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

Board |
RBSE |

Te×tbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Real Numbers |

E×ercise |
Additional Questions |

Number of Questions Solved |
20 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Additional Questions

**Objective Type Questions**

Question 1.

Prime factor (RBSESolutions.com) of 4050 is

(A) 2 x 3^{5} x 5

(B) 2 x 3^{4} x 5

(C) 2 x 3^{4} x 5^{2}

(D) 2 x 3^{4} x 5^{3}

Solution

4050 = 2 x 3^{4} x 5^{2}

Hence, correct choice is (C).

Question 2.

For any integer m, the form of each even integer is

(A) m

(B) m + 1

(C) 2m

(D) 2m + 1.

Solution

Correct choice (C).

Question 3.

Which of the following (RBSESolutions.com) rational number. **[CBSE 2012]**

Solution

The correct choice is (C).

Question 4.

If p is a prime number then LCM of p, p^{2}, and p^{3} **[CBSE 2012]**

(A) p

(B) p^{3}

(C) p^{2}

(D) p^{6}

Solution

The correct choice is (B).

Question 5.

If p^{2} is an even number (RBSESolutions.com) then p also a/an

(A) odd number

(B) even number

(C) even and odd number

(D) none of there.

Solution

The correct choice is (B).

Question 6.

If number n^{2} – 1, divides by 8 then n is an **(NCERT Exemplar Problem)**

(A) integer

(B) natural number

(C) odd number

(D) even number

Solution

The correct choice is (C).

Question 7.

If the HCF of 65 and 117 can be (RBSESolutions.com) express in the form 65m -117, the value of m in **(NCERT Exemplar Problem)**

(A) 4

(B) 2

(C) 1

(D) 3

Solution

65 = 5 x 13

117 = 3 x 3 x 13

H.C.F. = 13

⇒ 65m – 117 = 13 [∴ H.C.F. = 65m – 117]

⇒ 65m = 13 + 117

⇒ 65m = 130

⇒ m = 2

Hence, correct choice is (B).

Question 8.

The product of a non-zero rational number (RBSESolutions.com) and an irrational number is **(NCERT Exemplar Problem)**

(A) always an irrational number

(B) always a rational number

(C) a rational or irrational number

(D) one

Solution

The correct choice is (A).

Question 9.

If p is a positive prime number then √p.

(A) irrational number

(B) rational number

(C) integer number

(D) none of these

Hence, the correct choice is (A).

Question 10.

If L.C.M and H.C.F of two rational numbers (RBSESolutions.com) are the same, the number will be

(A) prime

(B) co-prime

(C) composite

(D) equal

Solution

The correct choice is (D)

Question 11.

If two positive intergersp and q are (RBSESolutions.com) expressible in the form p = a^{2}b^{3} and q = ab, where a and b are co-prime, then LCM (p, q) will be

(A) ab

(B) a^{2}b^{2}

(C) a^{2}b^{3}

(D) a^{3}b^{3}

Solution

p = a^{2}b^{3}

q = ab

L.C.M. = a^{2}b^{3}

So, correct choice is (C)

Question 12.

If HCF and LCM of two numbers are 6 and 60 respectively, (RBSESolutions.com) then those numbers are

(A) 10 and 36

(B) 24 and 15

(C) 12 and 30

(D) 18 and 20.

Solution

(i) HCF of 10 and 36 = 2

(ii) HCF of 24 and 15 = 3

(iii) HCF of 12 and 30 = 6

(iv) HCF of 18 and 20 = 2.

Hence correct choice is (C).

So, L.C.M. x H.C.F. = 60 x 6 = 360

First number x second number = 12 x 30 = 360.

Question 13.

HCF of three (RBSESolutions.com) numbers 17, 23 and 29

(A) 1

(B) 2

(C) 3

(D) 4

Solution

Given numbers are 17, 23 and 29 by prime factorisation.

17 = 1 x 17, 23 = 1 x 23 and 29 = 1 x 29

Product of common prime factor (in minimum powers)

H.C.F. = 1.

So, the correct choice is (A).

Question 14.

If a rational number x = \(\frac { p }{ q }\) such as(RBSESolutions.com) a prime factor of q is not the form of 2^{n} x 5^{m} where n, m is a non-zero integer, then the decimal expansion of x is

(A) terminating decimal expansion

(B) non-terminating repeating decimal

(C) choice (A) and (B) both correct

(D) none of there

Solution

Hence, the correct choice is (B).

Question 15.

First Greek mathematician was

(A) AL-knwarijmi

(B) Aryabhata

(C) Einstein

(D) Euclid

Solution

The correct choice is (D).

**Very Short/Short Answer Type Questions**

Question 1.

Find the HCF of 96 and 404 by (RBSESolutions.com) prime factorisation method. [Raj. 2015]

Solution

96 = 2 x 2 x 2 x 2 x 2 x 3

404 = 2 x 2 x 101

Common prime factor

H.C.F. = 2 x 2 = 4.

Question 2.

Find the HCF of numbers 44 and 99 [Raj. 2013]

Solution

44 = 2 x 2 x 11= 2^{2} x 11

and 99 = 3 x 3 x 11 = 3^{2} x 11

Now product of prime factors of both numbers = 11

(H.C.F.) = 11.

Question 3.

