RBSE Solutions for Class 10 Maths Chapter 2 Real Numbers Ex 2.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Exercise 2.2.

Board |
RBSE |

Te×tbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Real Numbers |

E×ercise |
Exercise 2.2 |

Number of Questions Solved |
5 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Ex 2.2

Question 1.

E×press each number as a product (RBSESolutions.com) of its prime factors:

(i) 468

(ii) 945

(iii) 140

(iv) 3825

(v) 20570

Solution

(i) 468

By prime factorization of number 468.

So, 468 = 2 × 2 × 3 × 3 × 13 = 2^{2} × 3^{2} × 13

Second method: By prime factorisation we get

So, 468 = 2 × 2 × 3 × 3 × 13 = 2^{2} × 3^{2} × 13

(ii) 945

By prime factorisation, (RBSESolutions.com) we get

So, 945 = 3 × 3 × 3 × 5 × 7 = 3^{3} × 5 × 7

(iii) 140

By prime factorisation, we get

So, 140 = 2 × 2 × 5 × 7 = 2^{2} × 5 × 7

(iv) 3825

By prime factorisation, we get

So, 3825 = 3 × 3 × 5 × 5 × 17 = 3^{2} × 5^{2} × 17

(v) 20570

By prime factorisation, (RBSESolutions.com) we get

So, 20570 = 2 × 5 × 11 × 11 × 17 = 2 × 5 × 11^{2} × 17

In this quick guide we’ll describe what the factor pairs of 96 are, how you find them and list them out for you to prove the calculation works.

Question 2.

Find the LCM and HCF of the following pairs of integers and verify that HCF × LCM = Product of the two numbers

(i) 96 and 404

(ii) 336 and 54

(iii) 90 and 144

Solution

(i) 96 and 404

By prime factorization of (RBSESolutions.com) numbers 96 and 404, we get.

Hence 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2^{5} × 3

and 404 = 2 × 2 × 101 = 2^{2} × 101

For HCF, the smallest power of prime factor = (2)

Hence, HCF = 2 × 2 = 4

For LCM, take the highest power of prime factor

Hence L.C.M. = 2^{5} × 3^{1} × 101^{1} = 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696

**Verification:**

Product of both (RBSESolutions.com) integers = 96 × 404 = 38784

H.C.F. × L.C.M. = 4 × 9696 = 38784

Hence, H.C.F. × L.C.M. = product of both integers numbers.

(ii) 336 and 54

By prime factorisation of numbers 336 and 54.

So 336 = 2 × 2 × 2 × 2 × 3 × 7 = 2^{4} × 3 × 7

and 54 = 2 × 3 × 3 × 3 = 2 × 3^{3}

For H.C.F., we find minimum power of common prime factor.

H.C.F. = 2 × 3 = 6

For L.C.M., we find the highest power of the prime factor.

So, L.C.M. = 2^{4} × 3^{3} × 7 = 16 × 27 × 7 = 3024

**Verification:**

Product of both (RBSESolutions.com) integers numbers. = 336 × 54= 18144

H.C.F. × L.C.M. = 6 × 3024 = 18144

So H.C.F. × L.C.M. = Product of both integer numbers

(iii) 90 and 144

By prime factorisation of numbers 90 and 144.

So 90 = 2 × 3 × 3 × 5 = 2 × 3^{2} × 5

and 144 = 2 × 2 × 2 × 2 × 3 × 3 = 2^{4} × 3^{2}

For H.C.F. we find minimum power of common prime factor

H.C.F. = 2 × 3^{2} = 2 × 9 = 18

For L.C.M. we find highest power of common prime factor

So L.C.M. = 2^{4} × 3^{2} × 5 = 16 × 9 × 5 = 720

**Verification:**

Product of both integer (RBSESolutions.com) number = 90 × 144 = 12960

HCF × LCM = 18 × 720 = 12960

So HCF × LCM = product of both integers numbers

Question 3.

