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NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

Edited By Ramraj Saini | Updated on Sep 10, 2023 06:53 PM IST

NCERT solutions for Class 10 Maths chapter 1 Real Numbers are discussed here. In Mathematics, there are two kinds of numbers - Real number and Imaginary number. Real numbers are those numbers that can be represented on the number line. NCERT solutions for Class 10 Maths Chapter 1 Real Numbers will help the students to prepare for the exams in a better way. The problems and solutions are prepared by expert teachers keeping in mind the Latest updated CBSE syllabus 2023. Students can practice these NCERT solutions for class 10 to get good hold on the concepts.

This Story also Contains
  1. NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers PDF Free Download
  2. NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers - Important Formulae
  3. NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Intext Questions and Exercise)
  4. Summary Of NCERT solutions for Real Numbers Class 10 Maths
  5. NCERT Books and NCERT Syllabus
  6. Real Numbers Class 10 Solutions - Chapter Wise
  7. NCERT solutions of class 10 - subject wise
  8. NCERT Exemplar solutions - Subject wise
NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers
NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

In this Class 10 Maths Chapter 1, we are going to talk about real numbers. The real numbers class 10 carry detailed explanations for each and every question in simple, comprehensive, and step by step explanation. NCERT solutions for other subjects and classes can be downloaded by clicking on the above link.

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers PDF Free Download

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers - Important Formulae

Types of Numbers:

  • Natural Numbers: N = {1, 2, 3, 4, 5, ...}
  • Whole Numbers: W = {0, 1, 2, 3, 4, 5, ...}
  • Integers: {..., -4, -3, -2, -1, 0, 1, 2, 3, 4, ...}
  • Positive Integers: Z+ = {1, 2, 3, 4, 5, ...}
  • Negative Integers: Z– = {-1, -2, -3, -4, -5, ...}
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Rational and Irrational Numbers:

  • Rational Numbers: p/q where p and q are integers and q is not zero. (e.g., 3/7)
  • Irrational Numbers: Cannot be expressed as p/q (e.g., π, √5)

Real Numbers: Numbers located on the number line, used in real-world applications.

Euclid's Division Algorithm (Lemma):

  • For positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r ≤ b.
  • (a is the dividend, b is the divisor, q is the quotient, and r is the remainder.)
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Fundamental Theorem of Arithmetic:

  • Composite Numbers can be expressed as the product of prime numbers.

HCF and LCM by Prime Factorization:

  • Highest Common Factor (HCF): Product of smallest power of common factors in numbers.
  • Least Common Multiple (LCM): Product of greatest power of prime factors in numbers.

HCF (a, b) × LCM (a, b) = a × b

Free download NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers PDF for CBSE Exam.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Intext Questions and Exercise)

Real Numbers class 10 Exercise: 1.1

Q1 (1) Use Euclid’s division algorithm to find the HCF of 135 and 225

Answer:

225 > 135. Applying Euclid's Division algorithm we get

225=135\times 1+90

since remainder \neq 0 we again apply the algorithm

135=90\times 1+45

since remainder \neq 0 we again apply the algorithm

90=45\times 2

since remainder = 0 we conclude the HCF of 135 and 225 is 45.

Q1 (2) Use Euclid’s division algorithm to find the HCF of 196 and 38220

Answer:

38220 > 196. Applying Euclid's Division algorithm we get

38220=196\times 195+0

since remainder = 0 we conclude the HCF of 38220 and 196 is 196.

Q1 (3) Use Euclid’s division algorithm to find the HCF of 867 and 255

Answer:

867 > 225. Applying Euclid's Division algorithm we get

867=255\times 3+102

since remainder \neq 0 we apply the algorithm again.

since 255 > 102

255=102\times 2+51

since remainder \neq 0 we apply the algorithm again.

since 102 > 51

102=51\times 2+0

since remainder = 0 we conclude the HCF of 867 and 255 is 51.

Q2 Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Answer:

Let p be any positive integer. It can be expressed as

p = 6q + r

where q\geq 0 and 0\leq r< 6

but for r = 0, 2 or 4 p will be an even number therefore all odd positive integers can be written in the form 6q + 1, 6q + 3 or 6q + 5.

Q3 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?

Answer:

The maximum number of columns in which they can march = HCF (32, 616)

Since 616 > 32, applying Euclid's Division Algorithm we have

616=32\times 19+8

Since remainder \neq 0 we again apply Euclid's Division Algorithm

Since 32 > 8

32=8\times 4+0

Since remainder = 0 we conclude, 8 is the HCF of 616 and 32.

The maximum number of columns in which they can march is 8.

