Real Numbers Exercise – 1.2 NCRT Download pdf || UP Board

 Real Numbers
Exercise – 1.2

 

1. Express each number as product of its prime factors:

(i) 140

Solution:

Real Numbers
The prime factor of 140 = 2 × 2 × 5 × 7
                                = 22 × 5 × 7


(ii) 156

 

Solution:

 

(Solve as first question)

The prime factor of 156 = 2 × 2 × 3 × 13

                    = 22 × 3 × 7

 

(iii) 3825

 

Solution:

 

 The prime factor of 3825 = 3 × 3 × 5 × 5 × 17

                    = 32 × 52 × 17

 

(iv) 5005

 

Solution:

 

 The prime factor of 5005 = 5 × 7 × 11 × 13

                   

(v) 7429

 

Solution:

 

 The prime factor of 7429 = 17 × 19 × 23

 

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

Solution:

The prime factor of 26 = 2 × 13

The prime factor of 91 = 7 × 13

HCF = 13

LCM = 2×7×13 = 182

product of the two numbers = 26×91 = 2366

HCF×LCM = 13×182 = 2366

product of the two numbers = LCM × HCF

 

(ii) 510 and 92

Solution:

510 = 2×3×5×17

92 = 2×2×23

HCF = 2

LCM = 2×2×3×5×17×23 = 23460

product of the two numbers = 510×92 = 46920

HCF×LCM = 2×23460 = 46920

product of the two numbers = LCM × HCF

 

(iii) 336 and 54

हल:

336 = 2×2×2×2×3×7

54 = 2×3×3×3

LCM = 24×33×7 = 3024

HCM = 2×3 = 6

product of the two numbers = 336×54 = 18144

HCF×LCM = 6×3024 = 18144

product of the two numbers = LCM × HCF

 

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

Solution:

12 = 2×2×3

15 = 3×5

21 = 3×7

HCF = 3

LCM = 22×3×5×7 = 420

 

(ii) 17, 23 and 29

Solution:

17 = 17×1

23 = 23×1

29 = 29×1

HCF = 1

LCM = 17×23×29 = 11339

 

(iii) 8, 9 and 25

Solution:

8 = 2×2×2×1

9 = 3×3×1

25= 5×5×1

HCF = 1

LCM = 2×2×2×3×3×5×5 = 1800

 

4. Given that HCF(306, 657) = 9, find LCM(306, 657).

Solution:

HCF(306, 657) = 9

We know that,

LCM × HCF = product of the two numbers

Real Numbers

LCM(306, 657) = 22338



5. Check whether 6n can end with the digit 0 for any natural number n.

Solution:

If a number ends on 0, it is divisible by 2 and 5 (eg- 10)

The prime factor of 6n = (2×3)n

There is no 5 in the prime factorization of 6n. So 6n will not be divisible by 5.

Hence, for any natural number n, the number 6n cannot end with the digit 0.

 

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

Solution:

First Number = 7×11×13+13

= 13×(7×11×1+1)

= 13×(77+1)

= 13×78

= 13×13×6

This number has more than two factors (1, 6, 13) so it is a composite number, because composite numbers have more than two factors.

Second Number = 7×6×5×4×3×2×1+5

= 5×(7×6×4×3×2×1+1)

= 5×(1008+1)

= 5×1009

This number has more than two factors (1, 5, 1009) so it is a composite number.

 

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

Solution:

Sonia takes 18 minutes while Ravi takes 12 minutes, after a time both will be at their starting positions that time will be the LCM of 12 and 18.

18 = 2×3×3

12 = 2×2×3

LCM(18, 12) = 2×2×3×3 = 36

Hence, Ravi and Sonia will meet again at the starting place after 36 minutes.


Real Numbers Exercise – 1.2 NCRT Download pdf