Real number EXERCISE 1.2

  1. Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution:

 140 = 2 x 2 x 5 x 7 = 2² x 5 x 7

156 = 2 x 2 x 3 x 13 = 2² x 3 x 13

3825 = 3 x 3 x 5 x 5 x 17 = 3² x 5² x 17

5005 = 5 x 7 x 11 x 13

7429 = 17 x 19 x 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 (ii) 510 and 92 (iii) 336 and 54

Solution:

(i) 26 and 91

26 = 2 x 13

91 = 7 x 13

HCF = 13

LCM = 2 x 7 x 13 = 182

The product of two numbers = 91 x 26 = 2366

LCM x HCF = 2366

 

(ii) 510 and 92

510 = 2 x 3 x 5 x 17

92 = 2 x 2 x 23

HCF = 2

LCM = 2 x 2 x 3 x 5 x 17 x 23 = 23460

The product of two numbers = 510 x 92 = 46920

LCM x HCF = 46920

 

(iii) 336 and 54

336 = 2 x 2 x 2 x 2 x 3 x 7

91 = 2 x 3 x 3 x 3

HCF = 2 x 3 = 6

LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7 = 3024

The product of two numbers = 336 x 91 = 30576

LCM x HCF = 30576


3. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

Solution:

(i) 12, 15 and 21

12 = 2 x 2 x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 3 x 2 x 2 x 5 x 7 = 420

 

(ii) 17, 23 and 29

17 = 1 x 17

23 = 1 x 23

29 = 1 x 29

HCF = 1

LCM = 17 x 23 x 29 =11339

 

(iii) 8, 9 and 25

8 = 2 x 2 x 2 x 1

9 = 3 x 3 x 1

25 = 5 x 5 x 1

HCF = 1

LCM = 8 x 9 x 25 = 1800


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

Solution:

HCF(306, 657) = 9.

We know, the product of two numbers = LCM x HCF

LCM X HCF = 306 x 657

Given HCF = 9

LCM = (306×657)/HCF = (306×657)/9 = 22338.


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

Solution:

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of 6n = (2 ×3)n

It can be observed that 5 is not in the prime factorisation of 6n . Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.


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

Solution:

Numbers are of two types – prime and composite.

Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 13 ×13 × 6

The given expression has 6 and 13 as its factors.

Therefore, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 ×1009 1009 cannot be factorised further.

Therefore, the given expression has 5 and 1009 as its factors. Hence, 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:

It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path.

As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e.,LCM of 18 minutes and 12 minutes.

18 = 2 × 3 × 3 And, 12 = 2 × 2 × 3 LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36 Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes. 



 

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