Class 6: Maths Chapter 8 solutions. Complete Class 6 Maths Chapter 8 Notes.
Contents
Selina Class 6 ICSE Solutions Mathematics : Chapter 8- HCF And LCM
Selina 6th Maths Chapter 8, Class 6 Maths Chapter 8 solutions
Exercise 8(A) page No: 63
1. Write all the factors of:
(i) 15
(ii) 55
(iii) 48
(iv) 36
(v) 84
Solution:
(i) The factors of 15 are 1, 3, 5 and 15
∴ F(15) = 1, 3, 5 and 15
(ii) The factors of 55 are 1, 5, 11 and 55
∴F(55) = 1, 5, 11 and 55
(iii) The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48
∴F (48) = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48
(iv) The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36
∴F (36) = 1, 2, 3, 4, 6, 9, 12, 18 and 36
(v) The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84
∴ F (84) = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84
2. Write all prime numbers:
(i) less than 25
(ii) between 15 and 35
(iii) between 8 and 76
Solutions:
(i) 2, 3, 5, 7, 11, 13, 17, 19 and 23 are the prime numbers less than 25
(ii) 17, 19, 23, 29 and 31 are the prime numbers between 15 and 35
(iii) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71 and 73 are the prime numbers between 8 and 76
3. Write the prime numbers from:
(i) 5 to 45
(ii) 2 to 32
(iii) 8 to 48
(iv) 9 to 59
Solutions:
(i) The prime numbers from 5 to 45 are 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41 and 43
(ii) The prime numbers from 2 to 32 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31
(iii) The prime numbers from 8 to 48 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47
(iv) The prime numbers from 9 to 59 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53 and 59
4. Write the prime factors of:
(i) 16
(ii) 27
(iii) 35
(iv) 49
Solution:
(i)
∴Prime factor of 16 is 2
(ii)
∴Prime factor of 27 is 3
(iii)
∴Prime factors of 35 are 5 and 7
(iv)
∴Prime factor of 49 is 7
5. If Pn means prime factors of n, find:
(i) P6
(ii) P24
(iii) P50
(iv) P42
Solution:
(i)
∴The prime factors of 6 are 2 and 3
(ii)
∴The prime factors of 24 are 2 and 3
(iii)
∴The prime factors of 50 are 2 and 5
(iv)
∴The prime factors of 42 are 2, 3 and 7
Exercise 8(B) Page no: 66
1. Using the common factor method, find the H.C.F. of:
(i) 16 and 35
(ii) 25 and 20
(iii) 27 and 75
(iv) 8, 12 and 18
(v) 24, 36, 45 and 60
Solution:
(i) Common factors of 16 and 35 are as follows:
F (16) = 1, 2, 4, 8, 16
F (35) = 1, 5, 7, 35
The common factors between 16 and 35 = 1
∴The H.C.F. of 16 and 35 = 1
(ii) Common factors of 25 and 20 are as follows:
F (25) = 1, 5, 25
F (20) = 1, 2, 4, 5, 10, 20
The common factors between 25 and 20 = 1, 5
∴The H.C.F. of 25 and 20 = 5
(iii) Common factors between 27 and 75 are as follows:
F (27) = 1, 3, 9, 27
F (75) = 1, 3, 5, 15, 25, 75
The common factors between 27 and 75 = 1, 3
∴The H.C.F. of 27 and 75 = 3
(iv) Common factors between 8, 12 and 18 are as follows:
F (8) = 1, 2, 4, 8
F (12) = 1, 2, 3, 4, 6, 12
F (18) = 1, 2, 3, 6, 9, 18
Common factors between 8, 12 and 18 = 1, 2
∴The H.C.F. of 8, 12 and 18 = 2
(v) Common factors between 24, 36, 45 and 60 are as follows:
F (24) = 1, 2, 3, 4, 6, 8, 12, 24
