NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers
The NCERT Solutions for Class 10 Maths, Chapter 1, is "Real Numbers." The solutions in almost all questions are based on the important concepts concerning real numbers. This chapter is important for laying a solid mathematics base because it introduces some of the critical concepts, such as the Euclidean Division Lemma, the Fundamental Theorem of Arithmetic, and the properties of irrational numbers. The students can learn how to find the HCF of two given numbers by using the Euclidean algorithm. Skill in this algorithm is very important because it can solve many mathematical problems. This chapter deals with the decimal expansions of rational numbers and also discusses when the decimal expansions are terminating or non-terminating recurring decimals. In higher classes, grasping these concepts are important to solve complex problems.
Download PDF For NCERT Solutions for Maths Real Numbers
The NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers
Students can access the NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 1.1
Use Euclid’s Division Algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.
Exercise 1.2
Express each number as product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
(i) 140 = 2 x 2 x 5 x 7 = 22 x 5 x 7
(ii) 156 = 2 x 2 x 3 x 13 = 22 x 3 x 13
(iii) 3825 = 3 x 3 x 5 x 17 = 32 x 52 x x 17
(iv) 5005 = 5 x 7 x 11 x 13
(v) 7429 = 17 x 19 x 23
Find the LCM and HCF of the following pairs of integers and verify that LCM xHCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
(i) 26 and 91
a = 26, b = 91
∴ H.C.F = 13
L.C.M = 2 × 7 × 13
= 14 × 13 = 182
∴ H.C.F × L.C.M = a × b
13 × 182 = 26 × 96
2366 = 2366
(ii) 510 and 92
a = 510, b = 92
(iii) 336 and 54
a = 336, b = 54
336= 2 x 2 x 2 x 2 x 3 x 7 = 24 x 3 x 7
54 = 2 x 3 x 3 x 3 = 2 x 33
HCF = 2 x 3 = 6
LCM = 24 x 33 x 7= 3024
Product of two numbers 336 and 54 = 336 x 54= 18144
3024 x 6= 18144
Hence, product of two numbers = 18144
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
(i) 12, 15 and 21
12 = 2 × 2 × 3 = 22× 3
15 = 3 × 5
and 21 = 3 × 7
For HCF, we find minimum power of prime factor
H.C.F. = (3)1= 3
For LCM, taking maximum power of prime factors
L.C.M. = 22 × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420
So,H.C.F. = (3)1= 3
and L.C.M. = 420
(ii) 17, 23 and 29
17 = 1 × 17
23 = 1 x 23
29 = 1 x 29
For HCF, common factor is 1
HCF = 1
For LCM taking maximum power of prime factor.
L.C.M. = 1 × 17 × 23 × 29 = 11339
So H.C.F. = 1
L.C.M. = 11339
(iii) 8, 9 and 25
8 = 2 × 2 × 2 × 1 = 23× 1
9 = 3 × 3 = 32
and 25 = 5 × 5 = 52
For HCF common factor is 1
H.C.F. = 1
For LCM, taking maximum power of prime factors
L.C.M. = 23× 32 × 52
= 8 × 9 × 25 = 1800
So H.C.F. = 1
L.C.M. = 1800
Given that HCF (306, 657) = 9, find LCM (306, 657).
HCF (306, 657) = 9
We know that, LCM × HCF = Product of two numbers
L.C.M x H.C.F = first Number x Second Number
L.C.M x 9 = 306 x 657
LCM = 22338
Check whether 6n can end with the digit 0 for any natural number n.
TO CHECK: Whether 62can end with the digit 0 for any natural number n.
We know that
62 = (2 × 3)n
62 = (2)n ×(3)n
Therefore, prime factorization of 6ndoes not contain 5 and 2 as a factor together.
Hence 6ncan never end with the digit 0 for any natural number n
Explain why 7 x 11 x 13 and 7 x 6 x 5 x 3 x 2 x 1 + 5 are composite numbers.
So, the given expression has 6 and 13 as its factors. Therefore, we can conclude that it is a composite number.
Similarly,
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5
= 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1) [taking 5 out- common]
= 5 x (1008 + 1)
= 5 x 1009
Since, 1009 is a prime number the given expression has 5 and 1009 as its factors other than 1 and the number itself.
Hence, it is also a composite number.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
LCM of 12 and 18= 2 x 2 x 3 x 3 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.
