NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.1.
- Playing with Numbers Class 6 Ex 3.2
- Playing with Numbers Class 6 Ex 3.3
- Playing with Numbers Class 6 Ex 3.4
- Playing with Numbers Class 6 Ex 3.5
- Playing with Numbers Class 6 Ex 3.6
- Playing with Numbers Class 6 Ex 3.7
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 6 |
| Subject | Maths |
| Chapter | Chapter 3 |
| Chapter Name | Playing with Numbers |
| Exercise | Ex 3.1 |
| Number of Questions Solved | 55 |
| Category | NCERT Solutions |
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.1
Question 1.
Write all the factors of the following numbers :
(a) 24
(b) 15
(c) 21
(d) 27
(e) 12
(f) 20
(g) 18
(h) 23
(i) 36
Solution:
(a) We have,
24 = 1 x 24
24 = 2 x 12
24 = 3 x 8
24 = 4 x 6
24 = 6 x 4
Stop here, because 4 and 6 have occurred earlier.
Thus, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
(b) We have,
15 = 1 x 15
15 = 3 x 5
15 = 5 x 3
Stop here, because 3 and 5 have occurred earlier.
Thus, all the factors of 15 are 1, 3, 5 and 15.
(c) We have,
21 = 1 x 21
21 = 3 x 7
21 = 7 x 3
Stop here, because 3 and 7 have occurred earlier.
Thus, all the factors of 21 are 1, 3, 7 and 21.
(d) We have,
27 = 1 x 27
27 = 3 x 9
27 = 9 x 3
Stop here, because 3 and 9 have occurred earlier.
Thus, all the factors of 27 are 1, 3, 9 and 27.
(e) We have,
12 = 1 x 12
12 = 2 x 6
12 = 3 x 4
12 = 4 x 3
Stop here, because 3 and 4 have occurred earlier.
Thus, all the factors of 12 are 1, 2, 3, 4, 6 and 12.
(f) We have,
20 = 1 x 20
20 = 2 x10
20 = 4 x 5
20 = 5 x 4
Stop here, because 4 and 5 have occurred earlier.
Thus, all the factors of 20 are 1, 2, 4, 5, 10 and 20.
(g) We have,
18 = 1 x 18
18 = 2 x 9
18 = 3 x 6
18 = 6 x 3
Stop here, because 3 and 6 have occurred earlier.
Thus, all the factors of 18 are 1, 2, 3, 6, 9 and 18.
(h) We have,
23 = 1 x 23
23 = 23 x 1
Stop here, because 1 and 23 have occurred earlier.
Thus, all the factors of 23 are 1 and 23.
(i) We have,
36 = 1 x 36
36 = 2 x 18
36 = 3 x 12
36 = 4 x 9
36 = 1 x 6
Stop here, because both the factors (6) are same.
Thus, all the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.
Question 2.
Write first five multiples of:
(a) 5
(b) 8
(c) 9
Solution:
(a) In order to obtain first five multiples of 5, we multiply it by 1, 2, 3, 4 and 5 respectively. ,r
We have,
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4=20
5 x 5 = 25
Hence, the first five multiples of 5 are 5, 10, 15, 20 and 25 respectively.
(b) In order to obtain first five multiples of 8, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
8 x 1=8
8 x 2=16
8 x 3=24
8 x 4 = 32
8 x 5 = 40
Hence, the first five multiples of 8 are 8, 16, 24, 32 and 40 respectively,
(c) In order to obtain first five multiples of 9, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
Hence, the first five multiples of 9 are 9, 18, 27, 36 and 45 respectively.
Question 3.
Match the items in column 1 with the items in column 2.
Solution:
Matching is as under :
(i) → (c) ∵ 35 x 2 = 70
(ii) →(d) ∵ 30 = 15 = 2
(iii) → (a) ∵ 8 x 2 = 16
(iv)→ (f) ∵ 20 = 20 = 1
(v) → (e) ∵ 50 = 25 = 2
Question 4.
Find all the multiples of 9 upto 100.
Solution:
All the multiples of 9 upto 100 are
9 x 1, 9 x 2, 9 x 3, 9 x 4, 9 x 5, 9 x 6, 9 x 7, 9 x 8, 9 x 9, 9 x 10 and 9 x 11.
i. e., 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.
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