F (36) = 1, 2, 3, 4, 6, 12, 18, 36
F (45) = 1, 3, 5, 9, 15, 45
F (60) = 1, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common factors between 24, 36, 45, and 60 = 1, 3
∴The H.C.F. of 24, 36, 45 and 60 = 3
2. Using the prime factor method, find the H.C.F. of:
(i) 5 and 8
(ii) 24 and 49
(iii) 40, 60 and 80
(iv) 48, 84 and 88
(v) 12, 16 and 28
Solution:
(i) The prime factors of 5 and 8 are as follows:
P5 = 5
P8 = 2 × 2 × 2
No common prime factors between 5 and 8
Hence, H.C.F. of 5 and 8 = 1
(ii) The prime factors of 24 and 49 are as follows:
P24 = 2 × 2 × 2 × 3
P49 = 7 × 7
No common prime factors between 24 and 49
Hence, H.C.F. of 24 and 49 = 1
(iii) The prime factors of 40, 60 and 80 are as follows:
P40 = 2 × 2 × 2 × 5
P60 = 2 × 2 × 3 × 5
P80 = 2 × 2 × 2 × 2 × 5
Common prime factors between 40, 60 and 80 = 2 × 2 × 5
Hence, H.C.F. of 40, 60 and 80 = 20
(iv) The prime factors of 48, 84 and 88 are as follows:
P48 = 2 × 2 × 2 × 2 × 3
P84 = 2 × 2 × 3 × 7
P88 = 2 × 2 × 2 × 11
Common prime factors between 48, 84 and 88 = 2 × 2
Hence, H.C.F. of 48, 84 and 88 = 2 × 2 = 4
(v) The prime factors of 12, 16 and 28 are as follows:
P12 = 2 × 2 × 3
P16 = 2 × 2 × 2 × 2
P28 = 2 × 2 × 7
Common prime factors between 12, 16 and 28 = 2 × 2
Hence, H.C.F. of 12, 16 and 28 = 2 × 2 = 4
3. Using the division method, find the H.C.F. of the following:
(i) 16 and 24
(ii) 18 and 30
(iii) 7, 14 and 24
(iv) 70, 80, 120 and 150
(v) 32, 56 and 46
Solution:
(i) 16 and 24
Using division method, we get
Here, the last division is 8
Hence, H.C.F. of 16 and 24 = 8
(ii) 18 and 30
Using division method, we get
Here, last division is 6
Hence, H.C.F. of 18 and 30 = 6
(iii) 7, 14 and 24
Using division method, we get
Here, last division is 1
Hence, H.C.F. of 7, 14 and 24 = 1
(iv) 70, 80, 120 and 150
Using division method, we get
Here, last division is 10
Hence, H.C.F. of 70, 80, 120 and 150 = 10
(v) 32, 56 and 46
Using division method, we get
Here, last division is 2
Hence, H.C.F. of 32, 56 and 46 = 2
4. Use a method of your own choice to find the H.C.F. of:
(i) 45, 75 and 135
(ii) 48, 36 and 96
(iii) 66, 33 and 132
(iv) 24, 36, 60 and 132
(v) 30, 60, 90, and 105
Solution:
(i) 45, 75 and 135
The prime factors of 45, 75 and 135 are as follows:
P45 = 3 × 3 × 5
P75 = 3 × 5 × 5
P135 = 3 × 3 × 3 × 5
The common factors of 45, 75 and 135 = 3 × 5
∴H.C.F. of 45, 75 and 135 = 15
(ii) 48, 36 and 96
The prime factors of 48, 36 and 96 are as follows:
P48 = 2 × 2 × 2 × 2 × 3
P36 = 2 × 2 × 3 × 3
P96 = 2 × 2 × 2 × 2 × 2 × 3
The common factors of 48, 36 and 96 = 2 × 2 × 3
∴H.C.F. of 48, 36 and 96 = 12
(iii) 66, 33 and 132
The prime factors of 66, 33 and 132 are as follows:
P66 = 2 × 3 × 11
P33 = 3 × 11
P132 = 2 × 2 × 3 × 11
The common factors of 66, 33 and 132 = 3 × 11
∴H.C.F. of 66, 33 and 132 = 33
(iv) 24, 36, 60 and 132
The prime factors of 24, 36, 60 and 132 are as follows:
P24 = 2 × 2 × 2 × 3
P36 = 2 × 2 × 3 × 3
P60 = 2 × 2 × 3 × 5
P132 = 2 × 2 × 3 × 11
The common factors of 24, 36, 60 and 132 = 2 × 2 × 3
∴H.C.F. of 24, 36, 60 and 132 = 12
(v) 30, 60, 90, and 105
The prime factors of 30, 60, 90 and 105 are as follows:
P30 = 2 × 3 × 5
P60 = 2 × 2 × 3 × 5
P90 = 2 × 3 × 3 × 5
P105 = 3 × 5 × 7
The common factors of 30, 60, 90 and 105 = 3 × 5
∴H.C.F. of 30, 60, 90 and 105 = 15
5. Find the greatest number that divides each of 180, 225 and 315 completely.
Solution:
The greatest number that divides each of 180, 225 and 315 will be the H.C.F. of 180, 225 and 315
Using division method, the H.C.F of 180, 225 and 315 are shown below
Since last division is 45
∴H.C.F. of 180, 225 and 315 = 45
6. Show that 45 and 56 are co-prime numbers.
Solution:
The H.C.F. of two co-prime numbers is always 1
Using division method, the H.C.F. of 45 and 56 are shown below
Here, last division is 1
H.C.F. of 45 and 56 = 1
∴45 and 56 are co-prime numbers
7. Out of 15, 16, 21 and 28, find out all the pairs of co-prime numbers.
Solution:
15 and 16, 15 and 21, 15 and 28, 16 and 21, 16 and 28 and 21 and 28 will be the possible pairs
We know that the H.C.F. of two co-prime numbers is always 1
Using division method, the H.C.F. of pair 15 and 16 is given below
Hence, H.C.F. of 15 and 16 = 1
Using division method, the H.C.F. of pair 15 and 21 is given below
Hence, H.C.F. of 15 and 21= 3
Using division method, the H.C.F. of pair 15 and 28 is given below
Hence, H.C.F. of 15 and 28 = 1
Using division method, the H.C.F. of pair 16 and 21 is given below
Hence, H.C.F. of 16 and 21 = 1
Using division method, the H.C.F. of 16 and 28 is given below
Hence, H.C.F. of 16 and 28 = 4
Using division method, the H.C.F. of 21 and 28 is given below
Hence, H.C.F. of 21 and 28 = 7
∴ The pairs 15 and 16, 15 and 28 and 16 and 21 are co-prime numbers
8. Find the greatest number that will divide 93, 111 and 129, leaving remainder 3 in each case.
Solution:
First, decrease the leaving remainder 3 from numbers 93, 111 and 129 to find the required number
93 – 3 = 90
111 – 3 = 108
129 – 3 = 126
In each case, the H.C.F. of 90, 108 and 126 will be the greatest number that will divide 93, 111 and 129 leaving remainder 3
Using division method, the H.C.F. of 90, 108 and 126 is given below
Since, last division is 18
∴H.C.F. of 90, 108 and 126 is 18
Hence, the greatest number that will divide 93, 111 and 129 is 18
Exercise 8(C) page no: 71
1. Using the common multiple method, find the L.C.M. of the following:
(i) 8, 12 and 24
(ii) 10, 15 and 20
(iii) 3, 6, 9 and 12
Solution:
(i) 8, 12 and 24
We get,
L.C.M = 4 × 3 × 2
= 24
Hence, L.C.M. of 8, 12 and 24 = 24
(ii) 10, 15 and 20
We get,
L.C.M = 2 × 2 × 3 × 5
= 60
Hence, L.C.M. of 10, 15 and 20 = 60
(iii) 3, 6, 9 and 12
We get,
L.C.M. = 2 × 2 × 3 × 3
= 36
Hence, L.C.M. of 3, 6, 9 and 12 = 36
2. Find the L.C.M. of each of the following groups of numbers, using (i) the prime factor method and (ii) the common division method:
(i) 18, 24 and 96
(ii) 100, 150 and 200
(iii) 14, 21 and 98
(iv) 22, 121 and 33
(v) 34, 85 and 51
Solution:
(i) 18, 24 and 96
By using prime factor method, L.C.M. of 18, 24 and 96 are given below
Prime factors of 18 = 2 × 3 × 3
Prime factors of 24 = 2 × 2 × 2 × 3
Prime factors of 96 = 2 × 2 × 2 × 2 × 2 × 3
∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3