Q4 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.

[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1]

Answer:

Let x be any positive integer.

It can be written in the form 3q + r where q\geq 0 and r = 0, 1 or 2

Case 1:

For r = 0 we have

x 2 = (3q) 2

x 2 = 9q 2

x 2 = 3(3q 2 )

x 2 = 3m

Case 2:

For r = 1 we have

x 2 = (3q+1) 2

x 2 = 9q 2 + 6q +1

x 2 = 3(3q 2 + 2q) + 1

x 2 = 3m + 1

Case 3:

For r = 2 we have

x 2 = (3q+2) 2

x 2 = 9q 2 + 12q +4

x 2 = 3(3q 2 + 4q + 1) + 1

x 2 = 3m + 1

Hence proved.

Q5 Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer:

Let x be any positive integer.

It can be written in the form 3q + r where q\geq 0 and r = 0, 1 or 2

Case 1:

For r = 0 we have

x 3 = (3q) 3

x 3 = 27q 3

x 3 = 9(3q 3 )

x 3 = 9m

Case 2:

For r = 1 we have

x 3 = (3q+1) 3

x 3 = 27q 3 + 27q 2 + 9q + 1

x 3 = 9(3q 3 + 3q 2 +q) + 1

x 3 = 3m + 1

Case 3:

For r = 2 we have

x 3 = (3q+2) 3

x 3 = 27q 3 + 54q 2 + 36q + 8

x 3 = 9(3q 3 + 6q 2 +4q) + 8

x 3 = 3m + 8

Hence proved.

Real Numbers class 10 Exercise: 1.2

Q1 (1) Express each number as a product of its prime factors: 140

Answer:

The number can be as a product of its prime factors as follows

\\140=2\times 2\times 5\times 7\\ 140=2^{2}\times 5\times 7

Q 1 (2) Express each number as a product of its prime factors: 156

Answer:

The given number can be expressed as follows

\\156=2\times 2\times 3\times 13\\ 156=2^{2}\times 3\times 13

Q1 (3) Express each number as a product of its prime factors: 3825

Answer:

The number is expressed as the product of the prime factors as follows

\\3825=3\times 3\times 5\times 5\times 17\\ 3825=3^{2}\times 5^{2}\times 17

Q1 (4) Express each number as a product of its prime factors: 5005

Answer:

The given number can be expressed as the product of its prime factors as follows.

5005=5\times 7\times 11\times 13

Q1 (5) Express each number as a product of its prime factors: 7429

Answer:

The given number can be expressed as the product of their prime factors as follows

7429=17\times 19\times 23

Q2 (1) Find the LCM and HCF of the following pairs of integers and verify that LCM \times HCF = product of the two numbers: 26 and 91

Answer:

26 = 2 x 13

91 = 7 x 13

HCF(26,91) = 13

LCM(26,91) = 2 x 7 x 13 = 182

HCF x LCM = 13 x 182 = 2366

26 x 91 = 2366

26 x 91 = HCF x LCM

Hence Verified

Q2 (2) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 510 and 92

Answer:

The number can be expressed as the product of prime factors as

510 = 2 x 3 x 5 x 17

92 = 2 2 x 23

HCF(510,92) = 2

LCM(510,92) = 2 2 x 3 x 5 x 17 x 23 = 23460

HCF x LCM = 2 x 23460 = 46920

510 x 92 = 46920

510 x 92 = HCF x LCM

Hence Verified

Q2 (3) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 336 and 54

Answer:

336 is expressed as the product of its prime factor as

336 = 2 4 x 3 x 7

54 is expressed as the product of its prime factor as

54 = 2 x 3 3

HCF(336,54) = 2 x 3 = 6

LCM(336,54) = 2 4 x 3 3 x 7 = 3024

HCF x LCM = 6 x 3024 = 18144

336 x 54 = 18144

336 x 54 = HCF x LCM

Hence Verified

Q3 (1) Find the LCM and HCF of the following integers by applying the prime factorization method. 12, 15 and 21

Answer:

The numbers can be written as the product of their prime factors as follows

12 = 2 2 x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 2 2 x 3 x 5 x 7 = 420

Q3 (2) Find the LCM and HCF of the following integers by applying the prime factorisation method. 17, 23 and 29

Answer:

The given numbers are written as the product of their prime factors as follows

17 = 1 x 17

23 = 1 x 23

29 = 1 x 29

HCF = 1

LCM = 17 x 23 x 29 = 11339

Q3 (3) Find the LCM and HCF of the following integers by applying the prime factorization method. 8, 9 and 25

Answer:

The given numbers are written as the product of their prime factors as follows

8 = 2 3

9 = 3 2

25 = 5 2

HCF = 1

LCM = 2 3 x 3 2 x 5 2 = 1800

Q4 Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer:

As we know the product of HCF and LCM of two numbers is equal to the product of the two numbers we have

HCF (306, 657) x LCM (306, 657) = 306 x 657

\\ LCM (306, 657) = \frac{306\times 657}{HCF (306, 657) }\\ LCM (306, 657) = \frac{306\times 657}{9}\\ LCM (306, 657) =22338

Q5 Check whether 6 ^n can end with the digit 0 for any natural number n.

