# Real Numbers – Exercise 3.2 – Class X

Previous Exercise – Real Numbers – Exercise 3.1 – Class X

#### Real Numbers – Exercise 3.2

1.  Express each number as a product of prime factors.

i)120

ii)3825

iii)6762

iv)32844

2.If 25025 = p1x1, p2x2, p3x3,p4x4 find the value of p1, p2, p3, p4 and x1, x2, x3, x4

3.Find LCM and HCF of the following integers by expressing them as product of primes.

(i) 12, 15 and 30

(ii) 18, 81 and 108

4) Find the HCF and LCM of the pairs of integers and verify that LCM(a, b) x HCF(a, b) = a x b

(i) 16 and 80

(ii)125 and 55

1. If HCF of 52 and 182 is 26, find their LCM.
2. Find the HCF of 105 and 1515 by prime factorization method and hence find its LCM
3. Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468
4. A rectangular hall is 18m 72cm long & 13m 20cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

#### Real numbers – Exercise 3.2  – Solutions:

1) Express each number as a product of prime factors.

i)120

ii)3825

iii)6762

iv)32844

Solution:

(i)

120 = 23 x 3 x 5

(ii)

3825 = 32 x 52 x 17

(iii)

6762  = 22 x 32 x 72

(iv)

32844 = 22 x 3 x 7 x 17 x 23

2) If 25025 = p1x1, p2x2, p3x3,p4x4 find the value of p1, p2, p3, p4 and x1, x2, x3, x4

Solution:

25025 = 52 x 71 x 111 x 131

Given, 25025 = p1x1, p2x2, p3x3,p4x4

Therefore, p1x1= 5 , p2x2= 71 , p3x3= 111 ,p4x4 = 131 and x1 = 2, x2 = 1, x3 = 1, x4 = 1.

3.Find LCM and HCF of the following integers by expressing them as product of primes.

(i) 12, 15 and 30

(ii) 18, 81 and 108

Solution:

(i) 12, 15 and 30

12 = 4 x 3 = 22 x 3

15 = 5 x 3

30 = 3 x 2 x 5

Therefore, LCM = 60 and HCF = 3

(ii) 18, 81 and 108

18 = 3 x 3 x 2 = 32 x 2

81 = 9 x 9 = 3 x 3 x 3 x 3 = 34

108 = 3 x 2 x 3 x 2 x 3 = 22 x 33

Therefore, LCM = 9 and HCF = 34 x 22 = 324

4) Find the HCF and LCM of the pairs of integers and verify that LCM(a, b) x HCF(a, b) = a x b

(i) 16 and 80

(ii)125 and 55

Solution:

(i) 16 and 80

a = 16 = 2 x 2 x 2 x 2 = 24

b = 80 = 2 x 2 x 2 x 2 x 15 = 24 x 5

LCM (16, 80) = 80

HCF (16, 80) = 16

LCM (16, 80) x HCF (16, 80) = a x b

80 x 16 = 16 x 80

1280 = 1280

(ii)125 and 55

a = 125 = 53

b = 55 = 5 x 11

LCM(125, 55) = 5 x 5 x 5 x 11 = 1375

HCF (125, 55) = 5

LCM(125, 55) x HCF (125, 55) = a x b

1375 x 5 = 125 x55

6875 = 6875

1. If HCF of 52 and 182 is 26, find their LCM.

Solution:

We know, HCF x LCM = a x b

We have, a = 52 and b = 182

26 x LCM = 52 x 182

26 x LCM = 9464

LCM = 9464/26 = 364

1. Find the HCF of 105 and 1515 by prime factorization method and hence find its LCM

Solution:

105 = 5  x 3 x  7

1515 = 5 x 3 x 101

HCF (105, 1515) = 3 x 5 = 15

HCF x LCM = a x b

15 x LCM = 105 x 1515

LCM = 10605

1. Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468

Solution:

520 = 52  X 10

468 = 52 X 9

LCM = 52 X 10

LCM  =  52 x  10 x 9 = 4680

The smallest number which when increased by 17 which is exactly divisible by 520 & 468 is 4680 – 17 = 4663.

1. A rectangular hall is 18m 72cm long & 13m 20cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

Solution:

Area of rectangle = l x b = 1872 X 1320 = 24 X 32 X 13 X  23 X 3  X 5  X 11

HCF = 23 X  3 = 24

The greatest possible length of square tiles = 24

The least possible lengths of such tiles = 78 X 55 = 4290

Next exercise Real Numbers – Exercise 3.3