= 288
By using common division method, L.C.M. of 18, 24 and 96 are given below
∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3
= 288
(ii) 100, 150 and 200
By using prime factor method, L.C.M. of 100, 150 and 200 are given below
Prime factor of 100 = 2 × 2 × 5 × 5
Prime factor of 150 = 2 × 3 × 5 × 5
Prime factor of 200 = 2 × 2 × 2 × 5 × 5
∴ L.C.M. = 2 × 2 × 2 × 3 × 5 × 5
= 600
By using common division method, L.C.M. of 100, 150 and 200 are given below
∴L.C.M. = 2 × 2 × 2 × 3 × 5 × 5
= 600
(iii) 14, 21 and 98
By using prime factor method, L.C.M. of 14, 21 and 98 are given below
Prime factor of 14 = 2 × 7
Prime factor of 21 = 3 × 7
Prime factor of 98 = 2 × 7 × 7
∴ L.C.M. = 2 × 3 × 7 × 7
= 294
By using common division method, L.C.M. of 14, 21 and 98 are given below
∴ L.C.M. = 2 × 3 × 7 × 7
= 294
(iv) 22, 121 and 33
By using prime factor method, L.C.M. of 22, 121 and 33 are given below
Prime factor of 22 = 2 × 11
Prime factor of 121 = 11 × 11
Prime factor of 33 = 3 × 11
∴ L.C.M. = 2 × 3 × 11 × 11
= 726
By using common division method, L.C.M. of 22, 121 and 33 are given below
∴ L.C.M. = 2 × 3 × 11 × 11
= 726
(v) 34, 85 and 51
By using prime factor method, L.C.M. of 34, 85 and 51 are given below
Prime factor of 34 = 2 × 17
Prime factor of 85 = 5 × 17
Prime factor of 51 = 3 × 17
∴ L.C.M. = 2 × 3 × 5 × 17
= 510
By using common division method, L.C.M. of 34, 85 and 51 are given below
∴ L.C.M. = 2 × 3 × 5 × 17
= 510
3. The H.C.F. and the L.C.M. of two numbers are 50 and 300 respectively. If one of the numbers is 150, find the other one.
Solution:
Given
H.C.F. = 50
L.C.M. = 300
One number = 150
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers
For other number,
50 × 300 = 150 × other number
15000 / 150 = other number
100 = other number
Hence, the other number is 100
4. The product of two numbers is 432 and their L.C.M. is 72. Find their H.C.F.
Solution:
Given
Product of two numbers = 432 and L.C.M.= 72
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers.
Now, to find H.C.F
H.C.F. × 72 = 432
H.C.F.= 432 / 72
H.C.F. = 6
Hence, H.C.F. = 6
5. The product of two numbers is 19,200 and their H.C.F. is 40. Find their L.C.M.
Solution:
Given
Product of two numbers = 19200 and H.C.F. = 40
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers
Now, to find L.C.M.
40 × L.C.M. = 19200
L.C.M. = 19200 / 40
L.C.M. = 480
Hence, L.C.M. = 480
6. Find the smallest number which, when divided by 12, 15, 18, 24 and 36 leaves no remainder.
Solution:
The given numbers L.C.M. will be the least number which is exactly divisible 12, 15, 18, 24 and 36 and leaves no remainder
L.C.M. = 2 × 2 × 2 × 3 × 3 × 5
= 360
Hence, smallest required number = 360
7. Find the smallest number which, when increased by one is exactly divisible by 12, 18, 24, 32 and 40.
Solution:
First, let us find out the L.C.M. of 12, 18, 24, 32 and 40
L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5
= 1440
This can be written as
= 1439 + 1
Hence, 1439 is the smallest number which, when increased by one is exactly divisible by the given numbers
