Page 164 - New Medley 5 Semester 1

2 110

5 55

11 11
1

∴ 110 = 2 × 5 × 11

Thus, HCF of 60, 90 and 110 = 2 × 5 = 10

Alternative Method Again,

60 90 1 30 110 3

– 60 – 90

30 60 2 20 30 1

– 60 – 20
0 10 20 2
HCF
– 20
HCF 0

∴ HCF of 60, 90 and 110 = 10

EXERCISE 3.4

A. Find the HCF of the following by the prime factorisation method.

1. 16, 24 2. 35, 45 3. 16, 40
4. 18, 60 5. 40, 60, 80 6. 45, 75, 90

7. 48, 112, 120 8. 64, 96, 144

B. Find the HCF of each of the following by the factor method.
1. 6, 9 2. 10, 25 3. 12, 20

4. 15, 30 5. 20, 45 6. 8, 12, 20

7. 16, 24, 40 8. 28, 42, 56

C. Find the HCF of the following by the division method.

1. 25, 45 2. 24, 60 3. 18, 45

4. 45, 105 5. 54, 72, 90 6. 63, 81, 105

7. 132, 180, 284 8. 120, 150, 486

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