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Frank Learning Maths Class 6 Solutions Chapter 3 Factors and Multiples
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Frank Learning Maths Class 6 Book, Chapter 3 Factors and Multiples. Here students can easily find step by step solutions of all the problems for Factors and Multiples. Here students will find solutions for Exercise 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, and 3.7 and also solutions for chapter Check-up question. Exercise wise proper solutions. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.
Factors and Multiples Exercise 3.1 Solution
Question no – (1)
Solution :
(a) Factors of 12 = 1, 2, 3, 4, 6, 12
(b) Factors of 20 = 1, 2, 4, 5, 10, 20
(c) Factors of 36 = 1, 2, 3, 4, 6, 9, 19
(d) Factors of 68 = 1, 2, 4, 17, 34, 68
Question no – (2)
Solution :
(a) 2, 4, 6, 8 10, 12, 14, 16
(b) 6, 12, 18, 24, 30, 36, 42, 48
(c) 5, 10, 15, 20, 25, 30, 35, 40
(d) 8, 24, 32, 40, 48, 56, 64
Question no – (3)
Solution :
(a) Multiplies of 7
= 7, 14, 21, 28, 35, 42, 49, 56, 63, 70
(b) Multiplies of 11
= 11, 22, 333, 44, 55, 66, 77, 88, 99, 110
(c) Multiplies of 15
= 15, 30, 45, 60, 75, 90, 105, 120, 135, 150
(d) Multiplies of 45
= 45, 90, 135, 180, 225, 270, 315, 360, 405, 450
Question no – (4)
Solution :
(a) Every number is factors
(b) The number of multiples of a given number is infinite.
(c) The first multiple of every number is the number itself.
(d) Every multiple of a number is greater than or equal to that number.
(e) If a number x is a factor of number y, then y ÷ x will have O as the remainder.
(f) A number which is twice the sum of all its factors is known as perfect number.
Question no – (6)
Solution :
(a) 108 ÷ 12 = 9
(b) 80 + 4 = 20
(c) 150 ÷ 15 = 10
(d) 117 ÷ 13 = 9
So, in every cases the first number is a multiple of the second number.
Question no – (7)
Solution :
(a) 121 ÷ 11 = 11
(b) 72 ÷ 6 = 12
(c) 13 ÷ 87 = 9
(d) 75 ÷ 5 = 15
So, here 87 is not a multiple of 13.
Question no – (8)
Solution :
(a) Multiples of 3 from 30 to 50 are – 30, 33, 36, 39, 42, 45, 48
(b) Factors of 75 are 1, 3, 19, 57
(c) Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
(d) 20 ÷ 2 = 10
So, there are 10 multiple of 2 upto 20.
Question no – (9)
Solution :
(a) Multiples of 14 are 14. 28, 42, 56, 70, 84, 98
(b) Multiples of 25 are 25, 50, 75
Factors and Multiples Exercise 3.2 Solution
Question no – (1)
Solution :
(a) 3, 17
(b) 13, 19
(c) 31, 47
(d) 43, 53
(d) 53 – 43 = 10,
∴ Here the difference is 10
Question no – (2)
Solution :
Given numbers,
(a) 7, 8, 9, 10, 11, 12, 13,
(b) 22, 23, 24, 25, 26, 27, 28
(c) 60, 61, 62, 63, 64, 65, 66
(d) 90, 91, 92, 93, 94, 95, 96
∴ (d) 90, 91, 92, 93, 94, 95, 96 are the 7 consecutive composite number such that are no prime numbers between them.
Question no – (3)
Solution :
(a) 1 and 30
= Prime numbers between 1 and 30 are 2, 3, 5, 7, 11, 17, 19, 23, 29
(b) 90 and 100
= Prime numbers between 90 and 100 are,
= 97
Question no – (7)
Solution :
The prime numbers adjacent to 50 are = 47, 53,
Their sum = 47, + 53
= 100
Question no – (8)
Solution :
Five prime numbers between 1 and 100 have their unit digit as 9.
They are = 19,29,59,979,89
Question no – (9)
Solution :
Such pairs of prime numbers between 1 and 100 are
= (13, 31), (17, 71) 937, 73), (79, 97)
Question no – (10)
Solution :
Five pairs of prime numbers less than 25 whose sum is divisible by 5 are
= (2, 3) (3, 7) (2, 3)
Question no – (11)
Solution :
(a) Even prime number
= 2
(b) Smallest odd composite number
= 9
(c) Smallest prime number and
= 2
(d) Smallest composite number
= 4
Question no – (12)
Solution :
No, every odd number not a prime number.
