CBSE Class 6

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4 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 3.4. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-3-ex-3-4/

• Playing With Numbers Class 6 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 3.5
• 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 3.4
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4

Question 1.
Find the common factors of:
(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120.
Solution :
(a) 20 and 28
Factors of 20 are 1, 2,4, 5, 10 and 20.
Factors of 28 are 1, 2,4, 7, 14 and 28.
Hence, the common factors of 20 and 28 are 1, 2 and 4.

(b) 15 and 25
Factors of 15 are 1, 3, 5 and 15.
Factors of 25 are 1, 5 and 25.
Hence, the common factors of 15 and 25 are 1 and 5.

(c) 35 and 50
Factors of 35 are 1, 5, 7 and 35.
Factors of 50 are 1, 5, 10, 25 and 50.
Hence, the common factors of 35 and 50 are 1 and 5.

(d) 56 and 120
Factors of 56 are 1, 2,4, 7, 8, 14, 28 and 56. Factors of 120 are 1, 2, 3,4, 5, 6, 8,10, 12,15, 20, 24, 30,40, 60 and 120. ,
Hence, the common factors of 56 and 120 are 1, 2,4 and 8.

Question 2.
Find the common factors of:
(a) 4, 8 and 12
(b) 5, 15 and 25.
Solution :
(a) 4,8 and 12 Factors of 4 are 1,2 and 4.
Factors of 8 are 1, 2,4 and 8.
Factors of 12 are 1, 2, 3,4,6 and 12.
Hence, the common factors of 4,8 and 12 are 1, 2 and 4.

(b) 5,15 and 25
Factors of 5 are 1 and 5.
Factors of 15 are 1, 3 and 5.
Factors of 25 are 1,5 and 25.
Hence, the common factors of 5,15 and 25 are 1 and 5.

Question 3.
Find the first three common multiples of:
(a) 6 and 8
(b) 12 and 18.
Solution :
(a) 6 and 8
Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,78, 84,90,96,
Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
Common multiples of 6 and 8 are 24, 48, 72, 96,
∴ First three common multiples of 6 and 8 are 24, 48 and 72.

(b) 12 and 18
Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, *
Multiples of 18 are 18,36,54,72,90,108,126, 144,
∴ Common multiples of 12 and 18 are 36, 72,108, 144,
∴ First three common multiples of 12 and 18 are 36, 72 and 108.

Question 4.
Write all the numbers less than 100 which are common multiples of 3 and 4.
Solution :
Multiples of 3 are : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108,
Multiples of 4 are : 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108,
Common multiples of 3 and 4 are : 12, 24, 36, 48, 60, 72, 84, 96, 108,
∴ All the numbers less than 100 which are common multiples of 3 and 4 are 12, 24, 36, 48, 60, 72, 84 and 96.

Question 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.
Solution :
(a) 18 and 35
Factors of 18 are 1, 2, 3, 6, 9 and 18.
Factors of 35 are 1, 5, 7 and 35.
∴ A common factor of 18 and 35 is 1.
∴ 18 and 3 5 have only 1 as the common factor
∴ 18 and 35 are co-prime numbers.

(b) 15 and 37
Factors of 15 are 1, 3, 5 and 15.
Factors of 37 are 1 and 37.
∴ A common factor of 15 and 37 is 1. v 15 and 37 have only 1 as the common factor ∴ 15 and 37 are co-prime numbers.

(c) 30 and 415
Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
Factors of 415 are 1, 5, 83 and 415.
∴ Common factors of 30 and 415 are 1 and 5. v
∴ 30 and 415 have two common factors
∴ 30 and 45 are not co-prime numbers.

(d) 17 and 68 Factors of 17 are 1 and 17.
Factors of 68 are 1, 2, 4, 17, 34 and 68.
∴ Common factors of 17 and 68 are 1 and 17.
∴ 17 and 68 have two common factors
∴ 17 and 68 are not co-prime numbers.

(e) 216 and 215
Factors of 216 are 1,2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108 and 216.
Factors of 215 are 1, 5 and 43
∴ A common factor of 216 and 215 is 1.
∴ 216 and 215 have only 1 as the common factor
∴ 216 and 215 are co-prime numbers.