Exercise 1.3
Prove that (3 + 2√5) is irrational.
We will prove this by contradiction.
Let us suppose that (3+2√5) is rational.
It means that we have co-prime integers aand b(b ≠ 0) such that
So, it can be written in the form a/b
3 + 2√5 = a/b
Here a and b are coprime numbers and b ≠ 0
Solving 3 + 2√5 = a/b we get,
=>2√5 = a/b – 3
=>2√5 = (a-3b)/b
=>√5 = (a-3b)/2b
This shows (a-3b)/2b is a rational number. But we know that √5 is an irrational number.
So, it contradicts our assumption. Our assumption of 3 + 2√5 is a rational number is incorrect.
3 + 2√5 is an irrational number
Hence proved
Solve the followings Questions.
1. Prove that √2 is irrational.
Let us prove √2 irrational by contradiction.
Let us suppose that √2 is rational.
So it can be expressed in the form p/q where p, q are co-prime integers and q≠0
√2 = p/q
Here p and q are coprime numbers and q ≠ 0
Solving
√2 = p/q
On squaring both the side we get,
=>2 = (p/q)2
=>2q2 = p2 ........(1)
=> p2 = q2
So 2 divides p and p is a multiple of 2.
⇒ p = 2m
⇒ p² = 4m² ………………………………..(2)
From equations (1) and (2), we get,
2q² = 4m²
⇒ q² = 2m²
⇒ q² is a multiple of 2
⇒ q is a multiple of 2
Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√2 is an irrational number.
Prove that the following are irrationals.
(i)
(ii)
(iii)
(i) We can prove
Let us suppose that
It means we have some co-prime integers a and b (b ≠ 0) such that
1/√2 = p/q
√2 = q/p
By Squaring on both sides
2 × p2= q2
2, divides q2
∴ 2, divides q
∵ q is an even number.
Similarly ‘p’ is an even number.
∴ p and q are even numbers.
∴ Common factor of p and q is 2.
This contradicts the fact that p and q also irrational.
∴ √2 is an irrational number.
∴ 
(ii) We can prove
Let us suppose that
It means we have some co-prime integers a and b (b ≠ 0) such that
It means √5 which is equal also a rational number.
This contradicts the fact that √5 is an irrational number.
This contradicts the fact that 7√5 is rational number.
∴ 7√5 is a rational number.
(iii) We will prove
Let us suppose that (
It means that we have co-prime integers aand b(b ≠ 0) such that
∴ √2 is also rational number.
This contradicts to the fact that √2 is an irrational number.
This contradicts to the fact that 6 + √2 is a rational number.
∴ 6 + √2 is an irrational number.
Exercise 1.4
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating decimal expansion.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(i)
q = 3125 = 5 x 5 x 5 x 5 x 5 = 55
Here, denominator is of the form 
It means rational number 
(ii)
q = 8 = 2 x 2 x 2 = 23
Here, denominator is of the form 
It means rational number 
(iii)
q =
Here, denominator is not of the
It means rational number 
(iv)
q = 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 =26 x 5 x 5
Here, denominator is of the form 
It means rational number 
(v)
q = 343 = 7 x 7 x 7
Here, denominator is not of the form 
It means rational number 
(vi)
q = 2 x 2 x 2 x 5 x 5
Here, denominator is of the form 
It means rational number 
(vii)
q = 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 7 x 7 x 7 x 7 x 7
Here, denominator is not of the form 
It means rational number 
(viii)
Here, denominator is of the form , where m = 1 and n = 0.
It means rational number 
(ix)
q = 10 = 2 x 5
Here, denominator is of the form 
It means rational number 
(x)
q = 30 = 2 x 3 x 5
Here, denominator is not of the form 
It means rational number 
Write down the decimal expansions of those rational numbers in Question 1 which have terminating decimal expansions.
(iv)
x
The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If, they are rational, and of the form 
(i) 43.123456789
(ii) 0.120112001200 0120000…
(iii)
(i) 43.123456789
It is rational because decimal expansion is terminating. Therefore, it can be expressed in 
(ii) 0.120112001200 0120000…
It is irrational because decimal expansion is neither terminating nor non-terminating repeating.
(iii)
It is rational because decimal expansion is non-terminating repeating. Therefore, it can be expressed in 
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