Example : 9 is odd but this is not prime because 3 is a factor of it.
Factors and Multiples Exercise 3.3 Solution
Question no – (2)
Solution :
(a) 43 432
(b) 9 914 4
(c) 39 3910 0
Question no – (3)
Solution :
(a) 932___
= 9327
(b) 53, 6___4
= 53, 694
(c) 2___ 713
= 28713
Question no – (5)
Solution :
(a) 4 ___6
= 416
(b) 48 __0
= 4800
(c) 58, 8 __2
= 58,812
Question no – (6)
Solution :
(a) 561___
= 5616
(b) 36, 79____
= 36, 792
(c) 18, 90, 21___4
= 18, 90, 264
Question no – (7)
Solution :
(a) 2__, 581
= 22, 581
(b) 7__, 223
= 74, 223
(c) 53, 2__, 891
= 53, 28, 891
Question no – (8)
Solution :
(a) 87,____62
= 87, 362
(b) 5, 27,____19
= 5, 27, 219
(c) 72, 73, ____45
= 72, 73, 145
Question no – (9)
Solution :
(a) iI the tens digit is 8.
= 81
(b) When both the digits are even and their difference is 0.
= 66
(c) When the tens digit is half the ones.
= 36
(d) When the ones digit is 5.
= 15
(e) When the ones digit is odd and the sum of the two digits is 6.
= 51
Question no – (10)
Solution :
(a) The sum of the two digits is 35.
(b) The ones digit is greater than the tens digit 75.
Question no – (11)
Solution :
(a) divisible by 3 but not by 9
= 102
(b) divisible by 2 but not by 4.
= 102
(c) divisible by 3 but not by 6.
= 105
(d) divisible by 4 but not by 8.
= 100
Factors and Multiples Exercise 3.4 Solution
Question no – (1)
Solution :
(a) 40, 48
= 40 = 1 × 2 × 2 × 2 × 5
= 48 = 1 × 2 × 2 × 2 × 2 × 3
∴ The common factor 40 and 48 is 1 × 2 × 2 × 2 = 8
∴ The common factor 40 and 48 is 8
(b) 55. 65
= 55 = 1 × 5 × 11
= 65 = 1 × 5 × 13
∴ The common factor of 45 and 60 is 1 × 3 × 5 = 15
∴ The common factor of 45 and 60 is 15
(c) 45, 60
= 45 = 1 × 3 × 5 × 3
= 60 = 1 × 2 × 3 × 4 × 2 × 5
∴ The common factor of 56 and 100 is 1 × 2 × 2 × = 4
∴ The common factor of 56 and 100 is 4
(d) 56, 100
= 56 = 1 × 2 × 2 × 2 × 7
= 100 = 1 × 2 × 2 5 × 5
∴ The common factor of 2, 50, 100 is 1 × 2 = 2
∴ The common factor of 2, 50, 100 is 2
Question no – (2)
Solution :
(a) 2, 50, 100
= 2 = 1 × 2
= 50 = 1 × 2 × 5 × 5
= 100 = 1 × 2 × 5 × 5
∴ The common factor of 2, 50, 100
= 1 × 2
= 2
∴ The common factor of 2, 50, 100 is 2
(b) 8, 24, 40
= 8 = 1 × 2 × 2
= 24 = 1 × 2 × 2 × 2 3
40 = 1 × 2 × 2 × 2 5
∴ The common factor of 8, 24, 40
= 1 × 2 × 2 × 2
= 8
∴ The common factor of 8, 24, 40 is 8
(c) 3, 30, 60
= 3 = 1 × 3
= 30 = 1 × 2 × 3 × 5
= 60 = 1 × 2 × 2 × 3 × 5
∴ The common factor of 3, 30, 60
= 1 × 3
= 3
∴ The common factor of 3, 30, 60 is 3
(d) 9, 36, 60
= 9 = 1 × 3 × 3
= 36 = 1 × 2 × 2 × 3 × 3
= 60 = 1 × 2 × 2 × 3 × 5
∴ The common factor of 9, 36, 60
= 1 × 3
= 3
∴ The common factor of 9, 36, 60 is 3
Question no – (3)
Solution :
(a) 3 and 4
Multiples of 3 + 3, 6, 9, 12, 15, 18, 21, 24
