Exercise 5
(1)
Choose the correct option.
i) What is the smallest prime factor of the number 75?
ii) What is the least common divisor of the numbers
17 and 18?
Ans. (i) (B) 3
(ii) (A) 1.
(2)
State True or False.
i) Every prime number has only two factors.
True
ii) The product of two prime numbers is always
a prime number.
False
iii) GCD is the divisor of LCM.
True
iv) LCM is greater than the given number or equal
to the smallest number between those numbers.
False
Ans. (i) True
(ii) False
(iii) True
(iv) False
(3)
Express the number as a product of prime factors.
(i) 91
| 7 | 91 |
| 13 | 13 |
| 1 |
∴ 91 = 7 × 13.
(ii) 160
| 2 | 160 |
| 2 | 80 |
| 2 | 40 |
| 2 | 20 |
| 2 | 10 |
| 5 | 5 |
| 1 |
∴ 160 = 2 × 2 × 2 × 2 × 2 × 5.
(iii) 89
| 89 | 89 |
| 1 |
∴ 89 = 89
(4)
Find GCD of the following numbers.
(i) 21, 42.
Solution :
Divisor of 21 : 1, 3, 7, 21.
Divisor of 42 : 1, 2, 3, 6, 7, 14, 21, 42.
Common divisor of 21 and 42 :
1, 3, 7, 21.
Largest common divisor : 21.
Ans. The GCD of 21 and 42 is 21.
(ii) 17, 9.
Solution :
Divisor of 17 : 1, 17.
Divisor of 9 : 1, 3, 9.
Common divisor of 17 and 9 : 1
Ans. The GCD of 17 and 9 is 1.
(iii) 120, 180, 240.
Solution :
Divisor of 120 :
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20,
24, 30, 40, 60, 120.
Divisor of 180 :
1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18,
20, 30, 36, 45, 60, 90, 180.
Divisor of 240 :
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16,
20, 24, 30, 40, 48, 60, 80, 120, 240.
Common divisor of 120, 180 and 240 :
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
Largest common divisor : 60
Or
GCD by prime factorisation method :
120 = 2 × 2 × 2 × 3 × 5
180 = 2 × 2 × 3 × 3 × 5
240 = 2 × 2 × 2 × 2 × 3 × 5
Common prime factors of 120, 180, 240 =
2, 2, 3, 5
GCD of the number 120, 180 and 240 =
Product of their common factors
= 2 × 2 × 3 × 5 = 60
Ans. The GCD of 120, 180 and 240 is 60.
(5)
Find LCM of the following numbers.
(i) 11, 16.
Solution :
Multiples of 11 :
11, 22, 33, 44, 55, 66, 77, 88, 99, 110,
132, 143, 154, 165, 176, 187, 198, 209, 220,
..., 352, 363, ...
Multiples of 16 :
16, 32, 48, 64, 80, 96, 112, 128, 144,
160, 176, 192, 208, 224, ..., 352, 368, ...
Common multiples of 11 and 16 :
176, 352, ...
Least common multiple : 176.
Or
LCM by prime factorisation method :
11 = 11
16 = 2 × 2 × 2 × 2
LCM = 11 × 2 × 2 × 2 × 2 = 176
∴ the LCM of 11 and 16 is 176.
Ans. The LCM of 11 and 16 is 176.
(ii) 7, 49.
Solution :
Multiples of 7 :
7, 14, 21, 28, 35, 42, 49, ...,
77, 84, 98, ..., 147, ...
Multiples of 49 :
49, 98, 147, ..., 245, ...
Common multiples of 7 and 49 :
49, 98, 147, ...
Least common multiple : 49
∴ LCM of 7 and 49 : 49.
Or
LCM by vertical layout method :
| 7 | 7 | 49 |
| 7 | 1 | 7 |
| 1 | 1 |
LCM = 7 × 7 = 49.
Ans. LCM of 7 and 49 is 49.
(iii) 16, 48, 6.
Solution :
Multiples of 16 :
16, 32, 48, 64, ..., 96, 112, 128,
144, ..., 192, 208, ...
Multiples of 48 :
48, 96, 144, 192, ...
Multiples of 6 :
6, 12, 18, ..., 48, 54, 60, ...,
90, 96, ..., 138, 144, ..., 192, 198, ...
Common multiples of 16, 48 and 6 :
48, 96, 144, 192, ...
Least common multiple : 48.
∴ LCM of 16, 48 and 6 is 48.
Or
LCM by vertical layout method :
| 2 | 16 | 48 | 6 |
| 2 | 8 | 24 | 3 |
| 2 | 4 | 12 | 3 |
| 2 | 2 | 6 | 3 |
| 3 | 1 | 3 | 3 |
| 1 | 1 | 1 |
LCM = 2 × 2 × 2 × 2 × 3 = 48.
Ans. The LCM of 16, 48 and 6 is 48.
(6)
Solve.
i) The product of two numbers is 120. If their LCM is
60, then what is their GCD?
Solution :
Product of two numbers = GCD × LCM
120 = GCD × 60
∴ GCD =
120
60
∴ GCD = 2
Ans. The GCD of two numbers is 2.
ii) The GCD of a number and 16 is 4 and their LCM is 48.
What is that number?
Solution :
Let the first number be x.
Product of two numbers = GCD × LCM
∴ x × 16 = 4 × 48
∴ x =
4 × 48
16
∴ x = 4 × 3
∴ x = 12
∴ first number : 12
Ans. First number is 12.
iii) A carpenter has two wooden strips. The length of one
strip is 54 inches and the length of the other strip is
72 inches. He wants to cut both the strips into equal
lengths so that no wooden strip is left over. What is the
maximum length of the strips that he can cut?
Solution :
We find the GCD of 54 and 72.
Divisor of 54 :
1, 2, 3, 6, 9, 18, 27, 54.
Divisor of 72 :
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
Common divisor of 54 and 72 :
1, 2, 3, 6, 9, 18.
Largest common divisor : 18.
Or
By Prime factorisation method :
54 = 2 × 3 × 3 × 3
72 = 2 × 2 × 2 × 3 × 3
GCD = 2 × 3 × 3 = 18
Ans. Maximum length of the strip is 18.
iv) Samip has 84 mangoes and 105 apples. He wants to
place these fruits in boxes, so that each box has the
same number of fruits and each box contains only one
type of fruit.
1) What is the maximum number of fruits he should put in a box?
2) How many boxes will he need to store the fruits?
1) What is the maximum number of fruits he should put in a box?
2) How many boxes will he need to store the fruits?
Solution :
(1) To find the maximum number of fruits that can be
placed in each box, we need to find the GCD of the
two numbers.
We find the GCD of 84 and 105.
Divisors of 84 :
1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84.
Divisors of 105 :
1, 3, 5, 7, 15, 21, 35, 105.
Common divisors of 84 and 105 :
1, 3, 7, 21.
Largest common divisor : 21.
Or
GCD by prime factorisation method :
84 = 2 × 2 × 3 × 7
105 = 3 × 5 × 7
GCD = 3 × 7 = 21
84 = 2 × 2 × 3 × 7
105 = 3 × 5 × 7
GCD = 3 × 7 = 21
Ans. He should put 21 fruits in a box.
(2) To find the minimum number of boxes needed to
store the fruits, we need to place 21 fruits in each
box.
∴ boxes needed for mangoes =
84
21
= 4
∴ boxes needed for apples =
105
21
= 5
∴ total number of boxes = 4 + 5 = 9
Ans. Samip will need a minimum of 9 boxes to store
the fruits.