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

(i) 12, 15 and 21

(ii) 24, 15 and 36

(iii) 17, 23 and 29

(iv) 6, 72 and 120

(v) 40, 36 and 126

(vi) 8, 9 and 25

Solution

(i) 12, 15 and 21

12 = 2 × 2 × 3 = 2^{2} × 3

15 = 3 × 5

and 21 = 3 × 7

For HCF, we find minimum (RBSESolutions.com) power of prime factor

H.C.F. = (3)^{1} = 3

For LCM, taking maximum power of prime factors

L.C.M. = 2^{2} × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420

So H.C.F. = 3

L.C.M. = 420

(ii) 24, 15 and 36

24 = 2 × 2 × 2 × 3 = 2^{3} × 3

15 =3 × 5

and 36 = 2 × 2 × 3 × 3 = 2^{2} × 3^{2}

For HCF, taking minimum power of common prime factor

H.C.F. = (3)^{1} = 3

For LCM, taking maximum (RBSESolutions.com) power of prime factor

L.C.M. = 2^{3} × 3^{2} × 5^{1} = 8 × 9 × 5 = 360

So H.C.F. = 3

L.C.M. = 360

(iii) 17, 23 and 29

17 = 1 × 17

23 = 1 × 23

and 29 = 1 × 29

For HCF, common factor is 1

HCF = 1

For LCM taking maximum power of prime factor.

L.C.M. = 1 × 17 × 23 × 29 = 11339

So H.C.F. = 1

L.C.M. = 11339

(iv) 6, 72 and 120

6 = 2 × 3

72 = 2 × 2 × 2 × 3 × 3 = 2^{3} × 3^{2}

and 120 = 2 × 2 × 2 × 3 × 5 = 2^{3} × 3 × 5

For HCF, taking minimum power (RBSESolutions.com) of common prime factor

HCF = 2 × 3

For LCM taking maximum power of prime factor

L.C.M. = 2^{3} × 3^{2} × 5 = 8 × 9 × 5 = 360

So H.C.F. = 6

L.C.M. = 360

(v) 40, 36 and 126

40 = 2 × 2 × 2 × 5 = 2^{3} × 5

36 = 2 × 2 × 3 × 3 = 2^{2} × 3^{2}

and 126 = 2 × 3 × 3 × 7 = 2 × 3^{2} × 7

For HCF, taking minimum power of common factor.

H.C.F. = (2)^{1} = 2

For LCM, taking maximum (RBSESolutions.com) power of prime factors.

L.C.M. = 2^{3} × 3^{2} × 5 × 7 = 8 × 9 × 5 × 7 = 2520

Hence, H.C.F. = 2

L.C.M. = 2520

(vi) 8, 9 and 25

8 = 2 × 2 × 2 = 2^{3}

9 = 3 × 3 = 3^{2}

and 25 = 5 × 5 = 5^{2}

For HCF common factor is 1

H.C.F. = 1

For LCM, taking maximum (RBSESolutions.com) power of prime factors

L.C.M. = 2^{3} × 3^{2} × 5^{2} = 8 × 9 × 25 = 1800

So H.C.F. = 1

L.C.M. = 1800

Question 4.

There is a circular path around a sports field. (RBSESolutions.com) Raman takes 18 minutes to drive one round of the field while Anupriya 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 is taken by Raman to complete one round of field = 18 min.

Time is taken by Anupriya to complete one round of field = 12 min.

They will meet again at staring point = L.C.M. (18, 12)

Prime factors of 18 and 12

18 = 2 × 3 × 3 = 2 × 3^{2}

12 = 2 × 2 × 3 = 2^{2} × 3

L.C.M. (18, 12) = 2^{2} × 3^{2} = 4 × 9 = 36

Hence, Raman and Anupriya will be meet after 36 minute at starting point.

Question 5.

In a seminar number of participants in Hindi, (RBSESolutions.com) English and Mathematics are 60, 84 and 108 respectively. If the equal number of participants of the same subject are sitting in each room, then find the least number of required rooms.

Solution

Since the least numbers of a room are required.

So, number of participants in each room must be the H.C.F. of 60, 84, and 108.

60 = 2^{2} × 3 × 5

84 = 2^{2} × 3 × 7

and 108 = 2^{2} × 3^{3}

HCF of 60, 84 and 108 = 2^{2} × 3 = 4 × 3 = 12

Therefore, in each room 12 participants can be seated

We hope the given RBSE Solutions for Class 10 Maths Chapter 2 Real Numbers Ex 2.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Exercise 2.2, drop a comment below and we will get back to you at the earliest.

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