Practice Set 5.2
(1) Find the GCD of the following numbers.
i) 36, 48
ii) 31, 2
iii) 65, 13
iv) 108, 132
v) 30, 24, 48
Three Methods of Finding GCD
The textbook uses three different methods for solving this question.
Select a method below to see its complete solution.
1
Divisor Method
i) 36, 48
The divisors of number 36 :
1,
2,
3,
4,
6,
9,
12,
18, 36.
The divisors of the number 48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
The greatest common divisor is 12.
The divisors of the number 48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
The common divisors of 36 and 48 :
1, 2, 3, 4, 6, 12
The greatest common divisor is 12.
∴ The GCD of numbers 36 and 48 is 12.
ii) 31, 2
The divisors of number 31 :
1, 31.
The divisors of the number 2 : 1, 2.
The greatest common divisor is 1.
The divisors of the number 2 : 1, 2.
The common divisors of 31 and 2 :
1
The greatest common divisor is 1.
∴ The GCD of numbers 31 and 2 is 1.
iii) 65, 13
The divisors of number 65 :
1,
5,
13,
65.
The divisors of the number 13 : 1, 13.
The greatest common divisor is 13.
The divisors of the number 13 : 1, 13.
The common divisors of 65 and 13 :
1, 13
The greatest common divisor is 13.
∴ The GCD of the numbers 65 and 13 is 13.
iv) 108, 132
The divisors of number 108 :
1,
2,
3,
4,
6,
9,
12,
18, 27, 36, 54, 108.
The divisors of the number 132 : 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
The greatest common divisor is 12.
The divisors of the number 132 : 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
The common divisors of 108 and 132 :
1, 2, 3, 4, 6, 12
The greatest common divisor is 12.
∴ The GCD of the numbers 108 and 132 is 12.
v) 30, 24, 48
The divisors of number 30 :
1,
2,
3,
5,
6,
10, 15, 30.
The divisors of the number 24 : 1, 2, 3, 4, 6, 8, 12, 24.
The divisors of the number 48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
The greatest common divisor is 6.
The divisors of the number 24 : 1, 2, 3, 4, 6, 8, 12, 24.
The divisors of the number 48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
The common divisors of 30, 24 and 48 :
1, 2, 3, 6
The greatest common divisor is 6.
∴ The GCD of the numbers 30, 24 and 48 is 6.
2
Prime factorisation Method
i) 36, 48
36
=
2 × 18
=
2 × 2 × 9
=
2 × 2 × 3 × 3
48
=
2 × 24
=
2 × 2 × 12
=
2 × 2 × 2 × 6
=
2 × 2 × 2 × 2 × 3
Common prime factors of the numbers 36 and 48 =
2, 2, 3
G.C.D. of the numbers 36 and 48 =
2 × 2 × 3 = 12
∴ The GCD of the numbers 36 and 48 is 12.
ii) 31, 2
31
=
31
2
=
2
Common prime factors of the numbers 31 and 2 =
None
Note: If there are no common prime factors, the G.C.D. is 1.
G.C.D. of the numbers 31 and 2 = 1
∴ The GCD of the numbers 31 and 2 is 1.
iii) 65, 13
65
=
5 × 13
13
=
13
Common prime factors of the numbers 65 and 13 =
13
G.C.D. of the numbers 65 and 13 = 13
∴ The GCD of the numbers 65 and 13 is 13.
iv) 108, 132
108
=
2 × 54
=
2 × 2 × 27
=
2 × 2 × 3 × 9
=
2 × 2 × 3 × 3 × 3
132
=
2 × 66
=
2 × 2 × 33
=
2 × 2 × 3 × 11
Common prime factors of the numbers 108 and 132 =
2, 2, 3
G.C.D. of the numbers 108 and 132 =
2 × 2 × 3 = 12
∴ The GCD of the numbers 108 and 132 is 12.
v) 30, 24, 48
30
=
2 × 15
=
2 × 3 × 5
24
=
2 × 12
=
2 × 2 × 6
=
2 × 2 × 2 × 3
48
=
2 × 24
=
2 × 2 × 12
=
2 × 2 × 2 × 6
=
2 × 2 × 2 × 2 × 3
Common prime factors of the numbers 30, 24 and 48 =
2, 3
G.C.D. of the numbers 30, 24 and 48 =
2 × 3 = 6
∴ The GCD of the numbers 30, 24 and 48 is 6.
3
Division method
Division method :
While finding the GCD of two numbers, divide the larger number by the smaller number.
If there is a remainder in the division, divide the divisor by that remainder.
This process should continue until the remainder is zero (0).