8. Find the smallest number which, on being decreased by 3, is completely divisible by 18, 36, 32 and 27.
Solution:
First, let us solve for L.C.M. of 18, 36, 32 and 27
L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 864
This can be written as
= 867 – 3
Hence, 867 is the smallest number which, when decreased by 3 is exactly divisible by the given numbers
Exercise 8(B)
Exercise 8(B) Page no: 66
1. Using the common factor method, find the H.C.F. of:
(i) 16 and 35
(ii) 25 and 20
(iii) 27 and 75
(iv) 8, 12 and 18
(v) 24, 36, 45 and 60
Solution:
(i) Common factors of 16 and 35 are as follows:
F (16) = 1, 2, 4, 8, 16
F (35) = 1, 5, 7, 35
The common factors between 16 and 35 = 1
∴The H.C.F. of 16 and 35 = 1
(ii) Common factors of 25 and 20 are as follows:
F (25) = 1, 5, 25
F (20) = 1, 2, 4, 5, 10, 20
The common factors between 25 and 20 = 1, 5
∴The H.C.F. of 25 and 20 = 5
(iii) Common factors between 27 and 75 are as follows:
F (27) = 1, 3, 9, 27
F (75) = 1, 3, 5, 15, 25, 75
The common factors between 27 and 75 = 1, 3
∴The H.C.F. of 27 and 75 = 3
(iv) Common factors between 8, 12 and 18 are as follows:
F (8) = 1, 2, 4, 8
F (12) = 1, 2, 3, 4, 6, 12
F (18) = 1, 2, 3, 6, 9, 18
Common factors between 8, 12 and 18 = 1, 2
∴The H.C.F. of 8, 12 and 18 = 2
(v) Common factors between 24, 36, 45 and 60 are as follows:
F (24) = 1, 2, 3, 4, 6, 8, 12, 24
F (36) = 1, 2, 3, 4, 6, 12, 18, 36
F (45) = 1, 3, 5, 9, 15, 45
F (60) = 1, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common factors between 24, 36, 45, and 60 = 1, 3
∴The H.C.F. of 24, 36, 45 and 60 = 3
2. Using the prime factor method, find the H.C.F. of:
(i) 5 and 8
(ii) 24 and 49
(iii) 40, 60 and 80
(iv) 48, 84 and 88
(v) 12, 16 and 28
Solution:
(i) The prime factors of 5 and 8 are as follows:
P5 = 5
P8 = 2 × 2 × 2
No common prime factors between 5 and 8
Hence, H.C.F. of 5 and 8 = 1
(ii) The prime factors of 24 and 49 are as follows:
P24 = 2 × 2 × 2 × 3
P49 = 7 × 7
No common prime factors between 24 and 49
Hence, H.C.F. of 24 and 49 = 1
(iii) The prime factors of 40, 60 and 80 are as follows:
P40 = 2 × 2 × 2 × 5
P60 = 2 × 2 × 3 × 5
P80 = 2 × 2 × 2 × 2 × 5
Common prime factors between 40, 60 and 80 = 2 × 2 × 5
Hence, H.C.F. of 40, 60 and 80 = 20
(iv) The prime factors of 48, 84 and 88 are as follows:
P48 = 2 × 2 × 2 × 2 × 3
P84 = 2 × 2 × 3 × 7
P88 = 2 × 2 × 2 × 11
Common prime factors between 48, 84 and 88 = 2 × 2
Hence, H.C.F. of 48, 84 and 88 = 2 × 2 = 4
(v) The prime factors of 12, 16 and 28 are as follows:
P12 = 2 × 2 × 3
P16 = 2 × 2 × 2 × 2
P28 = 2 × 2 × 7
Common prime factors between 12, 16 and 28 = 2 × 2
Hence, H.C.F. of 12, 16 and 28 = 2 × 2 = 4
3. Using the division method, find the H.C.F. of the following:
(i) 16 and 24
(ii) 18 and 30
(iii) 7, 14 and 24
(iv) 70, 80, 120 and 150
(v) 32, 56 and 46
Solution:
(i) 16 and 24
Using division method, we get
Here, the last division is 8
Hence, H.C.F. of 16 and 24 = 8
(ii) 18 and 30
Using division method, we get
Here, last division is 6
Hence, H.C.F. of 18 and 30 = 6
(iii) 7, 14 and 24
Using division method, we get
Here, last division is 1
Hence, H.C.F. of 7, 14 and 24 = 1
(iv) 70, 80, 120 and 150
Using division method, we get
Here, last division is 10
Hence, H.C.F. of 70, 80, 120 and 150 = 10
(v) 32, 56 and 46
Using division method, we get
Here, last division is 2
Hence, H.C.F. of 32, 56 and 46 = 2
4. Use a method of your own choice to find the H.C.F. of:
(i) 45, 75 and 135
(ii) 48, 36 and 96
(iii) 66, 33 and 132
(iv) 24, 36, 60 and 132
(v) 30, 60, 90, and 105
Solution:
(i) 45, 75 and 135
The prime factors of 45, 75 and 135 are as follows:
P45 = 3 × 3 × 5
P75 = 3 × 5 × 5
P135 = 3 × 3 × 3 × 5
The common factors of 45, 75 and 135 = 3 × 5
∴H.C.F. of 45, 75 and 135 = 15
(ii) 48, 36 and 96
The prime factors of 48, 36 and 96 are as follows:
P48 = 2 × 2 × 2 × 2 × 3
P36 = 2 × 2 × 3 × 3
P96 = 2 × 2 × 2 × 2 × 2 × 3
The common factors of 48, 36 and 96 = 2 × 2 × 3
∴H.C.F. of 48, 36 and 96 = 12
(iii) 66, 33 and 132
The prime factors of 66, 33 and 132 are as follows:
P66 = 2 × 3 × 11
P33 = 3 × 11
P132 = 2 × 2 × 3 × 11
The common factors of 66, 33 and 132 = 3 × 11
∴H.C.F. of 66, 33 and 132 = 33
(iv) 24, 36, 60 and 132
The prime factors of 24, 36, 60 and 132 are as follows:
P24 = 2 × 2 × 2 × 3
P36 = 2 × 2 × 3 × 3
P60 = 2 × 2 × 3 × 5
P132 = 2 × 2 × 3 × 11
The common factors of 24, 36, 60 and 132 = 2 × 2 × 3
∴H.C.F. of 24, 36, 60 and 132 = 12
(v) 30, 60, 90, and 105
The prime factors of 30, 60, 90 and 105 are as follows:
P30 = 2 × 3 × 5
P60 = 2 × 2 × 3 × 5
P90 = 2 × 3 × 3 × 5
P105 = 3 × 5 × 7
The common factors of 30, 60, 90 and 105 = 3 × 5
∴H.C.F. of 30, 60, 90 and 105 = 15
5. Find the greatest number that divides each of 180, 225 and 315 completely.
Solution:
The greatest number that divides each of 180, 225 and 315 will be the H.C.F. of 180, 225 and 315
Using division method, the H.C.F of 180, 225 and 315 are shown below
Since last division is 45
∴H.C.F. of 180, 225 and 315 = 45
6. Show that 45 and 56 are co-prime numbers.
Solution:
The H.C.F. of two co-prime numbers is always 1
Using division method, the H.C.F. of 45 and 56 are shown below
Here, last division is 1
H.C.F. of 45 and 56 = 1
∴45 and 56 are co-prime numbers
7. Out of 15, 16, 21 and 28, find out all the pairs of co-prime numbers.
Solution:
15 and 16, 15 and 21, 15 and 28, 16 and 21, 16 and 28 and 21 and 28 will be the possible pairs