(f) 81 and 16
Factors of 81 are 1, 3, 9, 27 and 81.
Factors of 16 are 1, 2, 4, 8 and 16.
∴ A common factor of 81 and 16 is 1.
∴ 81 and 16 have only 1 as the common factor
∴ 81 and 16 are co-prime numbers.

Question 6.
A number is divisible by both 5 and 12. By which other numbers will that number be always divisible?
Solution :
The number will be always divisible by 5 × 12 = 60.

Question 7.
A number is divisible by 12. By what other numbers will that number be divisible?
Solution :
Factors of 12 are 1, 2, 3, 4, 6 and 12. So, that number will be divisible by other numbers 1, 2, 3, 4 and 6.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3 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 3.3. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-3-ex-3-3/

• Playing With Numbers Class 6 Ex 3.1
• Playing With Numbers Class 6 Ex 3.2
• 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 3.3
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3

Question 1.
Using divisibility tests, determine which the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11; (say yes or no):

Solution :

Question 2.
Using divisibility tests, determine which of the following numbers are divisible by 4: by 8:
(a) 572
(b) 726352
(C) 5500
(d) 6000
(e) 12159
(f) 14560
(g) 21084
(h) 31795072
(i) 1700
(j) 2150.
Solution :
(a) 572
(i) Divisibility by 4
The number formed by last two digits = 72

∵ Remainder is 0
∴ 72 is divisible by 4
∴ 572 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 572

∵ Remainder = 4 ≠ 0
∴ 572 is not divisible by 8.

(b) 726352
(i) Divisibility by 4
The number formed by last two digits = 52

∵ Remainder is 0
∴ 52 is divisible by 4
∴ 726352 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 352

∵ Remainder is 0
∴ 352 is divisible by 8
∴ 726352 is divisible by 8.

(c) 5500
(i) Divisibility by 4
The number formed by last two digits = 00, which is divisible by 4 5500 is divisible by 4.
∴ 500 is Divisibility by 4

(ii)
The number formed by last three digits = 500

∵ Remainder 4 0
∴ 500 is not divisible by 8
∴ 5500 is not divisible by 8.

(d) 6000
(i) Divisibility by 4
The number formed by last two digits = 00, which is divisible by 4
6000 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 000, which is divisible by 8 6000 is divisible by 8.
6000 is divisible by 8

(e) 12159
(i) Divisibility by 4
The number formed by last two digits = 59

∵ Remainder = 3 ≠ 0
∴ 59 is not divisible by 4.
∴ 12159 is not divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits =159

∵ Remainder 7 ≠ 0
∴ 159 is not divisible by 8
∴ 12159 is not divisible by 8.

(f) 14560
(i) Divisibility by 4
The number formed by last two digits = 60

∵ Remainder is 0
∴ 60 is divisible by 4
∴ 14560 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 560

∵ Remainder is 0
∴ 560 is divisible by 8
∴ 14560 is divisible by 8.

(g) 21084
(i) Divisibility by 4
The number formed by last two digits = 84

∵ Remainder is 0
∴ 484 is divisible by 4
∴ 21084 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 84

∵ The remainder is not 0
∴ 84 is not divisible by 8
∴ 21084 is not divisible by 8.

(h) 31795072
(i) Divisibility by 4
The number formed by last two digits = 72

∵ Remainder is 0
∴ 72 is divisible by 4
∴ 31795072 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 72

∵ Remainder is 0
∴ 72 is divisible by 8
∴ 31795072 is divisible by 8.

(i) 1700
(i) Divisibility by 4
The number formed by last two digits = 00, which is divisible by 4
∴ 1700 is divisible by

(ii) Divisibility by 8
The number formed by last three digits = 700

∵ Remainder 4 ≠ 0
∴ 700 is not divisible by 8
∴ 1700 is not divisible by 8.

(j) 2150
(i) Divisibility by 4
The number formed by last two digits = 50

∵ Remainder = 2 ≠ 0
∴ 50 is not divisible by 4
∴ 2150 is not divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits =150

∵ Remainder 6 ≠ 0
∴ 150 is not divisible by 8
∴ 2150 is not divisible by 8.

