# Joy of Mathematics Class 6 Solutions Chapter 4

## Joy of Mathematics Class 6 Solutions Chapter 4 Factors and Multiples

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Joy of Mathematics Class 6 Book, Chapter 4 Factors and Multiples. Here students can easily find step by step solutions of all the problems for Factors and Multiples. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Students will get chapter 4 solutions. Here students will find exercise wise solutions for chapter 4 Exercise 4.1, 4.2, 4.3, 4.4 and 4.5

Factors and Multiples Exercise 4.1 Solution :

Question no – (1)

Solution :

(a) Factors are = 1, 13

(b) Factors are = 1, 3, 9, 27

(c) Factors are = 1, 5, 7, 35

(d) Factors are = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

(e) Factors are = 1, 2, 4, 8, 16, 32, 64

(f) Factors are = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

(g) Factors are = 1, 3, 5, 7, 15, 21, 35, 105

(h) Factors are = 1, 5, 7, 25, 35, 175

Question no – (2)

Solution :

(a) Six multiples 4 are = 4, 8, 12, 16, 20, 24

(b) Six multiples 5 are = 5, 10, 15, 20, 25, 30

(c) Six multiples 6 are = 6, 12, 18, 24, 30, 36

(d) Six multiples 8 are = 8, 16, 24, 32, 40, 48

(e) Six multiples 11 are = 11, 22, 33, 44, 55, 66

(f) Six multiples 12 are = 12, 24, 36, 48, 60, 72

(g) Six multiples 13 are = 13, 26, 39, 52, 65, 78

(h) Six multiples 16 are = 16, 32, 48, 64, 80, 96

Question no – (3)

Solution :

(a) Factor 19

= 1 × 19

99541 is divisible by 19.

So, the number is 99541.

Question no – (4)

Solution :

Multiplies of 5 less than 100 are

= 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95

Question no – (5)

Solution :

14th and 18th multiples of 16 is

= 16 × 14 = 224

= 16 × 18 = 288

Question no – (6)

Solution :

Multiplies of 15 less than 100 are 15, 30, 45, 60, 75, 90

Question no – (7)

Solution :

19038 = 19000 + 38

= 19 (1000 + 2)

Therefore, 19 is a factor of 19038

Factors and Multiples Exercise 4.2 Solution :

Question no – (1)

Solution :

(a) All prime numbers,

= 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

(b) All prime numbers are,

= 31, 37, 41, 43, 47

(c) All prime numbers are,

= 101, 103, 107, 109, 113, 127, 131, 137, 139, 149

(d) All prime numbers are,

= 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199

Question no – (2)

Solution :

(a) 37 = Prime no (1, 37)

(b) 39 = Composite no (1, 3, 39)

(c) 59 = Prime no (1, 59)

(d) 91 = Composite no (1, 91, 3, 17)

(e) 99 = Composite no (1, 3, 99)

(f) 103 = Prime no

Question no – (3)

Solution :

(a) Composite numbers are = 10, 12, 14, 15, 16, 18

(b) Composite numbers are = 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50

(c) Composite numbers are = 62, 63, 64, 65, 66, 68, 69

(d) Composite numbers are = 91, 92, 93, 94, 95, 96

Question no – (4)

Solution :

The formed number will be Composite number.

Question no – (5)

Solution :

The required numbers are,

Twin prime (3, 5), (5, 7), (11, 13)

Co-prime (4, 9), (5, 7), (11, 17)

Question no – (6)

Solution :

Possible digit are, = (1, 3, 7, 9)

Question no – (7)

Solution :

Twin prime = (41, 43), (71, 73)

Co-prime = (41, 43), (45, 47), (49, 51), (51, 53), (53, 55), (55, 57)

Question no – (8)

Solution :

(a) (29, 31)

(b) (41, 43)

(c) (71, 73)

(d) (101, 103)

Question no – (9)

Solution :

The numbers are,

= (23, 37, 53, 73)

Question no – (10)

Solution :

The three prime numbers are, where sum is 30,

= (2, 5, 23), (2, 11, 17)

Factors and Multiples Exercise 4.4 Solution :

Question no – (1)

Solution :

(a) 20 = 1, 2, 4, 5

28 = 1, 2, 4, 7

Common factor = (1, 2, 4)

(b) 42 = 1, 2, 3, 6, 7

60 = 1, 2, 3, 4, 5, 6, 12

Common factor = (1, 2, 3, 6)

(c) 63 = 1, 3, 7, 9, 63

91 = 1, 7, 13, 31

C.F. = (1, 7)

(d) 120, 160 Common factor = (1, 2, 3, 4, 6, 8, 12, 24)

Question no – (2)

Solution :

(a) 646 Prime factors

= (2 × 17 × 19)

(b) 848 Factors

= (2 × 2 × 2 × 2 × 58)

(c) 945 Factors are

= (3 × 3 × 3 × 5 × 7)

(d) 1226 Factors are

= (2 × 613)

(e) 1035 Factors are,

= (5 × 3 × 3 × 23)

Question no – (3)

Solution :

(a) P54 and  P66 54 = 2 × 3 × 3 × 3

66 = 2 × 3 × 11

Common factor is = (2, 3)

(b) Set of all factors are = (2, 3, 11)

Question no – (4)

Solution :

F48 and F60 Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Factor of 60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Largest element common factor = 12

Question no – (5)

Solution :

