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Brilliant’s Composite Mathematics Class 6 Solutions Chapter 4 Factors and Multiples
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Brilliant’s Composite Mathematics Class 6 Math Book, Chapter 4, Factors and Multiples. Here students can easily find step by step solutions of all the problems for Factors and Multiples, Exercise 4.1, 4.2, 4.3, 4.4 and 4.5 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Factors and Multiples Exercise 4.1 Solution
Question no – (1)
Solution :
(i) All the possible factors of 36
= 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6
∴ Factor are = 1, 2, 3, 4, 6, 9, 12, 18 26
(ii) All the possible factors of 25
= 25 × 1 = 5 × 5
∴ Factors are = 1, 5, 25
(iii) All the possible factors of 50
= 50 = 1 × 50 = 2 × 25 = 5 × 10
∴ Factors are = 1, 2, 5, 10, 25, 50
(iv) All the possible factors of 60
= 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10
∴ Factors are = 1, 2, 3, 4, 5, 6, 10, 15, 20, 30, 60
Question no – (2)
Solution :
(i) The first five multiples of 14 are 14, 28, 42, 56, 70.
(ii) The first five multiples of 22 are 22, 66, 44, 88, 110
(iii) The first five multiples of 31 are 31, 62, 93, 124, 155
(iv) The first five multiples of 38 are 38, 76, 114, 152, 190
(v) The first five multiples of 45 are 45, 90, 135, 225
(vi) The first five multiples of 62 are 62, 124, 186, 248, 310
Question no – (3)
Solution :
(i) 1625
= 25 is a factor of 1625
(ii) 18220
= 25 is not a factor of 18220
(iii) 167225
= 25 is a factor of 167225
(iv) 186550
= 25 is a factor of 186550
Question no – (4)
Solution :
(i) All prime numbers between 1 and 40 are 2, 3, 5, 7, 11, 13, 17, 19, 123, 29, 31, 37
(ii) All prime numbers between 50 and 100 are 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
(iii) All prime numbers between 87 and 100 are 89 and 97
(iv) All prime numbers between 55 and 75 are 59, 61, 67, 71, 73
Question no – (5)
Solution :
One even number is prime 2
Question no – (6)
Solution :
Prime Number :
A number which have exactly two factors is called a prime number.
Examples : 1, 2, 3, 5, 7 are prime numbers.
Question no – (7)
Solution :
Prime numbers are – (ii) 23 (iii) 31 (iv) 43
Composite. umber are – (i) 21 (v) 87 (vi) 105 (vii) 187
Question no – (8)
Solution :
Composite number :
A number which have more than two factors is called a composite number.
Yes, a composite number be odd.
Smallest odd composite number is 9
Question no – (9)
Solution :
The smallest even prime number is 2
Question no – (10)
Solution :
Two numbers are and said to be co-prime numbers if they do not have common factor other then 1.
Examples : (2, 3) (3, 4), (4, 5) (5, 6)
The pair of numbers 9, 10 are co-prime numbers while 9 is not a prime number.
Question no – (11)
Solution :
Twin Primes :
Two prime number which by 2 are called twin primes.
Question no – (12)
Solution :
(i) 40 as a sum of two odd primes,
= 3 + 37
(ii) 80 as a sum of two odd primes,
= 7 + 73
(iii) 100 as a sum of two odd primes,
= 100 = 3 + 97
(iv) 96 as a sum of two odd primes,
= 7 + 89
(v) 56 as a sum of two odd primes,
= 3 + 53
Question no – (13)
Solution :
(i) 35 as sum of three odd prime numbers,
= 23 + 7 + 5
(ii) 49 as sum of three odd prime numbers,
= 3 + 5 + 41
(iii) 31 as sum of three odd prime numbers,
= 19 + 5 + 7
(iv) 63 as sum of three odd prime numbers,
= 7 + 13 + 43
Question no – (14)
Solution :
list of five consecutive numbers, none of which is prime – 61, 62, 63, 64, 65,
– 61 is a prime number.
Question no – (15)
Solution :
(i) No, natural number having no factor at all.
(ii) Only One (1) natural numbers are there having exactly one factor.
(iii) The natural numbers which are having exactly one factor are = 4, 9, 25, 49
Question no – (16)
Solution :
(i) There is no natural number having no factor at all.
= Given statement is False
(ii) The sum of two prime numbers is not always a prime number.