Multiples of 4 = 4, 8, 12, 14, 20, 24
∴ The first four common multiples,
= 12, 24, 36, 48
Therefore, the first four common multiples of 3 and 4 are 12, 24, 36, 48
(b) 5 and 6
Multiples of 5 = 5, 10, 15, 20, 25, 30
Multiples of 6 = 6, 12, 18, 24, 30
∴ The first four common multiples,
= 30, 60, 90, 120
∴ The first four common multiples of 5 and 6 are 30, 60, 90, 120
(c) 6 and 8
Multiples of 6 = 6, 12, 18, 24, 30
Multiples of 8 = 8, 16, 24, 32, 40
∴ The first four common multiples,
= 24, 48, 72, 96
Hence, the first four common multiples of 6 and 8 are 24, 48, 72, 96
(d) 10 and 15
Multiples of 10 = 10, 20, 30, 40, 50, 60
Multiples 15 = 15, 30, 45, 60, 75, 90
∴ The first four common multiples,
= 30, 60, 90, 120
Therefore, the first four common multiples of 10 and 15 are 30, 60, 90, and 120
Question no – (4)
Solution :
The numbers less than 100 and which are common multiples of 4 and 5 are 20, 40, 60, 80
Factors and Multiples Exercise 3.5 Solution
Question no – (1)
Solution :
Given numbers are 60, 27, and 80
Factor tree for the given numbers
Question no – (2)
Solution :
The missing numbers are,
Question no – (3)
Solution :
(a) Given, 980
∴ 980 = 2 × 2 × 5 × 7 × 7
(b) Given, 252
∴ 252 = 2 × 2 × 3 × 3 × 7
(c) Given, 945
∴ 945 = 5 × 5 × 3 × 7
(d) Given, 1224
∴ 1224 = 2 × 2 × 2 × 3 × 3 × 17
For more better understanding,
Question no – (4)
Solution :
∴ 1729
= 7 × 13 × 19
Here the difference between two consecutive factor will be 6.
Question no – (5)
Solution :
As we know that,
The smallest 4-digit number is – 1000
∴ 1000 = 2 × 2 × 2 × 5 × 5 × 5
The greatest 4-digit number is -9999
∴ 9999 = 3 × 3 × 11 × 101
Question no – (6)
Solution :
(a) 160 = 2 × 2 × 8 × 5
Since 8 is not a prime, so as a prime factorization this is not correct.
(b) 76 = 2 × 2 × 19
This is a correct prime factorization.
(c) 150 = 2 × 3 × 5 × 5
This is a correct prime factorization.
(d) 220 = 2 × 2 × 5 × 11
This is a correct prime factorization.
Question no – (7)
Solution :
| Number | Divisibility by 3 | Divisibility by 5 | Divisibility by 15 | |
| (a) | 345 | √ | √ | √ |
| (b) | 14,040 | √ | √ | √ |
| (c) | 12,345 | √ | √ | √ |
| (d) | 5,43,210 | √ | √ | √ |
Question no – (9)
Solution :
We cannot say that.
Because, 24 is not the least common multiple of 6 and 4.
Question no – (10)
Solution :
(a) The number which is divisible by both 2 and 4 but not by 8 is 4.
(b) The number which is divisible by 4 and 8 but not by 32 is 16.
(c) The number which is divisible by 6 and 15 but not by 90 is 30.
Question no – (11)
Solution :
If a number is divisible by 5 and 12, then it will be divisible by 2, 3, 4, 6, 10, 15, 20, 30, 60.
Question no – (12)
Solution :
If a number is divisible by 18, it will be divisible by 2, 3, 6, 9.