(a) 144 and 18 144 = 2 × 2 × 2 × 3 × 2 × 3

180 = 2 × 2 × 3 × 3 × 5

H.C.F is = 2 × 2 × 3 = 12

(b) 120 and 204 120 = 2 × 2 × 2 × 3 × 5

204 = 2 × 2 × 3 × 17

H.C.F. is = (2 × 2 × 3)

= 12

(c) 72 and 108 72 = 2 × 2 × 2 × 3 × 3

108 = 2 × 2 × 3 × 3 × 3

H.C.F. is = (2 × 2 × 3 × 3)

= 36

(d) 44 and 110 44 = 2 × 2 × 11

110 = 2 × 5 × 11

H.C.F. is = 2 × 11

= 22

Question no – (6)

Solution :

(a) 615, 1599 H.C.F. is = 123

(b) 219, 1022 H.C.F. is = 73

(c) 882, 3150 H.C.F. is 126

(d) 513, 6384 H.C.F. is 57

Question no – (7)

Solution :

The H.C.F of any two consecutive odd numbers is 1

Question no – (8)

Solution : H.C.F. is 12

Question no – (9)

Solution :

Here, 285 – 9 = 276

1249 – 7 = 1242 The greatest no is 138

Question no – (10)

Solution :

Here, 2273 – 5 = 22.68

1823 – 5 = 1818

977 – 5 = 972 Greatest number is 18

Question no – (11)

Solution :

Here, 445 – 4 = 441

572 – 5 = 567

699 – 6 = 693

Now, Greatest no is 63

Question no – (12)

Solution :

Length = 21m 36cm = 2136cm

Broad = 17m 4cm = 1704cm

Area = 3639744cm2

Question no – (13)

Solution :

Taking H.C.F. H.C.F is 24

Number of square shaped PLACE

= 120 × 96/24 × 24

= 20

Question no – (14)

Solution :

Taking H.C.F. HCF is 525 cm

Maximum length of take 525 cm.

Factors and Multiples Exercise 4.5 Solution :

Question no – (1)

Solution :

(a) M16 multiples = 16, 32, 48, 64, 80, 96, 102, 128, 144, 160, 176, 192,..…..

M20 multiples = 20, 40, 60, 60, 100, 120, 140, 160, 180, 200,……..

(80, 160,…….)

(b) Multiple of M25 = 25, 50, 75, 100, 128, 150, 175, 200, 225, 250, 275,….,400,…….

Multiple of 40 = 40, 80, 120, 160, 200, 240, 280, 320, 360, 400,……

(200, 400, 600,…..)

Question no – (2)

Solutions :

(a) 9, 21

9 = 9, 18, 27, 36, 45, 54, 63,…..

21 = 21, 42, 63, 84, 105,…..

L.C.M. is 63

(b) 16, 36

16 = 16, 32, 48, 64, 80, 96, 102, 128, 144, 160

36 = 36, 72, 108, 144, 180,…..

L.C.M. is = 144

(c) 22, 42

22 = 22, 44, 66, 88, 110, 132, 154,…..

42 = 42, 84, 126, 168, 210 L.C.M. is = (2 × 11 × 21)

= (22 × 21)

= 462

Question no – (3)

Solutions :

(a) 9,  18 L.C.M. is = (3 × 3 × 2)

= 18

(b) 64, 128 L.C.M. is = (2 × 2 × 2 × 2 × 2 × 2 × 2)

= 128

(c) 9, 24, 36 L.C.M. is = (3 × 2 × 2 × 3 × 2)

= 72

(d) 6, 8, 45 L.C.M. is = (3 × 2 × 4 × 15)

= 360

(e) 15, 35, 45 L.C.M. is = 5 × 3 × 7 × 3

= 315

(f) 64, 72, 96, 192 L.C.M. is 576

(4) Find the LCM of the following by division method

Solutions :

(a) 48, 72, 80

L.C.M. = (3 × 2 × 2 × 2 × 2 × 3 × 5)

= 720

(b) 40, 48, 90 L.C.M. = 720

(c) 32, 48, 60, 72 L.C.M. = (2 × 2 × 2 × 2 × 3 × 2 × 5 × 3)

= 1440

(d) 32, 56, 80, 154 L.C.M. = (2 × 2 × 7 × 2 × 2 × 2 × 5 × 11)

= 12320

(e) 224, 280, 236 L.C.M. = 66080

Question no – (5)

Solution :

Other number,

= 131 × 8253/917

= 1179

Question no – (6)

Solution :

Other number,

= 23 × 1449/161

= 207

Question no – (7)

Solution :

L.C.M. is = 2160/12

= 180

Question no – (8)

Solution :

No, the product of three no is not equal to the product of their HCF and LCM.

Question no – (9)

Solution :

HCF is = 5

LCM is = 360

Multiple of 40 = 40, 80, 120, 160, 200, 240, 280, 320, 360,…. minutes

Multiple of 45 = 45, 90, 135, 180, 225, 270, 315, 360 minutes

L.C.M. = 360 minutes

= 6 hours

(10.15 AM + 6hour) = 4.15pm

Question no – (10)

Solution : L.C.M. = 120 second = 2 minute

In 30 minutes = 30/2

= 15 times

Question no – (11)

Solution :

Ajay save 9.25 Rs daily.

So, for exact no of Rs.

= (4 × 9)

= 36 Rs

and (4 × 0.25)

= 1 Rs

(36 + 1)

= 37 Rs in 4 days.

Next Chapter Solution :

Updated: June 23, 2023 — 5:53 am