We know that the H.C.F. of two co-prime numbers is always 1
Using division method, the H.C.F. of pair 15 and 16 is given below
Hence, H.C.F. of 15 and 16 = 1
Using division method, the H.C.F. of pair 15 and 21 is given below
Hence, H.C.F. of 15 and 21= 3
Using division method, the H.C.F. of pair 15 and 28 is given below
Hence, H.C.F. of 15 and 28 = 1
Using division method, the H.C.F. of pair 16 and 21 is given below
Hence, H.C.F. of 16 and 21 = 1
Using division method, the H.C.F. of 16 and 28 is given below
Hence, H.C.F. of 16 and 28 = 4
Using division method, the H.C.F. of 21 and 28 is given below
Hence, H.C.F. of 21 and 28 = 7
∴ The pairs 15 and 16, 15 and 28 and 16 and 21 are co-prime numbers
8. Find the greatest number that will divide 93, 111 and 129, leaving remainder 3 in each case.
Solution:
First, decrease the leaving remainder 3 from numbers 93, 111 and 129 to find the required number
93 – 3 = 90
111 – 3 = 108
129 – 3 = 126
In each case, the H.C.F. of 90, 108 and 126 will be the greatest number that will divide 93, 111 and 129 leaving remainder 3
Using division method, the H.C.F. of 90, 108 and 126 is given below
Since, last division is 18
∴H.C.F. of 90, 108 and 126 is 18
Hence, the greatest number that will divide 93, 111 and 129 is 18
Exercise 8(C) page no: 71
1. Using the common multiple method, find the L.C.M. of the following:
(i) 8, 12 and 24
(ii) 10, 15 and 20
(iii) 3, 6, 9 and 12
Solution:
(i) 8, 12 and 24
We get,
L.C.M = 4 × 3 × 2
= 24
Hence, L.C.M. of 8, 12 and 24 = 24
(ii) 10, 15 and 20
We get,
L.C.M = 2 × 2 × 3 × 5
= 60
Hence, L.C.M. of 10, 15 and 20 = 60
(iii) 3, 6, 9 and 12
We get,
L.C.M. = 2 × 2 × 3 × 3
= 36
Hence, L.C.M. of 3, 6, 9 and 12 = 36
2. Find the L.C.M. of each of the following groups of numbers, using (i) the prime factor method and (ii) the common division method:
(i) 18, 24 and 96
(ii) 100, 150 and 200
(iii) 14, 21 and 98
(iv) 22, 121 and 33
(v) 34, 85 and 51
Solution:
(i) 18, 24 and 96
By using prime factor method, L.C.M. of 18, 24 and 96 are given below
Prime factors of 18 = 2 × 3 × 3
Prime factors of 24 = 2 × 2 × 2 × 3
Prime factors of 96 = 2 × 2 × 2 × 2 × 2 × 3
∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3
= 288
By using common division method, L.C.M. of 18, 24 and 96 are given below
∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3
= 288
(ii) 100, 150 and 200
By using prime factor method, L.C.M. of 100, 150 and 200 are given below
Prime factor of 100 = 2 × 2 × 5 × 5
Prime factor of 150 = 2 × 3 × 5 × 5
Prime factor of 200 = 2 × 2 × 2 × 5 × 5
∴ L.C.M. = 2 × 2 × 2 × 3 × 5 × 5
= 600
By using common division method, L.C.M. of 100, 150 and 200 are given below
∴L.C.M. = 2 × 2 × 2 × 3 × 5 × 5
= 600
(iii) 14, 21 and 98
By using prime factor method, L.C.M. of 14, 21 and 98 are given below
Prime factor of 14 = 2 × 7
Prime factor of 21 = 3 × 7
Prime factor of 98 = 2 × 7 × 7
∴ L.C.M. = 2 × 3 × 7 × 7
= 294
By using common division method, L.C.M. of 14, 21 and 98 are given below
∴ L.C.M. = 2 × 3 × 7 × 7
= 294
(iv) 22, 121 and 33
By using prime factor method, L.C.M. of 22, 121 and 33 are given below
Prime factor of 22 = 2 × 11
Prime factor of 121 = 11 × 11
Prime factor of 33 = 3 × 11
∴ L.C.M. = 2 × 3 × 11 × 11
= 726
By using common division method, L.C.M. of 22, 121 and 33 are given below
∴ L.C.M. = 2 × 3 × 11 × 11
= 726
(v) 34, 85 and 51
By using prime factor method, L.C.M. of 34, 85 and 51 are given below
Prime factor of 34 = 2 × 17
Prime factor of 85 = 5 × 17
Prime factor of 51 = 3 × 17
∴ L.C.M. = 2 × 3 × 5 × 17
= 510
By using common division method, L.C.M. of 34, 85 and 51 are given below
∴ L.C.M. = 2 × 3 × 5 × 17
= 510
3. The H.C.F. and the L.C.M. of two numbers are 50 and 300 respectively. If one of the numbers is 150, find the other one.
