Table of Contents
HCF Division Method – Worked Solutions #
Question 1: Find the HCF of 56 and 42 using the Division Method.
| Step | Division | Answer |
|---|---|---|
| 1 | 56 ÷ 42 | 1 remainder 14 |
| 2 | 42 ÷ 14 | 3 remainder 0 |
Step 1: Divide 56 by 42
56 ÷ 42 = 1 remainder 14
Step 2: Divide 42 by 14
42 ÷ 14 = 3 remainder 0
Since the remainder is now 0, the process stops. The last divisor (14) is the HCF.
HCF of 56 and 42 = 14
Question 2: Calculate the HCF of 124 and 84 using the Division Method.
| Step | Division | Answer |
|---|---|---|
| 1 | 124 ÷ 84 | 1 remainder 40 |
| 2 | 84 ÷ 40 | 2 remainder 4 |
| 3 | 40 ÷ 4 | 10 remainder 0 |
Step 1: Divide 124 by 84
124 ÷ 84 = 1 remainder 40
Step 2: Divide 84 by 40
84 ÷ 40 = 2 remainder 4
Step 3: Divide 40 by 4
40 ÷ 4 = 10 remainder 0
Since the remainder is now 0, the process stops. The last divisor (4) is the HCF.
HCF of 124 and 84 = 4
Question 3: Find the HCF of 315 and 195 using the Division Method.
| Step | Division | Answer |
|---|---|---|
| 1 | 315 ÷ 195 | 1 remainder 120 |
| 2 | 195 ÷ 120 | 1 remainder 75 |
| 3 | 120 ÷ 75 | 1 remainder 45 |
| 4 | 75 ÷ 45 | 1 remainder 30 |
| 5 | 45 ÷ 30 | 1 remainder 15 |
| 6 | 30 ÷ 15 | 2 remainder 0 |
Step 1: Divide 315 by 195
315 ÷ 195 = 1 remainder 120
Step 2: Divide 195 by 120
195 ÷ 120 = 1 remainder 75
Step 3: Divide 120 by 75
120 ÷ 75 = 1 remainder 45
Step 4: Divide 75 by 45
75 ÷ 45 = 1 remainder 30
Step 5: Divide 45 by 30
45 ÷ 30 = 1 remainder 15
Step 6: Divide 30 by 15
30 ÷ 15 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (15) is the HCF.
HCF of 315 and 195 = 15
Question 4: A carpenter has two wooden planks of lengths 168 cm and 120 cm. He needs to cut them into pieces of equal length without any wastage. What is the maximum possible length of each piece? Use the Division Method to solve this problem.
To find the maximum possible length of each piece, we need to find the HCF of 168 and 120.
| Step | Division | Answer |
|---|---|---|
| 1 | 168 ÷ 120 | 1 remainder 48 |
| 2 | 120 ÷ 48 | 2 remainder 24 |
| 3 | 48 ÷ 24 | 2 remainder 0 |
Step 1: Divide 168 by 120
168 ÷ 120 = 1 remainder 48
Step 2: Divide 120 by 48
120 ÷ 48 = 2 remainder 24
Step 3: Divide 48 by 24
48 ÷ 24 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (24) is the HCF.
This means the carpenter can cut the planks into pieces of 24 cm each:
• 168 cm plank will yield 168 ÷ 24 = 7 pieces
• 120 cm plank will yield 120 ÷ 24 = 5 pieces
This means the carpenter can cut the planks into pieces of 24 cm each:
• 168 cm plank will yield 168 ÷ 24 = 7 pieces
• 120 cm plank will yield 120 ÷ 24 = 5 pieces
The maximum possible length of each piece is 24 cm.
Question 5: Find the HCF of 231 and 147 using the Division Method.
| Step | Division | Answer |
|---|---|---|
| 1 | 231 ÷ 147 | 1 remainder 84 |
| 2 | 147 ÷ 84 | 1 remainder 63 |
| 3 | 84 ÷ 63 | 1 remainder 21 |
| 4 | 63 ÷ 21 | 3 remainder 0 |
Step 1: Divide 231 by 147
231 ÷ 147 = 1 remainder 84
Step 2: Divide 147 by 84
147 ÷ 84 = 1 remainder 63
Step 3: Divide 84 by 63
84 ÷ 63 = 1 remainder 21
Step 4: Divide 63 by 21
63 ÷ 21 = 3 remainder 0
Since the remainder is now 0, the process stops. The last divisor (21) is the HCF.
HCF of 231 and 147 = 21
Question 6: A teacher wants to divide 170 pens and 255 pencils equally among students. Each student should get the same number of pens and the same number of pencils, with no supplies left over. What is the maximum number of students the teacher can accommodate? Use the Division Method to solve this problem.
To find the maximum number of students, we need to find the HCF of 170 and 255.
| Step | Division | Answer |
|---|---|---|
| 1 | 255 ÷ 170 | 1 remainder 85 |
| 2 | 170 ÷ 85 | 2 remainder 0 |
Step 1: Divide 255 by 170
255 ÷ 170 = 1 remainder 85
Step 2: Divide 170 by 85
170 ÷ 85 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (85) is the HCF.
This means the teacher can distribute the supplies to a maximum of 85 students:
• Each student will get 170 ÷ 85 = 2 pens
• Each student will get 255 ÷ 85 = 3 pencils
This means the teacher can distribute the supplies to a maximum of 85 students:
• Each student will get 170 ÷ 85 = 2 pens
• Each student will get 255 ÷ 85 = 3 pencils
The maximum number of students the teacher can accommodate is 85.
