The NCERT Solutions For Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.4 contains the answers to the problems present in the exercise. Common factors and multiples can be found easily for the given set of numbers, using the steps mentioned in NCERT Solutions, provided by BYJU’S. The problems in these exercises are solved in an easy way to help students boost their exam preparation. Students can refer to the answers for the questions present in the fourth exercise of the third chapter given in the NCERT Solutions Class 6 Maths to determine the factors and multiples.
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.4
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Access NCERT Solutions for Class 6 Chapter 3: Playing with Numbers Exercise 3.4
1. Find the common factors of:
(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120
Solutions:
(a) 20 and 28
1, 2, 4, 5, 10 and 20 are factors of 20
1, 2, 4, 7, 14 and 28 are factors of 28
Common factors = 1, 2, 4
(b) 15 and 25
1, 3, 5 and 15 are factors of 15
1, 5 and 25 are factors of 25
Common factors = 1, 5
(c) 35 and 50
1, 5, 7 and 35 are factors of 35
1, 2, 5, 10, 25 and 50 are factors of 50
Common factors = 1, 5
(d) 56 and 120
1, 2, 4, 7, 8, 14, 28 and 56 are factors of 56
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120 are factors of 120
Common factors = 1, 2, 4, 8
2. Find the common factors of:
(a) 4, 8 and 12
(b) 5, 15 and 25
Solutions:
(a) 4, 8 and 12
1, 2, 4 are factors of 4
1, 2, 4, 8 are factors of 8
1, 2, 3, 4, 6, 12 are factors of 12
Common factors = 1, 2, 4
(b) 5, 15 and 25
1, 5 are factors of 5
1, 3, 5, 15 are factors of 15
1, 5, 25 are factors of 25
Common factors = 1, 5
3. Find the first three common multiples of:
(a) 6 and 8
(b) 12 and 18
Solutions:
(a) 6 and 8
6, 12, 18, 24, 30 are multiples of 6
8, 16, 24, 32 are multiples of 8
Three common multiples are 24, 48, 72
(b) 12 and 18
12, 24, 36, 48 are multiples of 12
18, 36, 54, 72 are multiples of 18
Three common factors are 36, 72, 108
4. Write all the numbers less than 100 which are common multiples of 3 and 4.
Solutions:
Multiples of 3 are 3, 6, 9, 12, 15
Multiples of 4 are 4, 8, 12, 16, 20
Common multiples are 12, 24, 36, 48, 60, 72, 84 and 96
5. Which of the following numbers are co-prime?
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16
Solutions:
(a) 18 and 35
Factors of 18 are 1, 2, 3, 6, 9, 18
Factors of 35 are 1, 5, 7, 35
Common factor = 1
Since, their common factor is 1. Hence, the given two numbers are co-prime
(b) 15 and 37
Factors of 15 are 1, 3, 5, 15
Factors of 37 are 1, 37
Common factors = 1
Since, their common factor is 1. Hence, the given two numbers are co-prime
(c) 30 and 415
Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30
Factors of 415 are 1, 5, 83, 415
Common factors = 1, 5
Since, their common factor is other than 1. Hence, the given two numbers are not co-prime
(d) 17 and 68
Factors of 17 are 1, 17
Factors of 68 are 1, 2, 4, 17, 34, 68
Common factors = 1, 17
Since, their common factor is other than 1. Hence, the given two numbers are not co-prime
(e) 216 and 215
Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216
Factors of 215 are 1, 5, 43, 215
Common factors = 1
Since, their common factor is 1. Hence, the given two numbers are co-prime
(f) 81 and 16
Factors of 81 are 1, 3, 9, 27, 81
Factors of 16 are 1, 2, 4, 8, 16
Common factors = 1
Since, their common factor is 1. Hence, the given two numbers are co-prime
6. A number is divisible by both 5 and 12. By which other number will that number be always divisible?
Solutions:
Factors of 5 are 1, 5
Factors of 12 are 1, 2, 3, 4, 6, 12
Their common factor = 1
Since their common factor is 1. The given two numbers are co-prime and is also divisible by their product 60
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
7. A number is divisible by 12. By what other numbers will that number be divisible?
Solutions:
Since the number is divisible by 12. Hence, it also divisible by its factors i.e 1, 2, 3, 4, 6, 12
Therefore 1, 2, 3, 4, 6 are the numbers other than 12 by which this number is also divisible
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