Question 3.
Using divisibility tests, determine which of following numbers are divisible by 6 :
(a)297144
(b) 1258
(c) 4335
(d) 61233
(e) 901352
(f) 438750
(g) 1790184
(h) 12583
(i) 639210
(j) 17852.
Solution :
We know that number is divisible by 6 if it is divisible by 2 and 3 both.
(a) 297144
(i) Divisibility by 2
Unit’s digit = 4
297144 is divisible by 2.

(ii) Divisibility by 3
Sum of the digits = 2 + 9 + 7 + 1+ 4 + 4 = 27, which is divisible by 3 297144 is divisible by 3 Since, 297144 is divisible by 2 and 3 both, so it is divisible by 6.

(b) 1258
(i) Divisibility by 2 Unit’s digit = 8 1258 is divisible by 2.
(ii) Divisibility by 3 Sum of the digits =1+2 + 5 + 8=16 which is not divisible by 3 1258 is not divisible by 3. Since 1258 is divisible by 2 but not by 3. so 1258 is not divisible by 6.

(c) 4335
(i) Divisibility by 2
Unit’s digit = 5, which is not any of the digits 0, 2, 4, 6 or 8 4335 is not divisible by 2 4335 is not divisible by 6.

(d) 61233
(i) Divisibility by 2
Unit’s digit = 3, which is not any of the digits 0, 2, 4, 6 or 8

61233 is not divisible by 2
61233 is not divisible by 6.

(e) 901352
(i) Divisibility by 2 v Unit’s digit = 2
901352 is divisible by 2.
(ii) Divisibility by 3
Sum of the digits = 9 + 0+1+3 + 5 + 2 = 20, which is not divisible by 3
901352 is not divisible by 3 Since, 901352 is divisible by 2 but not by 3, so it is not divisible by 6.

(f) 438750
(i) Divisibility by 2 v Unit’s digit = 0
438750 is divisible by 2.

(ii) Divisibility by 3
Sum of the digits = 4 + 3 + 8 + 7 + 5+ 0 = 27, which is divisible by 3
438750 is divisible by 3 Since, 438750 is divisible by 2 and 3 both, so it is divisible by 6.

(g) 1790184
(i) Divisibility by 2 v Unit’s digit = 4
1790184 is divisible by 2.
(ii) Divisibility by 3
Sum of the digits =1+7 + 9 + 0+1 + 8 + 4=30, which is divisible by 3
1790184 is divisible by 3 Since, 1790184 is divisible by 2 and 3 both, so it is divisible by 6.

(h) 12583,
(i) Divisibility by 2
Unit’s digit = 3, which is not any of the digits 0, 2, 4, 6 or 8
12583 is not divisible by 2
12583 is not divisible by 6.

(i) 639210
(i) Divisibility by 2
Unit’s digit = 0
639210 is divisible by 2.
(ii) Divisibility by 3
Sum of the digits = 6 + 3 + 9 + 2+1+0 = 21, which is divisible by 3
639210 is divisible by 3 Since, 639210 is divisible by 2 and 3 both, so it is divisible by 6.

(j) 17852
(i) Divisibility by 2
Unit’s digit = 2
∴ 17852 is divisible by 2.
(ii) Divisibility by 3 Sum of the digits =1+7 + 8 + 5 + 2 = 23, which is not divisible by 3
17852 is not divisible by 3
Since 17852 is divisible by 2 but not by 3, so it is not divisible by 6.

Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 11:
(a) 5445
(b) 10824
(c) 7138965
(d) 70169308
(e) 10000001
(f) 901153.
Solution :
(a) 5445
Sum of the digits (at odd places) from the right =5+4=9
Sum of the digits (at even places) from the right =4+5=9
Difference of these sums = 9-9 = 0 v 0 is divisible by 11
5445 is divisible by 11.

(b) 10824
Sum of the digits (at odd places) from the right = 4 + 8+1 = 13
Sum of the digits (at even places) from the right = 2 + 0 = 2
Difference of these sums = 13 – 2 = 11
11 is divisible by 11
10824 is divisible by 11.

