Answers – Highest Common Factors – 2

Question 1: Find the HCF of 72 and 108.
Step Division Answer
1 108 ÷ 72 1 remainder 36
2 72 ÷ 36 2 remainder 0

Step 1: Divide 108 by 72

108 ÷ 72 = 1 remainder 36

Step 2: Divide 72 by 36

72 ÷ 36 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (36) is the HCF.
HCF of 72 and 108 = 36
Question 2: Calculate the HCF of 96 and 144.
Step Division Answer
1 144 ÷ 96 1 remainder 48
2 96 ÷ 48 2 remainder 0

Step 1: Divide 144 by 96

144 ÷ 96 = 1 remainder 48

Step 2: Divide 96 by 48

96 ÷ 48 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (48) is the HCF.
HCF of 96 and 144 = 48
Question 3: Find the HCF of 85 and 51.
Step Division Answer
1 85 ÷ 51 1 remainder 34
2 51 ÷ 34 1 remainder 17
3 34 ÷ 17 2 remainder 0

Step 1: Divide 85 by 51

85 ÷ 51 = 1 remainder 34

Step 2: Divide 51 by 34

51 ÷ 34 = 1 remainder 17

Step 3: Divide 34 by 17

34 ÷ 17 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (17) is the HCF.
HCF of 85 and 51 = 17
Question 4: A florist has 126 roses and 90 tulips. She wants to make identical bouquets using all the flowers, with each bouquet having the same number of roses and the same number of tulips. What is the maximum number of bouquets she can make?

To find the maximum number of bouquets, we need to find the HCF of 126 and 90.

Step Division Answer
1 126 ÷ 90 1 remainder 36
2 90 ÷ 36 2 remainder 18
3 36 ÷ 18 2 remainder 0

Step 1: Divide 126 by 90

126 ÷ 90 = 1 remainder 36

Step 2: Divide 90 by 36

90 ÷ 36 = 2 remainder 18

Step 3: Divide 36 by 18

36 ÷ 18 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (18) is the HCF.

This means the florist can make a maximum of 18 identical bouquets:
• Each bouquet will have 126 ÷ 18 = 7 roses
• Each bouquet will have 90 ÷ 18 = 5 tulips
The maximum number of bouquets she can make is 18.
Question 5: Find the HCF of 210 and 165.
Step Division Answer
1 210 ÷ 165 1 remainder 45
2 165 ÷ 45 3 remainder 30
3 45 ÷ 30 1 remainder 15
4 30 ÷ 15 2 remainder 0

Step 1: Divide 210 by 165

210 ÷ 165 = 1 remainder 45

Step 2: Divide 165 by 45

165 ÷ 45 = 3 remainder 30

Step 3: Divide 45 by 30

45 ÷ 30 = 1 remainder 15

Step 4: 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 210 and 165 = 15
Question 6: Calculate the HCF of 138 and 222.
Step Division Answer
1 222 ÷ 138 1 remainder 84
2 138 ÷ 84 1 remainder 54
3 84 ÷ 54 1 remainder 30
4 54 ÷ 30 1 remainder 24
5 30 ÷ 24 1 remainder 6
6 24 ÷ 6 4 remainder 0

Step 1: Divide 222 by 138

222 ÷ 138 = 1 remainder 84

Step 2: Divide 138 by 84

138 ÷ 84 = 1 remainder 54

Step 3: Divide 84 by 54

84 ÷ 54 = 1 remainder 30

Step 4: Divide 54 by 30

54 ÷ 30 = 1 remainder 24

Step 5: Divide 30 by 24

30 ÷ 24 = 1 remainder 6

Step 6: Divide 24 by 6

24 ÷ 6 = 4 remainder 0
Since the remainder is now 0, the process stops. The last divisor (6) is the HCF.
HCF of 138 and 222 = 6
Question 7: Find the HCF of 245 and 175.
Step Division Answer
1 245 ÷ 175 1 remainder 70
2 175 ÷ 70 2 remainder 35
3 70 ÷ 35 2 remainder 0

Step 1: Divide 245 by 175

245 ÷ 175 = 1 remainder 70

Step 2: Divide 175 by 70

175 ÷ 70 = 2 remainder 35

Step 3: Divide 70 by 35

70 ÷ 35 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (35) is the HCF.
HCF of 245 and 175 = 35
Question 8: Three ropes of lengths 72 meters, 108 meters, and 180 meters need to be cut into pieces of equal length. What is the maximum possible length of each piece if no rope is wasted?

We need to find the HCF of 72, 108, and 180. First, find the HCF of two numbers, then find the HCF of that result and the third number.

Step 1: Find the HCF of 72 and 108

Step Division Answer
1 108 ÷ 72 1 remainder 36
2 72 ÷ 36 2 remainder 0
HCF of 72 and 108 = 36

Step 2: Find the HCF of 36 and 180

Step Division Answer
1 180 ÷ 36 5 remainder 0
Since the remainder is 0, the HCF of 36 and 180 is 36.

Therefore, the HCF of all three numbers (72, 108, and 180) is 36.

