NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2 - Real Numbers

Q1 (1) Express each number as a product of its prime factors: 140

Answer:

By taking LCM of the number 140 we can express it as a product of its prime factors, like as follows:

$140=2\times 2\times 5\times 7$

$140=2^{2}\times 5\times 7$

Q 1 (2) Express each number as a product of its prime factors: 156

Answer:

By taking LCM of the number 156 we can express it as a product of its prime factors, like as follows:

$156=2\times 2\times 3\times 13$

$156=2^{2}\times 3\times 13$

Q1 (3) Express each number as a product of its prime factors: 3825

Answer:

By taking LCM of the number 3825 we can express it as a product of its prime factors, like as follows:

$3825=3\times 3\times 5\times 5\times 17$

$3825=3^{2}\times 5^{2}\times 17$

Q1 (4) Express each number as a product of its prime factors: 5005

Answer:

By taking LCM of the number 5005 we can express it as a product of its prime factors, like as follows:

$5005=5\times 7\times 11\times 13$

Q1 (5) Express each number as a product of its prime factors: 7429

Answer:

By taking LCM of the number 7429 we can express it as a product of its prime factors, like as follows:

$7429=17\times 19\times 23$

Q2 (1) Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers: 26 and 91

Answer:

Express the numbers as the product of their prime numbers.

26 = 2 × 13

91 = 7 × 13

HCF(26, 91) = 13

LCM(26, 91) = 2 × 7 × 13 = 182

Verification; HCF × LCM = 13 × 182 = 2366

26 × 91 = 2366

26 × 91 = HCF × LCM

Hence, verified.

Q2 (2) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 510 and 92

Answer:

The number can be expressed as the product of prime factors as

510 = 2 × 3 x 5 × 17

92 = 22 × 23

HCF (510, 92) = 2

LCM (510, 92) = 22 × 3 x 5 × 17 × 23 = 23460

Verification; HCF x LCM = 2 × 23460 = 46920

510 × 92 = 46920

510 × 92 = HCF × LCM

Hence, verified

Q2 (3) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 336 and 54

Answer:

Express the numbers as the product of their prime numbers.

336 = 24 × 3 × 7

54 = 2 × 33

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

LCM(336,54) = 24 × 33 × 7 = 3024

Verification; HCF × LCM = 6 × 3024 = 18144

336 × 54 = 18144

336 × 54 = HCF × LCM

Hence, verified

Q3 (1) Find the LCM and HCF of the following integers by applying the prime factorization method. 12, 15 and 21

Answer:

The numbers can be written as the product of their prime factors as follows

12 = 22 x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 22 x 3 x 5 x 7 = 420

Q3 (2) Find the LCM and HCF of the following integers by applying the prime factorisation method. 17, 23 and 29

Answer:

The given numbers are written as the product of their prime factors as follows

17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

HCF = 1

LCM = 17 × 23 × 29 = 11339

Q3 (3) Find the LCM and HCF of the following integers by applying the prime factorization method. 8, 9 and 25

Answer:

The given numbers are written as the product of their prime factors as follows

8 = 23

9 = 32

25 = 52

HCF = 1

LCM = 23 × 32 × 52 = 1800

Q4 Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer:

We know: HCF × LCM = a × b

So, HCF (306, 657) × LCM (306, 657) = 306 × 657

$LCM (306, 657) = \frac{306\times 657}{HCF (306, 657) }$

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

$LCM (306, 657) =22338$

Q5 Check whether $6 ^n$ can end with the digit 0 for any natural number n.

Answer:

By prime factorisation, we have

6n = 2n × 3n

If a number's prime factorisation includes at least 1 as the power of both 2 and 5, then the number will conclude with 0. We can infer that, for every value of n, 6n will terminate with the number 0 since the prime factorisation of 6n shows that the power of 5 equals 0.

Q6 Explain why 7 $\times$ 11 $\times$ 13 + 13 and 7 $\times$ 6 $\times$ 5 $\times$ 4 $\times$ 3 $\times$ 2 $\times$ 1 + 5 are composite numbers.

Answer:

7 × 11 × 13 + 13

= (7 × 11 + 1) × 13

= 78 × 13 or 2 × 3 × 132

Similarly, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= (7 × 6 × 4 × 3 × 2 × 1 + 1) × 5

= 5 × 1008

The number rule states that we can take at least two common numbers out of two, and after solving, we saw that both numbers were even, as a result of which the number is composite.

Q7 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?

Answer:

The time after which they meet again at the starting point will be equal to the LCM of the times they individually take to complete one round.

Time taken by Sonia = 18 = 2 × 32

Time taken by Ravi = 12 = 22 × 3

LCM (18, 12) = 22 × 32 = 36

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