(c) 7138965
Sum of the digits (at odd places) from the right =5+9+3+7=24
Sum of the digits (at even places) from the right = 6 + 8 + 1 = 15
Difference of these sums = 24 – 15 = 9
Y 9 is not divisible by 11
∴ 7138965 is not divisible by 11.

(d) 70169308
Sum of the digits (at odd places) from the right = 8 + 3+6 + 0=17
Sum of the digits (at even places) from the right = 0 + 9 + 1+7=17 Difference of these sums = 17-17=0
Y 0 is divisible by 11
∴ 70169308 is divisible by 11.

(e) 10000001
Sum of the digits (at odd places) from the right =l+0+0+0=l
Sum of the digits (at even places) from the right =0+0+0+1=1 Difference of these sums =1-1=0 0 is divisible by 11
∴ 10000001 is divisible by 11.

(f) 901153
Sum of the digits (at odd places) from the right = 3+ 1 +0 = 4
Sum of the digits (at even places) from the right = 5+1+9=15
Difference of these sums =15-4=11
Y 11 is divisible by 11
∴ 901153 is divisible by 11.

Question 5.
Write the smallest digit and the largest digit in the blank space of each of the following numbers so that the number is divisible by 3:
(a) __ 6724
(b) 4765 __ 2.
Solution :
(a) __ 6724
(i) Smallest digit
Sum of the given digits = 6 + 7 + 2 + 4=19
19 is not divisible by 3
∴ Smallest digit (non-zero) is 2.

(ii) Largest digit The largest digit is 8.

(b) 4765 _ 2
(i) Smallest digit
Sum of the given digits = 4 + 7 + 6 + 5 + 2 = 24 Y 24 is divisible by 3
∴ Smallest digit is 0.

(ii) Largest digit The largest digit is 9.

Question 6.
Write digit in the blank space of each of the following numbers so that the number is divisible by 11:
(a) 92 __ 389
(b) 8 __ 9484.
Solution :
(a) 92 __ 389
Sum of the given digits (at odd places) from the right = 9 + 3 + 2= 14
Sum of the given digits (at even places) from the right = 8 + required digit + 9 = required digit + 17
Difference of these sums = required digit + 3 For the above difference to be divisible by 11, required digit = 8.
Hence, the required number is 92 8 389.

(b) 8 __ 9484
Sum of the given digits (at odd places) from the right = 4 + 4 + required digit = 8 + required digit
Sum of the given digits (at even places) from the right = 8 + 9 + 8 = 25
Difference of the sums = 17 – required digit For the above difference to be divisible by 11, required digit = 6.
Hence, the required number is 8 6 9484.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-2-ex-2-3/

• Whole Numbers Class 6 Ex 2.1
• Whole Numbers Class 6 Ex 2.2

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 2
Chapter Name	Whole Numbers
Exercise	Ex 2.3
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3

Question 1.
Which of the following will not represent
(a) 1 + 0
(b) 0 × 0
(c) \(\frac { 0 }{ 2 }\)
(d) \(\frac { 10\quad 10 }{ 2 }\)
Solution :
(a) 1 + 0 = 1 ≠ 0
(b) 0 × 0 = 0
(c) \(\frac { 0 }{ 2 }\)
(d) \(\frac { 10-10 }{ 2 } =\frac { 0 }{ 2 } =0\)
Hence, (a) 1 + 0 will not represent zero.

Question 2.
If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Solution :
Yes! For example :
2 × 0 = 0
0 × 3 = 0
0 × 0 = 0.

Question 3.
If the product of two whole numbers is 1. can we say that one or both of them will be 1? Justify through examples.
Solution :
Both of them must be ‘ 1’ as 1 × 1 = 1.