When the remainder is zero, the last divisor is the GCD.
i) 36, 48
Here the remainder is 0 and the divisor is 12, so the GCD of the numbers 36 and 48 is 12.
ii) 31, 2
Here the remainder is 0 and the divisor is 1, so the GCD of the numbers 31 and 2 is 1.
iii) 65, 13
Here the remainder is 0 and the divisor is 13, so the GCD of the numbers 65 and 13 is 13.
iv) 108, 132
Here the remainder is 0 and the divisor is 12, so the GCD of the numbers 108 and 132 is 12.
v) 30, 24, 48
For three numbers, find the GCD of the first two numbers and then find the GCD of the result and the third number.
Here the remainder is 0 and the divisor is 6, so the GCD of the numbers 30 and 24 is 6.
Now find the GCD of 6 and 48.
Here the remainder is 0 and the divisor is 6, so the GCD of the numbers 30, 24 and 48 is 6.
2
The length and breadth of a room are 15 metres and 9 metres respectively.
What is the maximum length of each tile to be laid in square tiles in this room?
Three Methods of Finding GCD
Select any method below to view the complete solution.
Solution :
To find the maximum possible length of each tile,
we have to find the GCD of 15 and 9.
Divisor of 15 :
1,
3,
5, 15.
Divisor of 9 :
1,
3,
9.
Common Divisors of 15 and 9 :
1, 3.
The greatest common Divisor :
3.
Ans. The maximum possible length of each tile is 3 metres.
Solution :
To find the maximum possible length of each tile,
we have to find the GCD of 15 and 9.
15
=
3 × 5
9
=
3 × 3
Common prime factors of the numbers 15 and 9 =
3
G.C.D. = 3
Ans. The maximum possible length of each tile is 3 metres.
Solution :
To find the maximum possible length of each tile,
we have to find the GCD of 15 and 9.
GCD = 3
Ans. The maximum possible length of each tile is 3 metres.
3
A shopkeeper has two rolls of cloth of length 90 m and 120 m.
He wants to cut both of rolls into pieces of equal maximum length.
What can be the maximum length of each piece?
Three Methods of Finding GCD
Select any method below to view the complete solution.
Solution :
To find the maximum possible length of each piece of cloth,
we have to find the GCD of 90 and 120.
Divisor of 90 :
1,
2,
3,
5,
6,
9,
10,
15,
18,
30,
45, 90.
Divisor of 120 :
1,
2,
3,
4,
5,
6,
8,
10,
12,
15,
20, 24,
30,
40, 60, 120.
Common Divisor of 90 and 120 :
1, 2, 3, 5, 6, 10, 15, 30.
Greatest common divisor : 30.
Ans. The maximum length of each piece is 30 m.
Note : Students can use any method for finding GCD.
Solution :
To find the maximum possible length of each piece of cloth,
we have to find the GCD of 90 and 120.
90
=
2 × 45 = 2 × 3 × 15
=
2 × 3 × 3 × 5
120
=
2 × 60 = 2 × 2 × 30
=
2 × 2 × 2 × 15
=
2 × 2 × 2 × 3 × 5
GCD = 2 × 3 × 5 = 30
Ans. The maximum length of each piece is 30 m.
Note : Students can use any method for finding GCD.
Solution :
To find the maximum possible length of each piece of cloth,
we have to find the GCD of 90 and 120.
GCD = 30
Ans. The maximum length of each piece is 30 m.
Note : Students can use any method for finding GCD.
4
A garland seller has 48 red and 60 white roses.
He wants to make separate garlands of both types of flowers.
If the number of flowers in each garland is the same,
then what is the maximum number of flowers in each garland?
Three Methods of Finding GCD
Select any method below to view the complete solution.
Solution :
We find the GCD of 48 and 60.
Divisor of 48 :
1,
2,
3,
4,
6,
8,
12,
16, 24, 48.
Divisor of 60 :
1,
2,
3,
4,
5,
6,
10,
12,
15, 20, 30, 60.
Common Divisor of 48 and 60 :
1, 2, 3, 4, 6, 12.
Greatest common divisor = 12.
Ans. The maximum number of flowers in each garland is 12.
Solution :
We find the GCD of 48 and 60.
48
=
2 × 24 = 2 × 2 × 12
=
2 × 2 × 2 × 6
=
2 × 2 × 2 × 2 × 3
60
=
2 × 30 = 2 × 2 × 15
=
2 × 2 × 3 × 5
GCD = 2 × 2 × 3 = 12
Ans. The maximum number of flowers in each garland is 12.
Solution :
We find the GCD of 48 and 60.
GCD = 12
Ans. The maximum number of flowers in each garland is 12.
Good
Thanks