# SSC Real Numbers Exercise - 1.2 of 10th class

Class 10 Real Numbers Exercise 1.2

SSC Real Numbers :

• Positive or negative, Large or small, whole numbers or decimal numbers are all Real Numbers.

They are called “Real Numbers” because they are not Imaginary Numbers.

## EXERCISE – 1.2

### 1Q.  Express each of the following number as a product of its prime factors.

i)      140         ii) 156              iii) 3825                     iv) 5005                   v) 7429

Sol 140

140 = 2 X 2 X 5 X 7 = 22 X 5 X 7

Sol ii) 156

156 = 2 X 2 X 3 X 13 = 22 X 3 X 13

Sol :      iii)   3825

3825 = 3 X 3 X 5 X 5 X 17

= 32 X 52 X 17

Sol :      iv)   5005

👉 5005 = 5 X 7 X 11 X 13

Sol :         v)   7429

👉 7429 = 17 X 19 X 23

### 2Q. Find the L.C.M and H.C.F of the following integers by the prime factorization method.

Sol. i) 12, 15 and 21                ii) 17, 23 and 29

iii) 8, 9 and 25                            iv) 72 and 108

### Sol :      i) 12, 15 and 21

12 = 2 X 2 X 3 = 22 X 3

15 = 3 X 5

21 = 3 X 7

L.C.M = 22 X 3 X 5 X 7 = 420

H.C.F = 3

### Sol : ii) 17, 23 and 29

The given numbers 17, 23 and 29 are all primes.

L.C.M = their product

= 17 X 23 X 29 = 11339

H.C.F = 1

### iii) 8, 9 and 25

8 = 2 X 2 X 2 = 23

9 = 3 X 3 = 32

25 = 5 X 5 = 52

L.C.M = 23 X 32 X 52 = 1800

( OR )

8, 9 and 25 are relatively prime, there fore L.C.M is equal to their product. (i.e.) L.C.M = 8 X 9 X 25 = 1800

H.C.F = 1

### iv) 72 and 108

72 = 2 X 2 X 2 X 3 X 3

72 = 23 X 32

108 = 22 X 33

L.C.M = 23 X 33 = 8 X 27 = 216

H.C.F = 22 X 32 = 4 X 9 = 316

### v)   306 and 657

306 = 2 X 32 X 17

657 = 32 X 73

L.C.M = 2 X 32 X 17 X 73 = 22338

H.C.F = 32 = 9.

### SSC Real Numbers

Class 10 Real Numbers Exercise 1.2

### 3Q. Check whether 6n can end with the digit ‘0’ for any natural number n.

Sol : Given number = 6n = ( 2 X 3 )n

The prime factors here are 2 and 3 only.

To be end with 0; 6n should have a prime factor 5 and also 2. So,

6n can’t end with zero.

## 4Q. Explain why 7 X 11 X 13 + 13 and 7 X 6 X 5 X 4 X 3 X 2 X 1 + 5 are composite numbers.

Sol : Given numbers are

7 X 11 X 13 + 13 and

7 X 6 X 5 X 4 X 3 X 2 X 1 + 5

13 ( 7 X 11 + 1 ) and 5 ( 7 X 6 X 4 X 3 X 2 X 1 + 1 )

13 K and 5 L, where K = 78 and

L = 7 X 6 X 4 X 3 X 2 X 1 + 1 = 1009

As the given numbers can be written as product of two numbers, they are composite.

### 5Q. How will you show that ( 17 X 11 X 2 ) + ( 17 X 11 X 5 ) is a composite number? Explain.

Sol : ( 17 X 11 X 2 ) + ( 17 X 11 X 5 )

= ( 17 X 11 ) ( 2 + 5 )

= ( 17 X 11 ) ( 7 ) = 187 X 7

Now the given expression is written as a product of two integers and hence it is a composite number.

### Class 10 Real Numbers Exercise 1.2

For Real Number chapter of SSC Examination

For Job Informations