# 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

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.

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

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

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