= Given statement is True
(iii) If two numbers are co-primes, at least one of them must be a prime.
= Given statement is False
(iv) 6 is a perfect number.
= Given statement is True
(v) Numbers having more than two factors are known as composite numbers.
= Given statement is True
(vi) 1 is neither a prime nor a composite number.
= Given statement is True
(vii) 2 is not a the lowest prime number.
= Given statement is False
(viii) The only even prime number is 2.
= Given statement is True
Factors and Multiples Exercise 4.2 Solution
Question no – (1)
Solution :
As we know that,
If the last digit of the number is 0, 2, 4, 6, and 8 then the number is divisible by 2
A number is divisible by 4 if the last two digit of the number is divisible by 4 or the whole number is divisible by 4
A number is divisible by 8 if its last three digits are divisible by 8 then the number is divisible by 8
Hence,
(i), (ii) (iii), (iv) (iv), (iv) (vii) (viii) (ix), (x) all are divisible by 2
(iii), (iv) (v), (vi), (ix) are divisible by 4
(iv), (ix) divisible by 8
Question no – (2)
Solution :
As we know that,
A number is divisible by 3 if the sum of digits is divisible by 3
A number is divisible by 6 if the number is divisible by both the number 2 and 3
A number is divisible by 9 if the sum of all digits is divisible by 9
Hence,
(i), (ii) (iii), (iv), (v), (vi) (vii) (viii), are divisible by 3
(i), (ii) (iii), (v), (vi) are divisible by 6
(i), (ii) (iv) (v) are divisible 9
Question no – (3)
Solution :
As we know that,
A number is divisible by 5 if the last digit of the given number is 0 or 5
A number is divisible by 10 if its unit digit place is ‘0’
Therefore,
(i), (ii), (iii), (iv), (v), (vi) are divisible by 5
(i), (iii), (vi) are divisible by 10
Question no – (4)
Solution :
As we know that,
A number is divisible by 11 if the sum of odd place and the sum of even place. And then their difference is divisible by 11.
Therefore,
(i), (ii), (iii) (iv), (v) are divisible by 11.
Factors and Multiples Exercise 4.3 Solution :
Question no – (1)
Solution :
(i) 36
Now,
∴ 36 = 2 × 2 × 3 × 3
(ii) 48
Now,
∴ 48 = 2 × 2 × 2 × 2 × 3
(iii) 60
Now,
∴ 60 = 2 × 2 × 3 × 5
(iv) 84
Now,
∴ 84 = 2 × 2 × 3 × 7
(v) 132
Now,
∴ 132 = 2 × 2 × 3 × 11
(vi) 225
Now,
∴ Prime factorisation of 225 = 5 × 5 × 3 × 3
(vii) 441
Now,
∴ Prime factorisation of 441 = 3 × 3 × 7 × 7
(viii) 525
Now,
∴ Prime factorisation of 525 = 5 × 5 × 3 × 7
(ix) 900
Now,
∴ Prime factorisation of 900 = 3 × 2 × 2 × 2 × 5 × 5
(x) 1350
Now,
∴ Prime factorisation of 1350 = 3 × 3 × 3 × 2 × 5 × 5
Question no – (2)
Solution :
As we know,
Largest 4 digit number is 9999
Now,
∴ 9999 = 3 × 3 × 11 × 101
Question no – (3)
Solution :
As we know,
The Smallest 4 digit number is 1000
Now,
∴ 1000 = 2 × 2 × 2 × 5 × 5 × 5
Question no – (4)
Solution :
7 ; 13, 19
The difference between any two consecutive prime number is 6.
Question no – (5)
Solution :
1 and the number itself are not included in the prime factorisation of a composite number.
