RBSE Solutions for Class 10 Maths Chapter 2 Real Numbers Ex 2.1

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

RBSE Solutions For Class 10 Maths Chapter 2 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
RBSE Solutions For Class 10 Maths Chapter 2
Hence, a square of any positive odd integer is the form of 8q + 1.
Hence proved.

RBSE Class 10 Maths Chapter 2 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.
RBSE Class 10 Maths Chapter 2 Real Numbers
Exercise 2.1 Class 10 Maths RBSE Real Numbers
From equation (1), (2) and (3) it is clear that a3 is divisible by 9.
Then it can be written as follows
a3 = 9q
or a3 = 9q + 1,
or a3 = 9q + 8 cube
Hence the cube of any positive integer (RBSESolutions.com) is of the form 9q or (9q + 1) or (9q + 8).

Exercise 2.1 Class 10 Maths RBSE 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)

Class 10 Maths Chapter 2 RBSE 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
Class 10 Maths Chapter 2 RBSE Real Numbers
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)
RBSE Class 10 Maths Solutions Chapter 2 Real Numbers
Remainder ≠ 0
Step 3. For divisior 45 and (RBSESolutions.com) remainder 10 by Euclid division lemma
45 = 10 × 4 + 5 …(iii)
RBSE Class 10 Maths Chapter 2 Exercise 2.1 Real Numbers
remainder ≠ 0
Step 4. For new divisior 10 and remainder 5 by Euclid division lemma
10 = 5 × 2 + 0 …(iv)
Ex 2.1 Class 10 RBSE Real Numbers
Hence, remainder = 0
So, HCF of 210, 55 is 5.

(ii) 420,130
Step 1. For given integer 420 and 130.
RBSE Solutions For Class 10 Maths Chapter 2.1 Real Numbers
By Euclid division lemma
420 = 130 × 3 + 30 …(i)
Step 2. Here remainder is not zero.
So for divisior 130 and remainder 30
Class 10 RBSE Maths Chapter 2 Real Numbers
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
RBSE 10th Maths Chapter 2 Real Numbers
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
RBSE Solutions For Class 10 Maths English Medium Real Numbers
By Euclid division lemma
243 = 75 × 3 + 18 …(i)
Step 2. Here remainder ≠ 0
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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)
RBSE Solution.Com Class 10th Maths Real Numbers
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
Www RBSE Solutions Com Class 10 Maths Real Numbers
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
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Step 3. Here remainder is not zero. So for (RBSESolutions.com) divisor 18 and remainder 3.
90 = 45 × 2 + 0
RBSE Solution For Class 10th Maths Real Numbers
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
RBSE Solution 10th Class Maths Real Numbers

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
Www.RBSEsolutions.Com Class 10 Maths Real Numbers
Here, remainder ≠ 0
Divisor = 255, and remainder = 102
Step 3. Again, for divisor 255 and remainder 102 using Euclid Lemma.
255 = 102 × 2 + 51
RBSE Solutions For Class 10 Maths 2020 Real Numbers
Step 4. Here, remainder = 51 ≠ 0
again by using Euclid Divisor Lamma, for divisor 102 and remainder 51.
Real Numbers Class 10 Real Numbers
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
RBSE Solution Class 10 Maths English Medium Real Numbers
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)
RBSE Solution Class 10th Maths Real Numbers
Step 3. Remainder ≠ 0
For divisior 216 and remainder 192
Using Euclid division (RBSESolutions.com) lemma
RBSE Solutions Class 10 Maths English Medium
216 = 192 × 1 + 24 …(iii)
Step 4. Remainder ≠ 0
For division 192 and remainder 24
Using Euclid division lemma
Exercise 2.1 Class 10 Maths
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.