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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 :
👉 Chapter 5 👈