Answer:

By prime factorizing we have

6 n = 2 n x 3 n

A number will end with 0 if it has at least 1 as the power of both 2 and 5 in its prime factorization. Since the power of 5 is 0 in the prime factorization of 6 n we can conclude that for no value of n 6 n will end with the digit 0.

Q6 Explain why 7 \times 11 \times 13 + 13 and 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 are composite numbers.

Answer:

7 x 11 x 13 + 13

= (7 x 11 + 1) x 13

= 78 x 13

= 2 x 3 x 13 2

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

= (7 x 6 x 4 x 3 x 2 x 1 + 1) x 5

= 5 x 1008

After Solving we observed that both the number are even numbers and the number rule says that we can take atleast two common out of two numbers. So that the number is composite number.

Q7 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?

Answer:

The time after which they meet again at the starting point will be equal to the LCM of the times they individually take to complete one round.

Time taken by Sonia = 18 = 2 x 3 2

Time taken by Ravi = 12 = 2 2 x 3

LCM(18,12) = 2 2 x 3 2 = 36

Therefore they would again meet at the starting point after 36 minutes.

Class 10 Maths Chapter 1 Solutions Exercise: 1.3

Q1 Prove that \sqrt 5 is irrational.

Answer:

Let us assume \sqrt{5} is rational.

It means \sqrt{5} can be written in the form \frac{p}{q} where p and q are co-primes and q\neq 0

\\\sqrt{5}=\frac{p}{q}

Squaring both sides we obtain

\\\left ( \sqrt{5} \right )^{2}=\left (\frac{p}{q} \right )^{2}\\ 5=\frac{p^{2}}{q^{2}}\\ p^{2}=5q^{2}

From the above equation, we can see that p 2 is divisible by 5, Therefore p will also be divisible by 5 as 5 is a prime number. (i)

Therefore p can be written as 5r

p = 5r

p 2 = (5r) 2

5q 2 = 25r 2

q 2 = 5r 2

From the above equation, we can see that q 2 is divisible by 5, Therefore q will also be divisible by 5 as 5 is a prime number. (ii)

From (i) and (ii) we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that \sqrt{5} is rational was wrong. Hence proved that \sqrt{5} is irrational.

Q2 Prove that 3 + 2 \sqrt 5 is irrational.

Answer:

Let us assume 3 + 2 \sqrt 5 is rational.

This means 3 + 2 \sqrt 5 can be wriiten in the form \frac{p}{q} where p and q are co-prime integers.

\\3+2\sqrt{5}=\frac{p}{q}\\ 2\sqrt{5}=\frac{p}{q}-3\\ \sqrt{5}=\frac{p-3q}{2q}\\

As p and q are integers \frac{p-3q}{2q}\\ would be rational, this contradicts the fact that \sqrt{5} is irrational. This contradiction arises because our initial assumption that 3 + 2 \sqrt 5 is rational was wrong. Therefore 3 + 2 \sqrt 5 is irrational.

Q3 Prove that the following are irrationals :

(i) 1/ \sqrt 2

Answer:

Let us assume \frac{1}{\sqrt{2}} is rational.

This means \frac{1}{\sqrt{2}} can be written in the form \frac{p}{q} where p and q are co-prime integers.

\\\frac{1}{\sqrt{2}}=\frac{p}{q}\\ \sqrt{2}=\frac{q}{p}

Since p and q are co-prime integers \frac{q}{p} will be rational, this contradicts the fact that \sqrt{2} is irrational. This contradiction arises because our initial assumption that \frac{1}{\sqrt{2}} is rational was wrong. Therefore \frac{1}{\sqrt{2}} is irrational.

Q3 (2) Prove that the following are irrationals :

(ii) 7 \sqrt 5

Answer:

Let us assume 7 \sqrt 5 is rational.

This means 7 \sqrt 5 can be written in the form \frac{p}{q} where p and q are co-prime integers.