Solution:
Given
H.C.F. = 50
L.C.M. = 300
One number = 150
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers
For other number,
50 × 300 = 150 × other number
15000 / 150 = other number
100 = other number
Hence, the other number is 100
4. The product of two numbers is 432 and their L.C.M. is 72. Find their H.C.F.
Solution:
Given
Product of two numbers = 432 and L.C.M.= 72
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers.
Now, to find H.C.F
H.C.F. × 72 = 432
H.C.F.= 432 / 72
H.C.F. = 6
Hence, H.C.F. = 6
5. The product of two numbers is 19,200 and their H.C.F. is 40. Find their L.C.M.
Solution:
Given
Product of two numbers = 19200 and H.C.F. = 40
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers
Now, to find L.C.M.
40 × L.C.M. = 19200
L.C.M. = 19200 / 40
L.C.M. = 480
Hence, L.C.M. = 480
6. Find the smallest number which, when divided by 12, 15, 18, 24 and 36 leaves no remainder.
Solution:
The given numbers L.C.M. will be the least number which is exactly divisible 12, 15, 18, 24 and 36 and leaves no remainder
L.C.M. = 2 × 2 × 2 × 3 × 3 × 5
= 360
Hence, smallest required number = 360
7. Find the smallest number which, when increased by one is exactly divisible by 12, 18, 24, 32 and 40.
Solution:
First, let us find out the L.C.M. of 12, 18, 24, 32 and 40
L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5
= 1440
This can be written as
= 1439 + 1
Hence, 1439 is the smallest number which, when increased by one is exactly divisible by the given numbers
8. Find the smallest number which, on being decreased by 3, is completely divisible by 18, 36, 32 and 27.
Solution:
First, let us solve for L.C.M. of 18, 36, 32 and 27
L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 864
This can be written as
= 867 – 3
Hence, 867 is the smallest number which, when decreased by 3 is exactly divisible by the given numbers
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Selina Class 6 ICSE Solutions Mathematics : Chapter 8- HCF And LCM
Download PDF: Selina Class 6 ICSE Solutions Mathematics : Chapter 8- HCF And LCM PDF
Chapterwise Selina Publishers ICSE Solutions for Class 6 Mathematics :
- Chapter 1- Number System
- Chapter 2- Estimation
- Chapter 3- Numbers In Indian And International Systems
- Chapter 4- Place Value
- Chapter 5- Natural Numbers And Whole Numbers
- Chapter 6- Negative Numbers And Integers
- Chapter 7- Number Line
- Chapter 8- HCF And LCM
- Chapter 9- Playing With Numbers
- Chapter 10- Sets
- Chapter 11- Ratio
- Chapter 12- Proportion
- Chapter 13- Unitary Method
- Chapter 14- Fractions
- Chapter 15- Decimal Fractions
- Chapter 16- Percent (Percentage)
- Chapter 17- Idea of Speed, Distance and Time
- Chapter 18- Fundamental Concepts
- Chapter 19- Fundamental Operations
- Chapter 20- Substitution
- Chapter 21- Framing Algebraic Expressions (Including Evaluation)
- Chapter 22- Simple (Linear) Equations
- Chapter 23- Fundamental Concepts
- Chapter 24- Angles
- Chapter 25- Properties of Angles and Lines
- Chapter 26- Triangles
- Chapter 27- Quadrilateral
- Chapter 28- Polygons
- Chapter 29- The Circle
- Chapter 30- Revision Exercise Symmetry
- Chapter 31- Recognition of Solids
- Chapter 32- Perimeter and Area of Plane Figures
- Chapter 33- Data Handling
- Chapter 34- Mean and Median
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