This means the maximum possible length of each piece is 36 meters.
• The 72m rope will yield 72 ÷ 36 = 2 pieces
• The 108m rope will yield 108 ÷ 36 = 3 pieces
• The 180m rope will yield 180 ÷ 36 = 5 pieces
The maximum possible length of each piece is 36 meters.
Question 9: Find the HCF of 119 and 187.
Step Division Answer
1 187 ÷ 119 1 remainder 68
2 119 ÷ 68 1 remainder 51
3 68 ÷ 51 1 remainder 17
4 51 ÷ 17 3 remainder 0

Step 1: Divide 187 by 119

187 ÷ 119 = 1 remainder 68

Step 2: Divide 119 by 68

119 ÷ 68 = 1 remainder 51

Step 3: Divide 68 by 51

68 ÷ 51 = 1 remainder 17

Step 4: Divide 51 by 17

51 ÷ 17 = 3 remainder 0
Since the remainder is now 0, the process stops. The last divisor (17) is the HCF.
HCF of 119 and 187 = 17
Question 10: Calculate the HCF of 152 and 304.
Step Division Answer
1 304 ÷ 152 2 remainder 0

Step 1: Divide 304 by 152

304 ÷ 152 = 2 remainder 0
Since the remainder is 0 after the first division, the process stops. The smaller number (152) is the HCF.

Notice that 304 = 152 × 2, which means 152 is a factor of 304. When one number is a multiple of the other, the HCF is always the smaller number.
HCF of 152 and 304 = 152
Question 11: Find the HCF of 385 and 231.
Step Division Answer
1 385 ÷ 231 1 remainder 154
2 231 ÷ 154 1 remainder 77
3 154 ÷ 77 2 remainder 0

Step 1: Divide 385 by 231

385 ÷ 231 = 1 remainder 154

Step 2: Divide 231 by 154

231 ÷ 154 = 1 remainder 77

Step 3: Divide 154 by 77

154 ÷ 77 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (77) is the HCF.
HCF of 385 and 231 = 77
Question 12: A teacher has 204 red pens and 276 blue pens. The pens need to be packed into boxes, with each box containing the same number of red pens and the same number of blue pens. What is the maximum number of boxes that can be packed without any pens left over?

To find the maximum number of boxes, we need to find the HCF of 204 and 276.

Step Division Answer
1 276 ÷ 204 1 remainder 72
2 204 ÷ 72 2 remainder 60
3 72 ÷ 60 1 remainder 12
4 60 ÷ 12 5 remainder 0

Step 1: Divide 276 by 204

276 ÷ 204 = 1 remainder 72

Step 2: Divide 204 by 72

204 ÷ 72 = 2 remainder 60

Step 3: Divide 72 by 60

72 ÷ 60 = 1 remainder 12

Step 4: Divide 60 by 12

60 ÷ 12 = 5 remainder 0
Since the remainder is now 0, the process stops. The last divisor (12) is the HCF.

This means the teacher can pack the pens into a maximum of 12 boxes:
• Each box will have 204 ÷ 12 = 17 red pens
• Each box will have 276 ÷ 12 = 23 blue pens
The maximum number of boxes that can be packed is 12.
Question 13: Find the HCF of 224 and 168.
Step Division Answer
1 224 ÷ 168 1 remainder 56
2 168 ÷ 56 3 remainder 0

Step 1: Divide 224 by 168

224 ÷ 168 = 1 remainder 56

Step 2: Divide 168 by 56

168 ÷ 56 = 3 remainder 0
Since the remainder is now 0, the process stops. The last divisor (56) is the HCF.
HCF of 224 and 168 = 56
Question 14: Calculate the HCF of 323 and 437.
Step Division Answer
1 437 ÷ 323 1 remainder 114
2 323 ÷ 114 2 remainder 95
3 114 ÷ 95 1 remainder 19
4 95 ÷ 19 5 remainder 0

Step 1: Divide 437 by 323

437 ÷ 323 = 1 remainder 114

Step 2: Divide 323 by 114

323 ÷ 114 = 2 remainder 95

Step 3: Divide 114 by 95

114 ÷ 95 = 1 remainder 19

Step 4: Divide 95 by 19

95 ÷ 19 = 5 remainder 0
Since the remainder is now, the process stops. The last divisor (19) is the HCF.
HCF of 323 and 437 = 19
Question 15: Two metal rods measure 392 cm and 476 cm in length. They need to be cut into pieces of equal length, with no metal wasted. What is the maximum possible length of each piece?

To find the maximum possible length of each piece, we need to find the HCF of 392 and 476.

Step Division Answer
1 476 ÷ 392 1 remainder 84
2 392 ÷ 84 4 remainder 56
3 84 ÷ 56 1 remainder 28
4 56 ÷ 28 2 remainder 0

Step 1: Divide 476 by 392

476 ÷ 392 = 1 remainder 84

Step 2: Divide 392 by 84

392 ÷ 84 = 4 remainder 56

Step 3: Divide 84 by 56

84 ÷ 56 = 1 remainder 28

Step 4: Divide 56 by 28

56 ÷ 28 = 2 remainder 0
Since the remainder is now 0, the process stops. The last divisor (28) is the HCF.

This means the maximum possible length of each piece is 28 cm.
• The 392 cm rod will yield 392 ÷ 28 = 14 pieces
• The 476 cm rod will yield 476 ÷ 28 = 17 pieces
The maximum possible length of each piece is 28 cm.
What are your Feelings

Powered by BetterDocs

Leave a Reply

Your email address will not be published. Required fields are marked *

*