Exercise 1.2

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

(1) 140

Solution: by taking the LCM of 140, we will get the product of its prime factor.

Therefore, 140 = 2 x 2 x 5 x 7 x 1= 2² x 5 x 7

(2) 156

Solution: By taking the LCM of 156, we will get the product of its prime factor.

Hence, 156 = 2 x 2 x 13 x 3 x 1 = 2²x 13 x 2

(3) 3825

Solution: By taking the LCM of 38 25 we will get the product of its prime factor.

Hence, 3825 = 3 x 3 x 5 x 5 x 17 x 1= 3²x5²x 17

(4) 5005

Solution: By taking the LCM of 5005 we will get the product of its prime factor.

Hence, 5005 = 5 x 7 x 11 x 13 x 1

(5) 7429

Solution: By taking the LCM of 7429 we will get the product of its prime factor.

Hence, 7429 = 17 x 19 x 23 x 1 = 17 x 19 x 23

Question 2. Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = Product of the two numbers.

(1) 26 and 91

Solution: expressing 26 and 91 as product of its prime factors, we get,

26 = 2 x 13 x 1

91 = 7 x 13 x 1

therefore LCM 26 and 91= 2 x 7 x 13 x 1 = 182

And HCF = (26 , 91) = 13

Verification

Now , Product of 26 and 91 = 26 x 91 = 2366

And product of LCM and HCF = 182 x 13 = 2366

Hence, LCM x HCF = product of the 26 and 91.

(2) 510 and 92

Solution: Expressing 510 and 92 as product of its prime factors, we get,

510 = 2 x 3 x 17 x 5 x 1

92 = 2 x 2 x 23 x 1

LCM (510 and 92) = 2 x 2 x 3 x 5 x 17 x 23 = 23460

And HCF = (510, 92) = 2

Verification

Now, product of 510 and 92 = 46920

And product of LCM and HCF = 23460 x 2 = 46920

Hence, LCM x HCF = product of the 510 and 92.

(3) 336 and 54

Solution: Expressing 336 and 54 as product of its prime factors. we get,

336 = 2 x 2 x 2 x 2 x 7 x 3 x 1

54 = 2 x 3 x 3 x 3 x 1

Therefore , LCM (336 , 54) = 3024

And HCF (336, 54) = 2 x 3 = 6

Verification

Now product of 336 and 54 = 36 x 54= 18,144

Hence , LCM x HCF = Product of the 336 and 54.

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

(1) 12, 15 and 21

Solution:

12,15 and 21

12 = 2 x 2 x 3

15 = 5 x 3

21 = 7 x 3

Therefore, (12,15 and 21) = 3

LCM (12, 15 and 21) = 2 x 2 x 3 x 5 x 7 = 420

(2) 17,23 and 29

Solution:

17 = 17 x 1

23 = 23 x 1

29 = 29 x 1

Therefore,

HCF (17,23,29) = 1

LCM (17,23,29) = 17 x 23 x 29 = 11339

(3) 8,9 and 25

Solution:

8 = 2 x 2 x 2 x 1

9 = 3 x 3 x 1

25 = 5 x 5 x 1

Therefore

HCF (8,9,25) = 1

LCM (8,9,25) = 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800

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

Solution:

As we know that

LCM x HCF = product of the two given numbers

Therefore,

9 x LCM = 306 x 657

LCM = (306 x 657)/9

LCM = 22338.

Question 5. Check weather 6 power n end with the digit 0 for any natural number n.

Solution:

Is the number 6 power and and with dead target zero (0), then it should be divisible by 5 as we know any number with unit place as zero or 5 is divisible by 5.

Prime factorization of 60 power n = (2 x 3)power n

Therefore, the prime factorization of 6 power and doesn't contain prime number 5.

Hence it is clear that for any natural number n 6 power and is not divisible by 5 and thus it proves that 6 power n cannot end with the digit 0 for any natural number n.

Question 6. 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.

Solution:

By the definition of composite number we know if a number is composite then it means it has factors other than one and itself therefore for the given expression.

7 x 11 x 13 + 13

Taking 13 as common factor we get

= 13(7 x 11 x 1+ 1) = 13(77+1) = 13 x 78 = 13 x 2 x 3 x13

Hence 7 x 11 x 13 + 13 is a composite number.

Now let's take the other number

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

Taking five as a common factor we get

= 5 ( 7 multiply 6 x 4 x 3 x 2 x 1 + 1) = 5 ( 1008 + 1) = 5 x 1009

Hans 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 is a composite number.

Question 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 the 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:

since both Soni and Ravi move in the same direction and at the same time the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.

Therefore LCM (18,12) = 2 x 3 x 3 x 2 x 1 = 36

Hans Sonia and Ravi will meet again at the starting point after 36 minutes.

We hope CBSE/MP Board Solution of Class 10th Chapter 1 "Real Numbers" Exercise 1.2 will help you.

Written By - Himanshu Sharma