RBSE Solutions for Class 10 Maths Chapter 2 Real Numbers Ex 2.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Exercise 2.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Real Numbers |

Exercise |
Exercise 2.1 |

Number of Questions Solved |
5 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Ex 2.1

Question 1.

Show that square of any positive odd integer (RBSESolutions.com) is of the form 8q + 1, where q is any positive integer.

Solution

Let a is any odd positive integer number.

We know that an odd positive integer will be in the form of 4m + 1 or 4m + 3

Hence, a square of any positive odd integer is the form of 8q + 1.

Hence proved.

Question 2.

Use Euclid’s division lemma to show that the (RBSESolutions.com) cube of any positive integer is of the form 9q or 9q + 8, where q is some integer.

Solution

Let a is any positive integer, then it is in the form of 3b or 3b + 1 or 3b + 2.

Hence, following three cases are possible.

From equation (1), (2) and (3) it is clear that a^{3} is divisible by 9.

Then it can be written as follows

a^{3} = 9q

or a^{3} = 9q + 1,

or a^{3} = 9q + 8 cube

Hence the cube of any positive integer (RBSESolutions.com) is of the form 9q or (9q + 1) or (9q + 8).

Question 3.

Show that any positive odd integer is of the form 6q + 1, or 6q + 3 or 6q + 5, where q is some integer.

Solution

Let a is a positive odd integer and apply Euclid’s division algorithm

a = 6q + r, Where 0 ≤ r < 6

for 0 ≤ r < 6 probable remainders are 0, 1, 2, 3, 4 and 5.

a = 6q + 0

or a = 6q + 1

or a = 6q + 2

or a = 6q + 3

or a = 6q + 4

or a = 6q + 5 may be form

Where q is quotient and a = odd integer.

This cannot be in the form of 6q, 6q + 2, 6q + 4. [all divides by 2]

Hence, any positive odd integer is of the form 6q + 1, or 6q + 3 or (6q + 5)

Question 4.

Use Euclid’s division algorithm (RBSESolutions.com) to find the HCF of:

(i) 210, 55

(ii) 420, 130

(iii) 75, 243

(iv) 135, 225

(v) 196, 38220

(vi) 867, 255.

Solution

**(i) 210, 55**

**Step 1**. Given integers are 210 and 55 such as 210 > 55

by Euclid division lemma

210 = 55 × 3 + 45 …(i)

**Step 2.** Since remainder 45 ≠ 0. so for divisior 55 and remainder 45

by Euclid division lemma

55 = 45 × 1 + 10 …(ii)

Remainder ≠ 0

**Step 3.** For divisior 45 and (RBSESolutions.com) remainder 10 by Euclid division lemma

45 = 10 × 4 + 5 …(iii)

remainder ≠ 0

**Step 4.** For new divisior 10 and remainder 5 by Euclid division lemma

10 = 5 × 2 + 0 …(iv)

Hence, remainder = 0

So, HCF of 210, 55 is 5.

**(ii) 420,130**

**Step 1.** For given integer 420 and 130.

By Euclid division lemma

420 = 130 × 3 + 30 …(i)

**Step 2.** Here remainder is not zero.

So for divisior 130 and remainder 30

By Euclid division lemma

130 = 30 × 4 + 10 …(ii)

**Step 3.** Here remainder (RBSESolutions.com) is not zero.

So for divisior 30 and remiander 10

By Euclid division lemma

30 = 10 × 3 + 0 …(iii)

Here, remainder is zero.

Hence, HCF of 420 and 130 is 10.

**(iii) 75, 243**

**Step 1.** For given integer 75 and 243

243 > 75

By Euclid division lemma

243 = 75 × 3 + 18 …(i)

**Step 2.** Here remainder ≠ 0

So, using Euclid division lemma for divisior 75 and remainder 18.

75 = 18 × 4 + 3 …(ii)

**Step 3.** Here remainder ≠ 0 fo divisior 18 and (RBSESolutions.com) remainder 3

18 = 3 × 6 + 0 …(iii)

Since, remainder = 0

Hence, H.C.F. of 75 and 243 = 3

**(iv) 135, 225**

**Step 1.** For given integer 135 and 225

225 > 135

By Euclid division lemma

225 = 135 × 1 + 90

**Step 2.** Here remainder is not zero.

So, for divisior 135 and remainder 90.

135 = 90 × 1 + 45

**Step 3.** Here remainder is not zero. So for (RBSESolutions.com) divisor 18 and remainder 3.

90 = 45 × 2 + 0

remainder = 0

Hence HCF of 135 and 225 = 45

**(v) 196, 38220**

**Step 1.** For a given integer 196 and 38220

38220 > 196

**Step 2.** Using Euclid division lemma

38220 = 196 × 195 + 0

**Step 3.** Since remainder = 0 and divisor = 196

Hence, HCF of 196 and 38220 = 196

**(vi) 867, 255**

**Step 1.** For a given integer 867 and 255

867 > 255

**Step 2.** Using Euclid (RBSESolutions.com) division lemma

867 = 255 × 3 + 102

Here, remainder ≠ 0

Divisor = 255, and remainder = 102

**Step 3.** Again, for divisor 255 and remainder 102 using Euclid Lemma.

255 = 102 × 2 + 51

**Step 4.** Here, remainder = 51 ≠ 0

again by using Euclid Divisor Lamma, for divisor 102 and remainder 51.

**Step 5.** Here, the remainder is zero.

Hence, HCF (687, 255) = 51

Question 5.

If HCF of or numbers 408 and 1032 is (RBSESolutions.com) expressed in the form of 1032x – 408 × 5, then find the value of x.

Solution

**Step 1.** For a number 408 and 1032 as 1032 > 408

Using Euclid division lemma

1032 = 408 × 2 + 216 …(i)

**Step 2.** Remainder ≠ 0

For divisior 408 and remainder 216.

Using Euclid division lemma

408 = 216 × 1 + 192 …(ii)

**Step 3.** Remainder ≠ 0

For divisior 216 and remainder 192

Using Euclid division (RBSESolutions.com) lemma

216 = 192 × 1 + 24 …(iii)

**Step 4.** Remainder ≠ 0

For division 192 and remainder 24

Using Euclid division lemma

192 = 24 × 8 + 0 …(iv)

Hence, remainder = 0

Hence, HCF of 408 and 1032 = 29

H.C.F. = 1032x – 408 × 5

⇒ 24 = 1032x – 408 × 5 (∴ H.C.F. = 24)

⇒ 1032x = 24 + 408 × 5

⇒ 1032x = 24 + 2040

⇒ 1032x = 2064

⇒ x = 2

Hence, the value of x = 2

We hope the given RBSE Solutions for Class 10 Maths Chapter 2 Real Numbers Ex 2.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 2 Real Numbers Exercise 2.1, drop a comment below and we will get back to you at the earliest.

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