Factors and Multiples Exercise 3.6 Solution
Question no – (1)
Solution :
(a) 36 and 252
∴ 36 = 2 × 2 × 3 × 3
252 = 2 × 2 × 3 × 3 × 7
∴ HCF is = 2 × 2 × 3 × 3
= 36
So, the HCF of 36 and 252 is 36
(b) 18 and 60
∴ 18 = 2 × 3 × 3
0 = 2 × 2 × 3 × 5
∴ HCF is 2 × 3 = 6
Thus, the HCF of 18 and 60 is 6
(c) 170 and 238
∴ 170 = 2 × 5 × 17
238 = 2 × 7 × 17
∴ HCF is 2 × 17 = 34
Hence, the HCF of 170 and 238 is 34
(e) 106, 159 and 371
∴ 106 = 2 × 53
159 = 3 × 53
371 = 7 × 53
∴ HCF is 53
Therefore, the HCF of 106, 159 and 371 is 53
(f) 90, 140 and 210
∴ 90 = 2 × 3 × 3 × 5
140 = 2 × 2 × 3 × 5
210 = 2 × 3 × 5 × 7
∴ HCF is 2 × 5 = 10
Therefore, the HCF of 90, 140 and 210 is 10
Question no – (2)
Solution :
(a) 30 and 75
∴ LCM = 3 × 5 × 2 × 5
= 150
Hence, the LCM of 30 and 75 is 150
(b) 42 and 77
∴ LCM = 7 × 6 × 11
= 462
So, the LCM of 42 and 77 is 462
(c) 24, 36 and 40
∴ LCM = 4 × 2 × 3 × 3 × 5
= 360
Thus, the LCM of 24, 36 and 40 is 360
(d) 28, 36, 45, 60
∴ LCM = 2 × 2 ×3 × 3 × 5 × 7
= 1260
Hence, the LCM of 28, 36, 45, 60 is 1260
(e) 144, 180 and 384
∴ LCM = 4 × 3 × 2 × 2 × 3 × 5 × 8
= 5760
Therefore, the LCM of 144, 180 and 384 is 5760
Question no – (3)
Solution :
First we have to find the LCM of 6, 9, and 12
∴ LCM = 2 × 3 × 3 × 2
= 36
Now, We have to find the smallest number which is divisible by 36
= 36 ÷ 100
= 28
∴ The required number is,
= (100 – 28) + 36
= 72 + 36
= 108
Thus, the smallest three-digit number is 108.
Question no – (4)
Solution :
First we have to find the LCM of 15, 24, 30
∴ LCM = 2 × 3 × 5 × 4
= 120
Now,
∴ The required number is,
= 999999 – 39
= 999960
Therefore, the required number is 999960.
Question no – (5)
Solution :
First we have to find the LCM of 16, 18, 24, 36
∴ LCM = 2 × 2 × 2 × 3 × 3 × 2
= 144
Now,
∴ The required number is,
= [(10000 – 64) + 144]
= 10080
Therefore, the required least number of five digits is 10080.
Question no – (6)
Solution :
First we have to find the LCM of 16, 28, 40, 77
∴ LCM = 2 × 2 × 2 × 7 × 2 × 5 × 11
= 6160
∴ The required number is,
= 6160 + 7
= 6167
Therefore, the required number is 6167
Question no – (7)
Solution :
First,
7 m = 700 cm,
3 m 85 cm = 385 cm,
12 m 95 cm = 1295 cm.
Now,
700 = 2 × 2 × 5 × 5 × 7
385 = 5 × 7 × 11
1295 = 5 × 7 × 37
∴ The length of the largest tape is,
= (5 × 7) cm
= 35 cm.
Therefore, the length of the largest tape will be 35 cm.
Question no – (8)
Solution :
= 3 × 3 × 4 × 5
= 180 sec
= 180/60 min
= 3 min
Therefore, after 3 minutes, the bells will toll together.
Question no – (9)
Solution :
= 2 × 2 × 3 × 3 × 4 × 3
= 432 m
Hence, After 432 m of distance, they will take steps together.
Question no – (10)
Solution :
∴ 2 × 2 × 2 × 3 × 3
= 72 sec
= 72 ÷ 60
= 1 min 12 sec
Thus, at 6 : 1 : 12 a.m. the lights will change again simultaneously.
Question no – (11)
Solution :
As per the question we know,
There are 527 pencils, 646 erasers and 748 sharpners
Now,
= 527 = 17 × 31
= 646 = 17 × 2 × 19
= 748 = 2 × 2 × 11 × 17
Therefore, in each packet, there can be 17 items.
Question no – (12)
Solution :
We have to find the HCF of
= (80 – 8), (118 – 10), (60 – 12)
i.e 72, 108, 48
= 72 = 2 × 2 × 2 × 3 × 3
= 108 = 2 ×2 × 3 × 3 × 3
= 48 = 2 × 2 × 2 × 2 × 3
∴ The required number is,
= 2 × 2 × 3
= 12
Therefore, the required greatest number is 12.
Factors and Multiples Exercise 3.7 Solution
Question no – (1)
Solution :
As we know that,
HCM × LCM = 1st number × 2nd number
or, 5 × 280 = 35 × 2nd number
2nd number = 5 × 240/35
2nd number = 40
Therefore, the other number is 40.
Question no – (2)
Solution :
As we know that,
HCF × LCM = 1st number × 2nd number
or, 16 × LCM = 3072
or, LCM = 3072/16 = 192
Therefore, their LCM is 192.
Question no – (3)
Solution :
As we know that,
HCM × LCM = 1st number × 2nd number
or, = 23 × 1449/161
= 207
Hence, the other number will be 207
Question no – (4)
Solution :
As we know that,
HCF = product of numbers/LCM
or, HCF = 2560/320
or, HCF = 8
Therefore, their HCF will be 8.
