RS Aggarwal Solutions Class 10 Maths Chapter 1 – Real Numbers (Updated For 2024)

RS Aggarwal Solutions Class 10 Maths Chapter 1 Real Numbers: Appearing for the Class 10 Maths Exam? You must prepare from the best help book and here we have RS Aggarwal Solutions Class 10 Maths. All the solutions are prepared by the subject matter experts. You can download the free PDF of RS Aggarwal Solutions Class 10 Maths Chapter 1 from here.

Download the Free PDF of RS Aggarwal Solutions Class 10 Maths Chapter 1 Real Numbers using the download link given in this blog. To know more, read the whole blog. 

Download RS Aggarwal Solutions Class 10 Maths Chapter 1 Real Numbers PDF

RS Aggarwal Solutions Class 10 Maths Chapter 1 Real Numbers

Access RS Aggarwal Solutions Class 10 Maths Chapter 1 Real Numbers

Question 1.
Solution:
For any two given positive integers a and b, there exist unique whole numbers q and r such that
a = bq + r, when 0 ≤ r < b.
Here, a is called the dividend, b is a divisor, q is the quotient, and r is the remainder.
Dividend = (Divisor x Quotient) + Remainder.

Question 2.
Solution:
Using Euclid’s division Lemma
Dividend = (Divisor x Quotient) + Remainder
= (61 x 27) + 32
= 1647 + 32
= 1679
Required number = 1679

Question 3.
Solution:
Let the required divisor = x
Then by Euclid’s division Lemma,
Dividend = (Divisor x Quotient) + remainder
1365 = x x 31 + 32
=> 1365 = 31x + 32
=> 31x= 1365 – 32 = 1333
x = 133131 = 43
Divisor = 43

Question 4.
Solution:
Let n be a given positive odd integer.
On dividing n by 6, let m be the quotient and r be the remainder, then by Euclid’s division Lemma.
n = 6m + r, where 0 ≤ r < 6 => n = 6m + r, where r = 0, 1, 2, 3, 4, 5
=> n = 6m or (6m + 1) or (6m + 2) or (6m + 3) or (6m + 4) or (6m + 5)
But n = 6m, (6m + 2) and (6m + 4) are even.
Thus when n is odd, it will be in the form of (6m + 1) (6m + 3), or (6m + 5) for some integer m.

Question 5.
Solution:
Let n be an arbitrary odd positive integer.
On dividing by 4, let m be the quotient and r be the remainder.
So to Euclid’s division lemma,
n = 4m + r, where 0 ≤ r < 4
n = 4m or (4m + 1) or (4m + 2) or (4m + 3)
But 4m and (4m + 2) are even integers.
Since n is odd, so n ≠ 4m or n ≠ (4m + 2)
n = (4m + 1) or (4m + 3) for some integer m.
Hence any positive odd integer is of the form (4m + 1) or (4m + 3) for some integer m.

Question 6.
Solution:
Let a = n³ – n
=> a = n (n² – 1)
=> a = n (n – 1) (n + 1) [(a² – b²) = (a – b) (a + b)]
=> a = (n – 1 ) n (n + 1)
We know that,
(i) If a number is completely divisible by 2 and 3, then it is also divisible by 6.
(ii) If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
(iii) If one of the factors of any number is an even number, then it is also divisible by 2.
a = (n – 1) n (n + 1) [From Eq. (i)]
Now, the sum of the digits
= n – 1 + n + n + 1 = 3n
= Multiple of 3, where n is any positive integer.
and (n – 1) n (n +1) will always be even, as one out of (n – 1) or n or (n + 1) must be even.
Since, conditions (ii) and (iii) completely satisfy Eq. (i).
Hence, by condition (i) the number n3 – n is always divisible by 6, where n is any positive integer.
Hence proved.

Question 7.
Solution:
Let x = 2m + 1 and y = 2m + 3 are odd positive integers, for every positive integer m.
Then, x² + y² = (2m + 1)² + (2m + 3)²
= 4m² + 1 + 4 m + 4m² + 9 + 12m [(a + b)² = a² + 2ab + b²]
= 8m² + 16m + 10 = even
= 2(4m² + 8m + 5) or 4(2m² + 4m + 2) + 1
Hence, x² + y² is even for every positive integer m but not divisible by 4.

Access RS Aggarwal Solutions Class 10 Maths Chapter 1 Real Numbers Other Exercise 

RS Aggarwal Solutions Chapter 1 Exercise 1.1

RS Aggarwal Solutions Chapter 1 Exercise 1.2

RS Aggarwal Solutions Chapter 1 Exercise 1.3

RS Aggarwal Solutions Chapter 1 Exercise 1.4

This is the complete blog on the RS Aggarwal Solutions Class 10 Maths Chapter 1 Real Numbers. To know more about the CBSE Class 10 Maths exam, ask in the comments.

FAQs on RS Aggarwal Solutions Class 10 Maths Chapter 1 Real Numbers