\\7\sqrt{5}=\frac{p}{q}\\ \sqrt{5}=\frac{p}{7q}

As p and q are integers \frac{p}{7q}\\ would be rational, this contradicts the fact that \sqrt{5} is irrational. This contradiction arises because our initial assumption that 7 \sqrt 5 is rational was wrong. Therefore 7 \sqrt 5 is irrational.

Q3 (3) Prove that the following are irrationals : 6 + \sqrt 2

Answer:

Let us assume 6 + \sqrt 2 is rational.

This means 6 + \sqrt 2 can be written in the form \frac{p}{q} where p and q are co-prime integers.

\\6+\sqrt{2}=\frac{p}{q}\\ \sqrt{2}=\frac{p}{q}-6\\ \sqrt{2}=\frac{p-6q}{q}

As p and q are integers \frac{p-6q}{q} would be rational, this contradicts the fact that \sqrt{2} is irrational. This contradiction arises because our initial assumption that 6 + \sqrt 2 is rational was wrong. Therefore 6 + \sqrt 2 is irrational.

Class 10 Maths Chapter 1 Solutions Exercise: 1.4

Q1 (1) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:13/3125

Answer:

\frac{13}{3125}=\frac{13}{5^{5}}

The denominator is of the form 2 a x 5 b where a = 0 and b = 5. Therefore the given rational number will have a terminating decimal expansion.

Q1 (2) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:17 / 8

Answer:

\frac{17}{8}=\frac{17}{2^{3}}

The denominator is of the form 2 a x 5 b where a = 3 and b = 0. Therefore the given rational number will have a terminating decimal expansion.

Q1 (3) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 64 / 455

Answer:

\frac{64}{455}=\frac{64}{5\times 7\times 13}

The denominator is not of the form 2 a x 5 b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

Q 1 (4) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 15 / 1600

Answer:

\\\frac{15}{1600}=\frac{3\times 5}{2^{6}\times 5^{2}}\\ \frac{15}{1600}=\frac{3}{2^{6}\times 5}

The denominator is of the form 2 a x 5 b where a = 6 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

Q 1 (5) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 29 / 343

Answer:

\frac{29}{343}=\frac{29}{7^{3}}

The denominator is not of the form 2 a x 5 b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

Q1 (6) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: \frac{23}{2 ^3 5 ^ 2 }

Answer:

\frac{23}{2 ^3 \times 5 ^ 2 }

The denominator is of the form 2 a x 5 b where a = 3 and b = 2. Therefore the given rational number will have a terminating decimal expansion.

Q1 (7) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: \frac{129 }{2 ^ 2 7^5 5 ^ 7 }

Answer:

\frac{129 }{2 ^ 2\times 7^5 \times 5 ^ 7 }

The denominator is not of the form 2 a x 5 b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

Q1 (8) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 6 / 15

Answer:

\\\frac{6}{15}=\frac{2\times 3}{3\times 5}\\ \frac{6}{15}=\frac{2}{5}

The denominator is of the form 2 a x 5 b where a = 0 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

Q1 (9) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 35 / 50

Answer:

\\\frac{35}{50}=\frac{5\times 7}{2\times 5^{2}}\\\frac{35}{50}=\frac{7}{2\times 5}

The denominator is of the form 2 a x 5 b where a = 1 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

Q1 (10) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 77 / 210

Answer:

\\\frac{77}{210}=\frac{7\times 11}{2\times 3\times5\times 7 }\\ \frac{77}{210}=\frac{11}{2\times 3\times5}

The denominator is not of the form 2 a x 5 b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

Q2 Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Answer:

decimal expansions of rational numbers are

(i) \frac{13}{3125}=0.00416

(ii)&nbsnbsp; \frac{17}{8}=2.125

(iv) \frac{15}{1600}=0.009375

(vI) \frac{23}{2^{3}\times 5^{2}}=\frac{23}{200}=0.115

(viii) \frac{6}{15}=\frac{2}{5}=0.4

(ix) \frac{35}{50}=\frac{7}{2\times 5}=0.7

Q3 (1) 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, p / q what can you say about the prime factors of q? 43.123456789

Answer:

\\43.123456789=\frac{43123456789}{10^{9}}\\ 43.123456789=\frac{43123456789}{2^{9}\times 5^{9}}

The denominator is of the form 2 a x 5 b where a = 9 and b = 9. Therefore the given number is rational and has a terminating decimal expansion.

Q3 (2) 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, p / q what can you say about the prime factors of q? 0.120120012000120000

Answer:

Since the decimal part of the given number is non-terminating and non-repeating we can conclude that the given number is irrational and cannot be written in the form \frac{p}{q} where p and q are integers.

Q3 (3) 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, p/q what can you say about the prime factors of q?

