250 \\
\hline
28
\end{array}
$$

Explanation:

1. Take all three digits, 278.
2. Divide 278 by 50. The largest multiple of 50 less than 278 is 250 ($50 \times 5$). Write 5 above the 8.
3. Multiply $5 \times 50 = 250$. Write 250 below 278.
4. Subtract $278 – 250 = 28$. This is the remainder.

So, $278 \div 50 = 5$ with a remainder of 28.

Using Division Charts

A division chart is a tool that helps to quickly find the results of division problems, especially for basic facts. It is often a multiplication chart used in reverse.

How to Use a Division Chart

The procedure for using a division chart involves the following steps:

1. Locate the divisor: Find the divisor (the number you are dividing by) in the first column (or top row) of the chart.
2. Find the dividend: Move across the row (or down the column) of the divisor until you find the dividend (the number being divided).
3. Identify the quotient: The number at the top of the column (or beginning of the row) where you found the dividend is the quotient.
4. For remainders: If the exact dividend is not found, look for the largest number in that row/column that is less than the dividend. The corresponding number at the top/side is the quotient. Subtract this found number from the original dividend to get the remainder.

Observe the example division chart below. It shows multiplication facts, which can be used for division.

Example using the chart: Divide $24 \div 4$.

1. Find 4 in the first column.
2. Move across the row for 4 until you find 24.
3. Look up to the top row from 24. You will find 6.

So, $24 \div 4 = 6$.

Example with remainder: Divide $27 \div 4$.

1. Find 4 in the first column.
2. Move across the row for 4. You will not find 27.
3. The largest number less than 27 in that row is 24. The quotient for 24 is 6.
4. Subtract $27 – 24 = 3$. This is the remainder.

So, $27 \div 4 = 6$ with a remainder of 3.

Quantitative Aptitude Problems Involving Division

Quantitative aptitude problems are word problems that require mathematical operations to solve. For division, these problems often involve sharing, grouping, or finding how many times one quantity fits into another.

Worked Examples: Quantitative Aptitude

Example 11: A farmer has 18 yams. If he shares them equally among 3 friends, how many yams does each friend get?

Solution:

Total yams = 18

Number of friends = 3

Yams per friend = $18 \div 3 = 6$

Each friend gets 6 yams.

Example 12: A tailor has 25 metres of fabric. If each dress requires 4 metres of fabric, how many dresses can the tailor make, and how much fabric will be left over?

Solution:

Total fabric = 25 metres

Fabric per dress = 4 metres

Number of dresses = $25 \div 4$

$$
\begin{array}{r}
6 \\
4 \overline{)25} \\
-24 \\
\hline
1
\end{array}
$$

The tailor can make 6 dresses, and 1 metre of fabric will be left over.

Example 13: There are 80 oranges to be packed into bags. If each bag holds 10 oranges, how many bags are needed?

Solution:

Total oranges = 80

Oranges per bag = 10

Number of bags = $80 \div 10 = 8$

8 bags are needed.

Example 14: A school has 95 exercise books. If the books are distributed to classes in bundles of 20, how many bundles can be made, and how many books will be left?

Solution:

Total exercise books = 95

Books per bundle = 20

Number of bundles = $95 \div 20$

$$
\begin{array}{r}
4 \\
20 \overline{)95} \\
-80 \\
\hline
15
\end{array}
$$

4 bundles can be made, and 15 exercise books will be left.