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

Real Numbers Class 10 Chapter 1 - Exercise 1.2

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

Answer:

The number can be as a product of its prime factors as follows

$$\begin{gathered} 140=2\times 2\times 5\times 7 \\ 140=2^2\times 5\times 7 \end{gathered}$$

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

Answer:

The given number can be expressed as follows

$$\begin{gathered} 156=2\times 2\times 3\times 13 \\ 156=2^2\times 3\times 13 \end{gathered}$$

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

Answer:

The number is expressed as the product of the prime factors as follows

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

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

Answer:

The given number can be expressed as the product of its prime factors as follows.

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

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

Answer:

The given number can be expressed as the product of their prime factors 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:

26 = 2 x 13

91 = 7 x 13

HCF(26,91) = 13

LCM(26,91) = 2 x 7 x 13 = 182

HCF x LCM = 13 x 182 = 2366

26 x 91 = 2366

26 x 91 = HCF x 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 x 3 x 5 x 17

92 = 2 ² x 23

HCF(510,92) = 2

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

HCF x LCM = 2 x 23460 = 46920

510 x 92 = 46920

510 x 92 = HCF x 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:

336 is expressed as the product of its prime factor as

336 = 2 ⁴ x 3 x 7

54 is expressed as the product of its prime factor as

54 = 2 x 3 ³

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

LCM(336,54) = 2 ⁴ x 3 ³ x 7 = 3024

HCF x LCM = 6 x 3024 = 18144

336 x 54 = 18144

336 x 54 = HCF x 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 = 2 ² x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 2 ² 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 x 17

23 = 1 x 23

29 = 1 x 29

HCF = 1

LCM = 17 x 23 x 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 = 2 ³

9 = 3 ²

25 = 5 ²

HCF = 1

LCM = 2 ³ x 3 ² x 5 ² = 1800

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

Answer:

As we know the product of HCF and LCM of two numbers is equal to the product of the two numbers we have

HCF (306, 657) x LCM (306, 657) = 306 x 657

$$\begin{gathered} LCM(306,657)=\frac{306\times 657}{HCF(306,657)} \\ LCM(306,657)=\frac{306\times 657}{9} \\ LCM(306,657)=22338 \end{gathered}$$

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

Answer:

By prime factorizing we have

6 ⁿ = 2 ⁿ x 3 ⁿ

A number will end with 0 if it has at least 1 as the power of both 2 and 5 in its prime factorization. Since the power of 5 is 0 in the prime factorization of 6 ⁿ we can conclude that for no value of n 6 ⁿ will end with the digit 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 x 11 x 13 + 13

= (7 x 11 + 1) x 13

= 78 x 13

= 2 x 3 x 13 ²

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

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

= 5 x 1008

After Solving we observed that both the number are even numbers and the number rule says that we can take atleast two common out of two numbers. So that the number is composite number.

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 x 3 ²

Time taken by Ravi = 12 = 2 ² x 3

LCM(18,12) = 2 ² x 3 ² = 36

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