Factors and Multiples Exercise 4.4 Solution :
Question no – (1)
Solution :
(i) 12 and 16
∴ 12 = 2 × 2 × 3
∴ 16 = 2 × 2 × 2 × 2
Therefore, H.C.F = 2 × 2 = 4
Hence, the HCF of 12 and 16 is 4
(ii) 36 and 60
∴ 36 = 4 × 9 = 2 × 2 × 3 × 3
∴ 60 = 2 × 2 × 3 × 5
∴ H.C.F = 2 × 2 × 3 = 12
Hence, the HCF of 36 and 60 is 12
(iii) 16 and 24
∴ 16 = 2 × 2 × 2 × 2
∴ 24 = 2 × 2 × 2 × 3
∴ H.C.F = 2 × 2 × 2 = 8
Hence, the HCF of 16 and 24 is 8
(iv) 47 and 61
∴ 47 = 47 × 1
∴ 61 = 61 × 1
∴ H.C.F = 1
Thus, the HCF of 47 and 61 is 1
(v) 40 and 48
∴ 40 = 2 × 2 × 2 × 5
∴ 48 = 2 × 2 × 2 × 2 × 3
∴ H.C.F = 2 × 2 × 2 = 8
Thus, the HCF of 40 and 48 is 8
(viii) 60, 90 and 120
∴ 60 = 2 × 2 × 3 × 5
∴ 90 = 3 × 3 × 0 × 2 × 5
∴ 120 = 2 × 2 × 2 × 3 × 5
∴ H.C.F = 2 × 3 × 5 = 30
Hence, the HCF of 60, 90 and 120 is 30
(ix) 106, 159, 26
Now,
∴ 106 = 2 × 53
∴ 159 = 3 × 53
∴ 265 = 5 × 53
∴ H.C.F = 53
Thus, the HCF of 106, 159, 26 is 53
(x) Given numbers, 101, 573, 1079
Now,
= 101 = 1 × 10
= 573 = 3 × 191 × 1
= 1079 = 1 ×1079
Hence, the HCF of 101, 573, 1079 is 1
Question no – (3)
Solution :
Given numbers are = 84 and 90.
Now,
∴ The greatest number will be 6
Thus, the largest number will be 138
Question no – (4)
Solution :
H.C.F of any two consecutive number is 1
Question no – (5)
Solution :
(i) Given, 576/252
= 16/7
∴ [H.C.F = 36]
(ii) Given, 522/1276
= 9/22
[H.C.F = 58]
(iii) Given, 1379/357
= 197/57
∴ [H.C.F = 7]
(iv) Given, 4130/7021
= 590/1003
∴ [H.C.F = 7]
Question no – (7)
Solution :
In the question,
Two milk containers contain = 300 litres and 450 litres
Now,
= 300 = 3 × 2 × 2 × 5 × 5
= 450 = 3 × 3 × 5 × 2 × 5
∴ Maximum capacity
= 2 × 5 × 5 × 3
= 150 litres
Hence, the maximum capacity of container will be 150 litres
Question no – (8)
Solution :
We should find H.C.F of,
= (152 – 2) = 150,
= (144 – 4) = 140,
= (211 – 1) = 210
Now,
∴ The largest number will be 14
Question no – (9)
Solution :
According to the question,
= 9 m 75 cm = 975 cm
= 8 m 25 cm = 825 cm
= 5 m = 25 cm = 252 cm
Now,
∴ The largest tape which can measure the these dimensions of the room exactly = 75 cm.
Question no – (10)
Solution :
We find the H.C.F of,
= 245 – 5 = 240 ,
= 1029 – 5 = 1025
Now,
∴ HCF of 240, 1024 is 16
Question no – (11)
Solution :
First we find the HCF of –
= (398 – 7) = 391,
= (542 – 15) = 527,
= (436 + 1) = 425
Therefore,
∴ H.C.F will be = 17
Question no – (12)
Solution :
First we find the HCF of,
= (533 – 13) = 520,
= (1445 – 15) = 1430
Therefore,
∴ H.C.F = 130
Question no – (13)
Solution :
First we find the H.C.F of ,
= (365 – 8) = 357
= (512 – 8) = 540
= (323 – 8) = 35
Now,
Therefore, the H.C.F will be = 21
Factors and Multiples Exercise 4.5 Solution :
Question no – (1)
Solution :
(iii) 40, 36, 126
Now,
∴ LCM = 2 ×4 × 5 × 7 × 9
= 2520
So, the LCM of 40, 36 and 126 is 2520.