Is the HCF of two numbers 15 and LCM 175 possible? Give reason. **[CBSE 2012]**

Solution

We know that LCM divides by HCF.

But LCM 175 does not divide by HCF 15.

Hence, HCF and LCM of two numbers can not 15 and 175 respectively

Question 4.

Whether the decimal expansion of (RBSESolutions.com) number \(\frac { 3 }{ 625 }\) is terminating or non-terminating 625 repeating? write it in decimal form? **[Raj. 2014]**

Solution

Question 5.

In a school, there are two sections- (RBSESolutions.com) section A and section B of class X. There are 48 students in section A and 60 students in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of each section. **(CBSE 2012)**

Solution

Number of students in section A = 48

Number of students in section B = 60

Minimum number of books in library = LCM of 48 and 60

So LCM of 48, 60 = 2 x 2 x 2 x 2 x 3 x 5 = 240

So, number of required books = 240.

Question 6.

Prove that 3 – √5 is an irrational number? **(CBSE 2012)**

Solution

Let 3 – √5 is a rational number, (RBSESolutions.com) we find out such integers numbers a and b (b ≠ 0)

a and b are integer numbers So, \(\frac { 3b-a }{ b }\) is a rational number.

So, √5 be also a rational number

But this fact is contradicted that √5 is an irrational number.

So our hypothesis is wrong.

3 – √5 is an irrational number.

Hence proved.

Question 7.

Prove that 7 – √5 is an irrational (RBSESolutions.com) number.

Solution

Let 7 – √5 is a rational number.

7 + √5 = \(\frac { a }{ b }\), b ≠ 0 …(i)

Where a and b co-prime integer number.

Equation (i) can be written as

√5 = \(\frac { a }{ b }\) – 7

or √5 = \(\frac { a-7b }{ b }\) ….(ii)

Since, a and b are integers. So \(\frac { a-7b }{ b }\) will be rational number, so from equation (ii) we find that √5 is a rational number. But we know that √5 is a irrational number.

So this result is contradicted.

So our hypothesis is wrong.

Hence 7 + √5 is a rational number.

Hence proved.

Question 8.

Find the HCF and LCM of two (RBSESolutions.com) numbers 3364 and 54.

Solution

Number 3364 and 54 By prime factorization of 3364 and 54.

3364 = 2 x 2 x 29 x 29 = 2^{2} x 29^{2}

and 54 = 2 x 3 x 3 x 3 = 2 x 3^{3}

For HCF, taking maximum power of common factor.

H.C.F. = 2

For LCM, taking maximum power of common factor.

L.C.M. = 2^{2} x 3^{3} x 29^{2} = 4 x 27 x 841 = 90,828

**Long Answer Type Questions**

Question 1.

Prove that only one out of three (RBSESolutions.com) consecutive positive integers is divisible. **(NCERT Exemplar Problem)**

Solution

Let three consecutive positive integers are n, n + 1, n + 2 respectively.

We know that n is of the form 3q or 3q + 1 or 3 q + 2.

Now following cases are possible.

**Case 1.**

when n = 3q which is divisible by 3

n + 1 = 3q + 1, Not divisible by 3

n + 2 = 3q + 2, Not divisible by 3

In this case n is divisible by 3 but (n + 1) and (n + 2) are not divisible by 3

**Case 2.**

When n = 3q + 1

In this positive Which is divisible by 3 but n and n + 1 are not divisible by 3.

**Case 3.**

When n = 3q + 2

In this position divisible by 3 but or (n + 2) are not divisible by 3

Hence out of n, (n + 1) and (n + 2) one is divisible by 3.

Hence proved.

Question 2.

Show that cube of any positive (RBSESolutions.com) integer is either of the form 4m, 4m + 1 or 4m + 3 for some integer m. **(NCERT Exemplar Problem)**

Solution

Let x is a positive integer.

Applying Euclid division lemma it is in the form of 4q or 4q + 1 or 4q + 1 or 4q + 2 or 4q + 3.

Now following cases are

where m = 16q^{3} + 36q^{2} + 27q + 6

Hence cube of any positive integer is either (RBSESolutions.com) of the form 4m, 4m + 1, or 4m + 3 for some integer.

Hence proved.

Question 3.

Find the LCM and HCF of the given pair of integers and check product of two numbers = H.C.F x L. C.M

(i) 26 and 91,

(ii) 510 and 92,

Solution

(i) Given numbers are 26 and 91

Product of factors of common (RBSESolutions.com) primes (in minimum powers) 26 and 91 = (13)^{1} = 13

The product of all prime factors of 26 and 91 in maximum powers.

Hence, (H.CF.) = 13

and (L.C.M.) = 182

Product of both numbers = 26 x 91 = 2366

and H.C.F. x L.C.M. = 13 x 182 = 2366

Now product of both number = H.C.F. x L.C.M.

(ii) Given numbers are 510 and 92

510 = 2 x 3 x 5 x 17 = (2) x (3) x (5) x (17)

Product of common prime (RBSESolutions.com) factors of 92 and 510 = (2)^{1} = 2.

So, H.C.F. = 2

Product of both numbers = (2)^{2} x (3) x (5) x (17) x (23) = 23460

So, L.C.M. = 23460

Product of both numbers = 92 x 510 = 46920

and H.C.F. x L.C.M. = 2 x 23460 = 46920

So, product of both number = H.C.F. x L.C.M.

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