**KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2.

## Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2

Question 1.

Express each number as a product of its prime factors :

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solution:

Question 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

a = 26, b = 91

26 = 2 × 13

91 = 7 × 13

LCM of (26, 91) = 2 × 13 × 7 = 182

HCF of (26, 91) = 13

Verification

LCM of (26, 91) × H C F (26, 91) = 182 × 13 = 2366

product of two numbers = 26 × 91 = 2366

Hence LCM × HCF = Product of two numbers.

(ii) 510 and 92

a = 510, b = 92

(iii) 336 and 54

a = 336, b = 54

Question 3.

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

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

Solution:

(i) 12, 15 and 21

12 = 2 × 2 × 3

15 = 3 × 5

21 = 3 × 7

∴ H.C.F. = 3

L.C.M. = 3 × 2 × 2 × 5 × 7 = 420

∴ HCF × LCM = Product of three numbers

3 × 420 = 12 × 15 × 21

1260 = 1260

(ii) 17, 23 and 29

17 = 17 × 1

23 = 23 × 1

29 = 29 × 1

∴ H.C.F. = 1

L.C.M. = 17 × 23 × 29 = 11339

∴HCF × LCM = Product of three numbers

1 × 11339 = 17 × 23 × 29

11339 = 11339

(iii) 8 = 2 × 2 × 2 = 2^{3}

9 = 3 × 3 = 3^{2}

25 = 5 × 5 = 5^{2}

L C M of (8, 9, 25) = 2^{3} × 3^{2} × 5^{2}

= 8 × 9 × 25 = 1800

LCM of (8, 9, 25)= 1800

HCF of(8, 9, 25)= 1

(∵ 8, 9 & 25 havg no common Factor)

Question 4.

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

Solution:

Here, HCF of 306 and 657 = 9 and we have

LCM × HCF = Product of the numbers.

∴ LCM × 9 = 306 × 657

Thus, LCM of 306 and 657 is 22338.

Question 5.

Check whether 6^{n} can end with the digit 0 for any natural number ‘n’.

Solution:

If the number ends with digit 0, then the prime factorization of the number must contain the factors of 10 as 2 × 5.

Now the prime factorization of 6^{n} = (2 × 3)^{n} = 2^{n} × 3^{n}

Since it does not contain the factors of 10 as 2 × 5, thus ^{n} cannot end with the digit 0, for any natural number n.

Question 6.

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

Solution:

We have

7 × 11 × 13 + 13 = 13[(7 × 11) + 1] = 13[78]

i. e., 13 × 78 cannot be a prime number because it has factors 13 and 78.

∴ 13 × 78 is a composite number.

Also 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5[7 × 6 × 4 × 3 × 2 × 1 + 1] which is also not a prime number because it has a factor 5

Thus, 7 × 11 × 13 + 13 and

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

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 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

Solution:

Sonia takes 18 minutes to drive one round of the field.

Ravi takes 12 minutes for the same.

Suppose they both start at the same point and at the same time.

Time to meet both again at the starting point,

For this, we have to find out HCF of 18 and 12.

18 = 2 × 3 × 3 = 2 × 3^{2}

12 = 2 × 2 × 3 = 2^{2} × 3

∴ H.C.F. = 2 × 3 = 6

∴ After 6 minutes, they both meet at the same point.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2, drop a comment below and we will get back to you at the earliest.