Gujarat Board GSEB Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Question 1.

Express each number as a product of its prime

- 140
- 156
- 3825
- 5005
- 7429

Solution:

1.

Prime factorisation of

140 = 2 Ã— 2 Ã— 5 Ã— 7

2.

Prime factorisation of

156 = 2 Ã— 2 Ã— 3 Ã— 13

3.

Prime factorisation of

3825 = 3 Ã— 3 Ã— 5 Ã— 5 Ã— 17

4.

Prime factorisation of

5005 = 5 Ã— 7 Ã— 11 Ã— 13

5.

Prime factorisation of

7429 = 17 Ã— 19 Ã— 23

Question 2.

Find the LCM and HCF of the following pairs of integers and verify that LCM Ã— HCF = Product of the two numbers.

- 26 and 91
- 510 and 92
- 336 and 54

Solution:

1. Prime factorisation of

26 = 2 Ã— 13

Prime factorisation of

91 = 7 Ã— 13

HCF of 26 and 91 = 13

LCM = 2 Ã— 7 Ã— 13 = 182

LCM Ã— HCF =182 Ã— 13

= 2366

Product of two number

= 26 Ã— 91 = 2366

âˆ´ HCF Ã— LCM = Product of two number.

2. Prime factorisation of

510 = 2 Ã— 3 Ã— 5 Ã— 17

Prime factorisation of

92 = 2 Ã— 2 Ã— 23

HCF of 510 and 92 = 2

LCM of 510 and 92

= 2 Ã— 2 Ã— 3 Ã— 5 Ã— 17 Ã— 23

= 23460

LCM Ã— HCF = 23460 Ã— 2

= 46920

Product of two numbers

= 510 Ã— 92

= 46920 = LCM Ã— HCF

3. Prime factorisation of

336 = 2Ã— 2Ã— 2 Ã— 2 Ã— 3 Ã— 7

Prime factorisation of

54 = 2 Ã— 3 Ã— 3 Ã— 3

HCF of 336 and 54 = 2 Ã— 3 = 6

LCM 336 and 54

= 2Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 7

= 3024

LCM Ã— HCF = 3024 Ã— 6 = 18144

Product of two number 3 9

= 336 Ã— 54 = 18144

= LCM Ã— HCF

Question 3.

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

- 12, 15 and 21
- 17, 23 and 29
- 8, 9 and 25

Solution:

1. Prime factorisation of 12 = 2 Ã— 2 Ã— 3

Prime factorisation of 15 = 3 Ã— 5

Prime factorisation of 21 = 3 Ã— 7

HCF of 12, 15 and 21 = 3

LCM of 12, 15 and 21

= 2 Ã— 2 Ã— 3 Ã— 5 Ã— 7 = 420

2. Prime factorisation of 17 = 17 Ã— 1

Prime factorisation of 23 = 23 Ã— 1

Prime factorisation of 29 = 29 Ã— 1

HCF of 12, 15, 29 = 1

LCM of 17, 23, 29 = 17 Ã— 23 Ã— 29

= 11339

3. Prime factorisation of 8 = 2 Ã— 2 Ã— 2

Prime factorisation of 9 = 3 Ã— 3

Prime factorisation of 23 = 23 Ã— 1

HCF of 8, 9 and 23 = 1

LCM of 8, 9, 23 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 23

= 1656

Question 4.

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

Solution:

Given that HCF (306, 657) = 9

LCM Ã— HCF = Product of two numbers

LCM Ã— 9 = 306 Ã— 657

LCM = \(\frac{306 \times 657}{9}\)

LCM = 34 Ã— 657

LCM = 22338

âˆ´ LCM of (306, 657) = 22338.

Question 5.

Check whether 6^{n} can end with the digit 0 for any natural number n.(CBSE)

Solution:

6^{n} will end with the digit zero if 6^{n} is divisible by 2 and 5.

But 6^{n} = (2 Ã— 3)^{n} = 2^{n} Ã— 3^{n}

i.e. in the factorisation of 6^{n}, no factor is of 5. Therefore by the fundamental theorem of arithmetic every composite number can be expressed a product of primes and this factorisation is unique apart from the order in which the prime factors occur.

Therefore our assumption is wrong that 6^{n} ends in zero, thus there does not exist any natural number n for which 6^{n} ends with zero.

Question 6.

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

Solution:

Given that 7 Ã— 11 Ã— 13 + 13

= (7 Ã— 11 Ã— 1 + 1) Ã— 13

= (77 + 1) Ã— 13

= 78 Ã— 13

Composite number because it is product of more than two factors.

7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 + 5

= (7 Ã— 6 Ã— 4 Ã— 3 Ã— 2 Ã— 2 + 1) Ã— 5

= (1008 + 1) Ã— 5

= 1009 Ã— 5

Product of more than two factors, which 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 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:

Time taken in one round by Sonia = 18 minute Time taken by Ravi in one round = 12 minute Now we find LCM of 18 and 12, which gives exact number of minutes after which they meet at the starting point again.

Prime factorisation of

18 = 2 Ã— 3 Ã— 3

Prime factorisation of

12 = 2 Ã— 2 Ã— 3

LCM of 12 and 18

= 2 Ã— 2 Ã— 3 Ã— 3

= 36 minutes

Therefore they will meet again at the starting point after 36 minutes.