EXCELLENCE IN MATHEMATICS-IV
EXERCISE 6.6
Find the LCM of the following numbers using prime factorisation method:
MPLE 12: Find the lowest number which is exactly divisible by 15 and 2
1. 8, 12
2. 25, 30
3. 12, 18, 36
5. 24, 30
6. 48, 72
7. 15, 25, 30
PROBLEMS ON HCF AND LCM
TION :
le find the LCM of 15 and 20 to get
e required number
3 15
4. 18,30
8. 60, 75, 120
Answers
Answer:
(b) 15
15 = 1 x 15
15 = 3 x 5
Therefore, all the factors of 15 are: 1, 3, 5 and 15.
(c) 21
21 = 1 x 21
21 = 3 x 7
Therefore, all the factors of 21 are: 1, 3, 7 and 21.
(d) 27
27 = 1 x 27
27 = 3 x 9
Therefore, all the factors of 27 are: 1, 3, 9 and 27.
(e) 12
12 = 1 x 12
12 = 2 x 6
12 = 3 x 4
Therefore, all the factors of 12 are: 1, 2, 3, 4, 6 and 12.
(f) Factors of 20 are:
20 = 1 x 20
20 = 2 x 10
20 = 4 x 5
Therefore, all the factors of 20 are: 1, 2, 4, 5, 10 and 20.
Answer:
HCF of 30, 60 and 75 is 15.
LCM of 30, 60 and 75 is 300.
Step-by-step explanation:
To Find:
The HCF and LCM of 30, 60 and 75 by prime factorisation method.
We know that:
HCF = Highest Common Factor
LCM = Lowest Common Multiple
Finding the HCF and LCM:
By prime factorisation.
Factors of 30 = 2 × 3 × 5
Factors of 60 = 2² × 3 × 5
Factors of 75 = 3 × 5²
Here,
Highest Common Factor = 3 × 5 = 15
Lowest Common Multiple = 2² × 3 × 5² = 300
Hence,
HCF of 30, 60 and 75 is 15.
LCM of 30, 60 and 75 is 300.