43. \overline{123456789}

Answer:

As the decimal part of the given number is non-terminating and repeating, the number is rational but its denominator will have factors other than 2 and 5.

Summary Of NCERT solutions for Real Numbers Class 10 Maths

The following points have been examined by us:

  1. Euclid's Division Lemma states that for any positive integers a and b, there exist whole numbers q and r such that a = bq + r where 0 ≤ r < b.
  2. Euclid's Division Algorithm, which is based on Euclid's division lemma, outlines the procedure for finding the highest common factor (HCF) of two positive integers a and b with a > b as follows: Step 1: Apply the division lemma to find q and r where a = bq + r, 0 ≤ r < b. Step 2: If r = 0, then the HCF is b. If r ≠ 0, apply the Euclid Lemma to b and r. Step 3: Continue the process until the remainder is zero. The divisor at this stage will be the HCF (a, b). Moreover, HCF (a, b) = HCF (b, r).
  3. The Fundamental Theorem of Arithmetic affirms that any composite number can be represented (factorized) as a product of primes, and this factorization is unique, except for the order in which the prime factors appear.

NCERT Books and NCERT Syllabus

Real Numbers Class 10 Solutions - Chapter Wise

If interested students can also check exercises here:

How to Use NCERT Solutions for Maths Class 10 Chapter 1 Real Numbers?

  • Now you have gone through the NCERT solutions for Class 10 Maths Chapter 1 Real Numbers and have a good knowledge of answer writing in a structural manner. It's time to practice various kinds of problems based on real numbers.

  • After the completion of the NCERT syllabus, you can check past 5-year papers of board exams. Previous year papers will increase your dealing ability with a variety of questions.

  • NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After covering NCERT and the previous year papers, you can jump to the next chapters.

NCERT solutions of class 10 - subject wise

NCERT Exemplar solutions - Subject wise

Benefits of NCERT Solutions for Maths Class 10 Chapter 1 Real Numbers

  • Using the NCERT Class 10 Maths solutions chapter 1, students will be able to confirm the right answers once they are done solving the questions themselves.

  • Class 10 Maths Chapter 1 NCERT solutions are solved by the subject matter experts. Therefore, the answers to all the questions are reliable.

  • NCERT Class 10 solutions for Chapter 1 Maths gives the step by step explanations to all the questions which makes it easy for the students to understand.

Frequently Asked Question (FAQs)

1. What are the main topics covered in the NCERT Solutions for Class 10 Maths Chapter 1?

NCERT Solutions for Class 10 Maths Real Numbers includes topics like Euclid's Division Algorithm, Euclid's Division Algorithm to Find the Highest Common Factor,  Prime & Composite Numbers, and Rational & Irrational Numbers. NCERT Solutions for ncert class 10 maths chapter 1  is designed by our experts to provide students with a solid foundation in the fundamental concepts and techniques of mathematics. it will help students to score well on tests as well as in board exams. 

2. Are NCERT Solutions for Class 10 Maths Chapter 1 imprtant?

Yes, NCERT Solutions for maths class 10 chapter 1 are important for several reasons. Firstly, these solutions provide detailed explanations of the various concepts and principles covered in the chapter.

Secondly, the NCERT Solutions for chapter 1 maths class 10  often include a large number of examples and practice problems, which can help students apply their knowledge to solve real-world problems. 

Thirdly, the NCERT Solutions for class 10 chapter 1 maths are typically written by experienced teachers or subject matter experts.

Overall, real numbers class 10 solutions are essential resources for students who want excel in maths.

3. What is the weightage of the chapter Real Numbers for CBSE board exam?

As per the CBSE, the total weightage of class 10 maths ch 1  Real Number is 4 marks in the final board exam. To get a good score follow the NCERT syllabus and problems from the NCERT book and the previous year's papers. Interested students can use class 10 maths chapter 1 solutions to command the concepts.

4. Where can I find the complete solutions of NCERT Class 10 Maths?

The NCERT solutions for all 15 chapters of Class 10 Mathematics can be downloaded from the careers360 website. these solutions are listed chapter-wise in tabulate format above in this article. students can visit maths ch 1 class 10 solutions if they are facing any issues while solving the maths ch 1 class 10 problems.

5. How many chapters are there in the Class 10 Maths?

There are 15 chapters in Class 10 Maths NCERT syllabus. it includes Real Numbers, Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmetic Progressions, Triangles, Coordinate Geometry, Introduction to Trigonometry, Some Applications of Trigonometry, Circles, Constructions, Area Related to Circles, Surface Areas and Volumes, Statistics, and Probability.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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