(vi) 108, 135, 162
Now,
= LCM = 2 × 2 × 3 × 3 × 5 × 9
= 1620
Thus, the LCM of 108, 135, and 162 is 1620
(vii) 112, 168, 266
Now,
= LCM = 2 × 2 ×3 × 4 × 7 × 19
= 6384
Thus, the LCM of 112, 168, 266 is 6384
(viii) 180, 384, 144
Now, LCM,
= LCM = 4 × 4 × 5 × 9 × 24 = 5760
Thus, the LCM of 180, 384 and 144 is 5760
(ix) 24, 36, 60, 84, 135
Now the LCM,
∴ LCM = 2 × 3 × 3 × 3 × 4 × 5 × 7 = 7560
Thus, the LCM of 24, 36, 60, 84 and 135 is 7560
Question no – (2)
Solution :
(i) 35, 40;
= 35 = 5 × 7
40 = 5 × 2 × 2 × 2
HCF = 5
LCM = 5 × 2 × 2 × 2 × 7 = 280
Product of numbers = 35 × 40
= 1400
LCM × HCF = 280 × 5 = 1400
(ii) 24, 40;
= 24 = 4 × 2 × 3
40 = 4 × 5 × 2
HCF = 4 × 2 = 8
LCM = 4 × 2 × 3 × 5 = 120
24 × 40 = 960
120 × 8 = 960
(iii) 50, 60;
= 50 = 5 × 5 × 2
60 = 6 × 5 × 2
HCF = 2 × 50 = 10
LCM = 5 × 2 × 5 × 6 = 300
50 × 60 = 3000
and 300 × 10 = 3000
(iv) 45, 105;
= 45 = 5 × 3 × 3
105 = 5 × 3 × 7
HCF = 5 × 3 = 15
LCM = 5 × 3 × 3 × 7 = 315
45 × 105 = 4725
1315 × 15 = 4725
(v) 117, 221.
= 117 = 3 × 3 × 13
221 = 13 × 17
LCM = 3 × 3 × 13 × 17 = 1989
117 × 221 = 25857
and 13 × 1989 = 25857
Question no – (3)
Solution :
(i) 4/55, 9/110, 13/165
Now the LCM,
∴ LCM = 11 × 5 × 2 × 3 = 330
= 4 × 6/55 × 6, 9 × 3/110 × 3, 13 × 2/165 × 2
= 24/330, 27/330, 26/330
(ii) 4/15, 7/30, 8/45
Now the LCM,
∴ LCM = 15 × 2 × 3 = 90
= 4 × 6/15 × 6, 7 × 3/30 × 3, 8 × 2/45 × 2
= 24/90, 21/90, 16/90
Question no – (4)
Solution :
In the given question,
HCF and LCM of two numbers are 25 and 450
One of the numbers is = 75
Other number = ?
∴ Other number = LCM × HCF/one number
= 450 × 25/75
= 150
Thus, the other number will be 150
Question no – (5)
Solution :
According to the question,
HCF of two numbers is = 24
Product is = 7200,
LCM = ?
∴ LCM = product of two number/HCF
= 7200/24
= 300
Hence, their LCM will be 300
Question no – (6)
Solution :
Given in the question,
LCM of two numbers is 720
Product of the numbers is 2160
HCF = ?
∴ HCF = Product of two number/LCM
= 2160/720
= 3
Hence, their HCF will be 3
Question no – (7)
Solution :
In the given question,
Smallest number which when divided by = 25, 40 and 60
leaves remainder = 7 in each case
Now, the LCM,
∴ LCM = 5 × 4 × 5 × 2 × 3 = 600
∴ Required number,
= 600 + 7
= 607
Hence, the smallest number will be 607.
Question no – (8)
Solution :
According to the question,
D to A = 40
A to D = 45 min
Now,
∴ Time take = 5 × 8 × 9 in
= 360 min = 6 hr
∴ (10 : 15 + 6) = 4 : 15 p.m. two buses pass the same bridge.
Question no – (9)
Solution :
= 204/14
= 102/7
∴ HCF of any number must divided their LCM. Here 14 does not divided 204.
Question no – (10)
Solution :
In the question we get,
A boy saves daily = Rs 4.65
∴ 4.65 = 465/100
in 2 days = 4.65 × 2 = 9.30
in 20 days = 4.65 × 20/100 = 93
Thus, in 20 days he will able to save an exact number of rupees.
Question no – (11)
Solution :
∴ LCM = 3 × 3 × 4 × 5 = 180 min
∴ 180 min = 3 hour after they next toll together.
Thus, they will next toll together in 3 hour.
Question no – (12)
Solution :
73 = 73 × 1
79 = 79 × 1
∴ The HCF = 1
∴ The LCM = 73 × 79 = 9761
Thus, their HCF is 1 and LCM is 9761
Question no – (13)
Solution :
In the question we get,
The HCF and LCM of two numbers are 19 and 114.