Question 4.
Find using distributive property :
(a) 728 × 101
(b) 5437 × 1001
(c) 824 × 25
(d) 4275 × 125
(e) 504 × 35
Solution :

Question 5.
Study the pattern :

Write the next two steps. Can you say how the pattern works ?
(Hint : 12345 =11111 + 1 1 11 + 111 + 11 + 1). Sol. Next two steps are as follows :
Solution :
Next two steps are as follows :
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543.
Working of the pattern

We hope the NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6 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 3.6. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-3-ex-3-6/

• Playing With Numbers Class 6 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 3.7

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 3
Chapter Name	Playing With Numbers
Exercise	Ex 3.6
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6

Question 1.
Find the H.C.F. of the following numbers,
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12 45 75
Solution :
(a) 18,48
Factors of 18 are 1, 2, 3, 6, 9 and 18.
Factors of 48 are 1, 2, 3,4, 6, 8, 12, 16, 24 and
∴ Common factors of 18 and 48 are 1, 2, 3
Highest of these common factors is 6. ∴ H.C.F. of 18 and 48 is 6.

(b) 30,42
Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
Factors of 42 are 1, 2, 3. 6. 7, 14, 21 and 42. ,
∴ Common factors of 30 and 42 are 1,2, 3 and 6.
Highest of these common factors is 6.
∴ H.C.F. of 30 and 42 is 6.

(c) 18,60
Factors of 18 are 1, 2, 3, 6, 9 and 18. Factors of 60 are 1,2, 3,4, 5,6,10 12,15, 20, 30 and 60.
∴ Common factors of 18 and 60 are 1, 2, 3 and 6.
Highest of these common factors is 6.
∴ H.C.F. of 18 and 60 is 6.

(d) 27,63
Factors of 27 are 1, 3, 9 and 27.
Factors of 63 are 1, 3, 7, 9, 21 and 63.
Common factors of 27 and 63 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 27 and 63 is 9.

(e) 36,84
Factors of 36 are 1, 2, 3,4, 6, 9,12, 18 and 36.
Factors of 84 are 1,2, 3,4,6,7, 12, 14,21, 28, 42 and 84.
Common factors of 36 and 84 are 1,2, 3,4, 6 and 12.
Highest of these common factors is 12.
∴ H.C.F. of 36 and 84 is 12. if) 34,102 • ‘
Factors of 34 are 1,2, 17 and 34.
Factors of 102 are 1, 2, 3,6,17, 34. 51 and 102.
∴ Common factors of 34 and 102 are 1, 2, 17 and 34.
Highest of these common factors is 34.
∴ H.C.F. of 34 and 102 is 34.

(g) 70,105,175
Factors of 70 are 1. 2, 5. 7, 10. 14, 35 and 70.
Factors of 105 are 1, 3, 5. 7. 15. 21. 35 and 105.
Factors of 175 are 1. 5, 7. 25. 35 and 175. .’. Common factors of 70, 105 and 175 are 1, 5 and 35.
Highest of these common factors is 35.
∴ H.C.F. of 70. 105 and 175 are 35.

(h) 91,112,49
Factors of 91 are 1,7, 13 and 91.
Factors of 112 are 1,2. 4. 7, 8, 14. 16. 28, 56 and 112.
Factors of 49 are 1.7 and 49.
Common factors of 91,112 and 49 are 1 and 7.
Highest of these common factors is 7.
∴ H.C.F. of 91, 112 and 49 is 7.

(i) 18,54,81
Factors of 18 are 1. 2, 3. 6, 9 and 18. Factors of 54 are 1, 2. 3, 6. 9, 18. 27 and 54.
Factors of 81 are 1. 3, 9, 27 and 81.
∴ Common factors of 18,54 and 81 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 18, 54 and 81 is 9.
(j) 12, 45, 75
Factors of 12 are 1, 2, 3, 4, 6 and 12. Factors of 45 are 1, 3, 5, 9, 15 and 45.
: Factors of 75 are 1, 3, 5, 15, 25 and 75.
∴ Common factors of 12,45 and 75 are 1 and 3.
Highest of these common factors is 3.
H.C.F. of 12. 45 and 75 is 3.

Question 2.
What is the H.C.F. of two consecutive :
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution :
(a) The H.C.F. of two consecutive numbers is 1.
(b) The H.C.F. of two consecutive even numbers is 2.
(c) The H.C.F. of two consecutive odd numbers is 1.

Question 3.
H. C.F. of co-prime numbers 4 and 15 was found as follows factorization: 4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F?
Solution :
No, the answer is not correct. The correct answer is as follows :
H.C.F. of 4 and 15 is 1.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6, drop a comment below and we will get back to you at the earliest.