Question no – (5)
Solution :
No, because LCM of two numbers must be divisible by their HCF. But, 16 does not divide 130. Therefore two numbers cannot have 16 as their HCF and 130 as their LCM.
Question no – (6)
Solution :
Since, 18 divides 162, two numbers can have 18 as their HCF and 162 as their LCM.
Question no – (7)
Solution :
(a) Given Numbers 5 and 8
HCF = 1
LCM = 40
(b) Given Numbers 21 and 10
HCF = 1
LCM = 210
(c) Given Numbers 24 and 7
HCF = 1
LCM = 168
Question no – (8)
Solution :
(a) 15, 45
LCM is = 45
(b) 10, 30
LCM is = 30
(c) 12, 60
LCM is = 60
(d) 9, 72
LCM is = 72
Factors and Multiples Chapter Check-up Solution :
Question no – (1)
Solution :
(a) If a number is divisible by 3, it must be divisible by 9.
The statement is False.
(b) 6 is divisible by 3 but not by 9. Therefore its false.
If a number is divisible by 9, it must be divisible by 3.
This statement is True.
(c) A number is divisible by 18, if it is divisible by both 3 and 6
This statement is True.
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90
This statement is True.
(e) If two numbers are co primes, at least one of them must be prime
This statement is False.
Explanation :
9 and 10 are co-primes, but they both are composite.
(f) All numbers which are divisible by 4 must also be divisible by 8
This statement is False.
Explanation :
12 is divisible by 4 but not by 8
(g) All numbers which are divisible by 8 must also be divisible by 4 –
This statement is True.
(h) If a number exactly divides two numbers separately, it must exactly divide their sum
This statement is True.
Question no – (2)
Solution :
1 is neither prime nor composite.
Hence, only one number is there.
Question no – (3)
Solution :
The required number is = 1324.
Question no – (5)
Solution :
First, we have to find the LCM of 3, 6, 8
∴ LCM = 2 × 3 × 4
= 24
Now,
∴ The required number is,
= 999 – 15
= 984
Question no – (6)
Solution :
First,
= 2 × 2 × 2 × 3 ×2 × 3 ×7
= 1002
Now,
= 16 min 42 sec
Therefore, after 7 a.m. the traffic lights will simultaneously change at 7 : 16 : 20 am.
Question no – (7)
Solution :
First, we have to find the LCM,
∴ LCM = 2 × 3 × 5 × 3
= 90
∴ The required number is,
= 90 + 5
= 95
Hence, the lest number will be 95
Question no – (9)
Solution :
Lets take 1, 2, 3 then 1 × 2 × 3 = 6, divisible by 6
2, 3, 4 then 2 × 3 ×4 = 24 divisible by 6
3, 4, 5 then 3 ×4 ×5 = 60, divisible by 6
4, 5, 6 then 4 × 5 × 6 = 120, divisible by 6
So, in every example we can see that the given statement is True.
Question no – (11)
Solution :
From the given numbers the co-prime number are –
(a) 18 and 35
(b) 15 and 37
(e) 216 and 215
(f) 81 and 16
Question no – (12)
Solution :
= 20 = 2 × 2 × 5
= 28 = 2 × 2 × 7
= 36 = 2 × 2 × 3 × 3
∴ The greatest denomination of stamps will be,
= 2 × 2
= 4 rupees.
Question no – (14)
Solution :
Numbers |
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | |
| (a) | 7132 | ✔ | × | × | × | × | × | × | × | × | |
| (b) | 3009 | × | ✔ | × | × | × | × | × | × | × | × |
| (c) | 12330 | ✔ | ✔ | × | ✔ | ✔ | × | × | ✔ | ✔ | × |
| (d) | 76505 | × | × | × | ✔ | × | × | × | × | × | ✔ |
| (e) | 335786 | ✔ | × | × | × | × | × | × | × | × | ✔ |
Question no – (15)
Solution :
The required number is the HCF of (1280 – 6), (1371 – 6)
Or, 1274, 1365
∴ 1274 = 2 × 7 × 7 × 13
∴ 1365 = 3 × 5 × 7 × 13
∴ The required number is,
= 7 × 13
= 91
Question no – (16)
Solution :
(a) Sum of any two odd numbers is an even number.
(b) Sum of an odd number and even number is an odd number.
(c) Product of any odd number and even number an even number.
(d) Product of two odd number is always an odd number.
(e) Every whole number is greater than the 0
Previous Chapter Solution :