One of the number is = 38,
Other number = ?
∴ Other number = LCM × HCF/one number
= 19 × 114/38
= 57
Hence, the other number will be 57
Question no – (14)
Solution :
In the question we get,
There are four bells which toll at intervals = 3, 7, 12 and 14 second
Now the LCM,
∴ LCM = 2 × 2 × 3 × 7 = 84 = 1 min 24 sec
∴ The time they will strike together next,
= 12 : 01 : 24 in 14 min
= 14 × 60/84
= 10 times
So, they will next toll together in 1 min 24 sec and 10 times.
Question no – (15)
Solution :
First, LCM of 40 and 30,
∴ LCM = 5 × 2 × 4 × 3
= 120 sec
= 2 min
Hence, they will change in 2 minutes.
Question no – (16)
Solution :
First the LCM,
∴ LCM = 5 × 2 × 8 × 9 × 17 = 12240
= 122 m 40 on
∴ At 122 m 40 cm from the starting point they again step off together.
Question no – (17)
Solution :
First, the LCM of 220 and 300,
∴ LCM = 10× 2 × 11 × 15 = 3300
Thus, 3300 metre far from the road.
Question no – (18)
Solution :
First, we find LCM of 27, 35, 25 and 21
∴ L.C.M = 3 × 5 × 5 × 7 × 9 = 4725
∴ The smallest number,
= 4725 – 3
= 4722
Question no – (19)
Solution :
First we find the LCM of = 48, 60 and 64.
∴ L.C.M = 2 × 2 × 2 × 2 × 3 × 4 × 5
= 960
Hence, the required greatest number will be 960.
Question no – (20)
Solution :
From the question we get,
= 7m 80 cm = 780 cm
= 6 m 60 cm = 660
Now, the LCM of 780 and 660
∴ LCM = 6 × 10 × 13 × 11 = 8580
Hence, Number of such tile will be 8580
Question no – (21)
Solution :
From the question we get,
= 6 m 50 cm = 650 cm
= 3 m 50 cm = 350 cm
= 3 m 50 cm = 300 cm
Now, The LCM of 650, 350, 300,
∴ L.C.M = 10 × 13 × 7 × 6 × 5 = 27300
Hence, the largest tape = 27300 cm = 273 m
Question no – (22)
Solution :
First we find the LCM of 2, 3, 4, 5, 6 and 7
∴ L.C.M = 2 × 2 × 3 × 5 × 7 = 920
We know, Greatest 4 digit number = 9999
∴ Required number,
= 9999 – 339
= 9660
Thus, the greatest number will be 9660
Question no – (23)
Solution :
L.C.M. of 777, 819 and 4329
∴ L.C.M = 3 × 3 × 7 × 13 × 37 = 30303
Thus, the LCM of 777, 819 and 4329 is 30303
Question no – (24)
Solution :
GCM + LCM = 375
GCM + 24 × GCM = 375
= 25 GCM = 375
∴ GCM = 375/25 = 15
∴ LCM = 15 × 24 = 3160
∴ Second number,
= LCM × GCM/First number
= 360 × 15/45
= 120
Hence, the other number will be 120
Question no – (25)
Solution :
Here we have to find the least square number,
Which is exactly divisible by 4 times 5, 6, 7, 8, 9, 10
= 20, 24, 28, 32, 36, 40
Now the LCM of 20, 24, 28, 32, 36, 40
∴ L.C.M = 2 × 4 × 4 × 3 × 3 5 × 7
= 22 × 22 × 32 × 70
= 10080
∴ The least square number,
= 10080 × 70
= 705600
Question no – (26)
Solution :
First LCM of 2, 3, 4, 5 and 6
∴ L.C.M = 2 × 3 2 × 5 = 60
∴ Least number of 4 digit = 1000
∴ The least number of 4 digit divisible,
= 100 + (60 – 40)
= 100 + 20
= 1020
∴ Required number,
= 1020 + 1
= 1021
Question no – (27)
Solution :
First we need to find the LCM of 12, 15, 20 or 54
∴ L.C.M = 2 × 2 × 3 × 5 × 9 = 540
∴ Required number,
= 540 + 4
= 544
Thus, the required number will be 544
Next Chapter Solution :
