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

Board |
RBSE |

Te×tbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Real Numbers |

E×ercise |
Miscellaneous Exercise |

Number of Questions Solved |
20 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Miscellaneous Exercise

Question 1.

Sum of the power of prime (RBSESolutions.com) factors of 196

(a) 1

(b) 2

(c) 4

(d) 6

Solution

196 = 2^{2} x 7^{2}

Sum of powers =2 + 2 = 4

So, correct choice is (c).

Question 2.

If two numbers are written in the (RBSESolutions.com) form m = pq^{3} and n = p^{3}q^{2} then HCF of m, n whereas p, q prime numbers are:

(a) pq

(b) pq^{2}

(c) p^{2}q^{2}

(d) p^{3}q^{3}

Solution

Question 3.

HCF of 95 and 152 is:

(a) 1

(b) 19

(c) 57

(d) 38

Solution

95 = 5 x 19

and 152 = 2^{3} x 19

H.C.F. = 19

Hence, correct choice is (b).

Question 4.

Product of two is 1080 and (RBSESolutions.com) their HCF is 30 then their LCM is:

(a) 5

(b) 16

(c) 36

(c) 108

Solution

Product of both numbers = H.C.F. x L.C.M.

1080 = 30 x L.C.M

L.C.M = 36

Hence, correct choice is (c)

Question 5.

Decimal expansion of number \(\frac { 441 }{ { 2 }^{ 2 }\times { 5 }^{ 7 }\times { 7 }^{ 2 } }\) will be:

(a) Terminating

(b) Non-terminating repeating

(c) Terminating and non-terminating both

(d) Non-rational

Solution

Since denominator is a factor the form of 2^{m} x 5^{n}

So correct choice is (a)

Question 6.

In the decimal expansion of rational (RBSESolutions.com) number \frac { 43 }{ { 2 }^{ 2 }\times { 5 }^{ 3 } }, after how many digits decimal will end?

(a) 1

(b) 2

(c) 3

(d) 4

Solution

Question 7.

The least number, which multiplies by √27 gives a natural number, will be :

(a) 3

(b) √3

(c) 9

(d) 3√3

Solution

√27 = 3√3

After multiple √3 use finds a (RBSESolutions.com) natural number So, the correct choice is (b).

Question 8.

If HCF = LCM for two rational numbers, the numbers should be

(a) Composite

(b) Equal

(c) Prime

(d) Co-prime

Solution

The correct choice is (b)

Question 9.

If LCM of a and 18 is 36 and HCF of a and 18 is 2 then value of a will be:

(a) 1

(b) 2

(c) 5

(d) 4

Solution

Since two numbers (RBSESolutions.com) are and 18

H.C.F. = 2

L.C.M. = 36

Product of both = H.C.F. x L.C.M.

a x 18 = 2 x 36

a = 4

Hence, correct choice is (d).

Question 10.

If n is a natural number, then unit digit in 6^{n} – 5^{n} is:

(a) 1

(b) 6

(c) 5

(d) 9

Solution

The correct choice is (a).

Question 11.

If \(\frac { p }{ q }\) (q ≠ 0) is a rational (RBSESolutions.com) number then what condition apply for q whereas \(\frac { p }{ q }\) is a terminating decimal ? Is a rational number is terminating decimal.

Solution

\(\frac { p }{ q }\) (q ≠ 0) is a rational number.

and \(\frac { p }{ q }\) is terminating decimal.

Prime factor of q will be in the form 2^{m} x 5^{n} where m, n is non-negative integers.

Question 12.

Simplify \(\frac { 2\surd 45+3\surd 20 }{ 2\surd 5 }\) and clear whether it is rational or irrational number.

Solution

So, the given number is a rational number.

Question 13.

Prove that any positive odd integer is (RBSESolutions.com) of the form 4q + 1 or 4q + 3, where q is any integer.

Solution

Let a is any positive odd integer and b = 4.

Applying Euclid division lemma in a, b

Where o ≤ r < 4 and q is any integer.

r = 0, 1, 2, 3 put

a = 4q + 0 ⇒ a = 4q

a = 4q + 1

a = 4q + 2

a = 4q + 4

For positive odd integer.

a ≠ 4 q, a ≠ 4q + 2

Hence, any odd integer is of the form 4q + 1, or 4q + 3.

Hence proved.

Question 14.

Prove that the product of two (RBSESolutions.com) consecutive positive integers is divisible by 2.

Solution

Let two consecutive positive integers is n, and (n + 1)

Product of both integers = n(n + 1) = n^{2} + n

We know that any positive integer is in the form 2q and 2q + 1. where q is an integer.

Here two cases are possible

Case. I. when n = 2q then

⇒ n^{2} + n = (2q)^{2} + (2q)

⇒ n^{2} + n = 4q^{2} + 2q

⇒ n^{2} + n = 2q(2q + 1) [Let r = q(2q + 1)]

⇒ n^{2} + n = 2r

⇒ n^{2} + n, 2 can be divided by 2

⇒ n(n + 1), also divided by 2

So, product of two consecutive positive integer is divided by 2

Hence proved.

Question 15.

Find the largest number which is (RBSESolutions.com) divided by 2053 and 967, left the remainder as 5 and 7 respectively.

Solution

Given that on dividing 2053, there is a remainder of 5.

This means that 2053 – 5 = 2048 is exactly divisible by the required number.

Similarly, 967 – 7 = 960 is exactly divisible by the required number.

Question 16.

Explain, why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 x 5 are (RBSESolutions.com) composite numbers?

Solution

7 x 11 x 13 + 13 = 13[7 x 11 + 1]

= 13(77 + 1)

= 13 x 78

= 13 x 2 x 3 x 13

= 2 x 3 x 13 x 13

2, 3 and 13 are prime numbers.

So according to the fundamental theorem of arithmetic, every composite number can be uniquely expressed as a product of prime numbers.

So, this is a composite number

Similarly

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

= 5[7 x 6 x 4 x 3 x 2 x 1 + 1]

= 5(1008 + 1)

= 5 x 1009

5 and 1009 are prime numbers.

So according to the fundamental theorem of arithmetic is a composite number.

Question 17.

If HCF of two numbers 306 and 657 is 9, then (RBSESolutions.com) find their LCM.

Solution

We know that

Question 18.

A rectangular veranda is of dimension 18 m 72 cm x 13 m 30 cm. Squared tiles (RBSESolutions.com) are used to cover them. Find the least number of such tiles.

Solution

length of varanda = 18 m 72 cm = (1800 + 72) cm = 1872 cm

and breadth = 13 m 20 cm = (1300 + 20) cm = 1320 cm

For finding the area of one tile we find the HCF of length and breadth.

Question 19.

Prove that following numbers (RBSESolutions.com) are irrational numbers

(i) 5√2

(ii) \(\frac { 2 }{ \surd 7 }\)

(iii) \(\frac { 3 }{ 2\surd 5 }\)

(iv) 4 + √2

Solution

a, b, ab and 4 all are integers.

\(\frac { a-4b }{ b }\) is a (RBSESolutions.com) rational number.

√2 will be a rational number

This contradicts.

our hypothesis is wrong.

So, 4 + √2 is an irrational number.

Hence proved.

Question 20.

What can you say about the prime factors of (RBSESolutions.com) denominator of the following rational numbers.

(i) 34.12345

(ii) \(43.\bar { 123456789 }\)

Solution

(i) 34.12345 = \(\frac { 3412345 }{ 100000 }\)

This number is in the form of \(\frac { p }{ q }\)

This is a rational number.

q = 100000 = (10)^{5} = (2 x 5)^{5} = 2^{5} x 5^{5}

So, prime factor of q are in the form of = 2 or 5.

(ii) \(43.\bar { 123456789 }\)

= 43.123456789 123456789 123456789 …

The decimal expansion of this number is nonterminating and recurring and it is possible to write in the form \(\frac { p }{ q }\).

This is a rational number.

So, besides 2 and 5 the prime factor of q, there (RBSESolutions.com) are other prime positive integers possible.

Hence, the given number is rational

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

## Leave a Reply