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This chapter is the foundation of all mathematics as it deals with the fundamental concepts of numbers, including rational and irrational numbers, prime numbers, and composite numbers. It is essential to have a clear understanding of these concepts to understand and solve more advanced mathematical problems.
In this chapter, you will learn about Euclid’s division algorithm, the fundamental theorem of arithmetic, and the properties of real numbers such as commutativity, associativity, and distributivity. You will also learn about the decimal expansion of real numbers, and how to differentiate between rational and irrational numbers.
This chapter has a total of 4 exercises, which cover various concepts related to real numbers. The NCERT Solutions for Class Xth Maths Chapter 1 – Real Numbers provide step-by-step solutions to all the exercises, which will help you to understand the concepts and solve the problems effectively. These solutions are designed to be easy to understand and follow, helping you to build a strong foundation in mathematics.
Overall, this chapter is an essential part of the mathematics curriculum and will lay the groundwork for more advanced mathematical topics.

NCERT Solutions for Class Xth Maths Chapter 1. Real Numbers PDF DOWNLOAD

Answers of Class Xth Maths Chapter 1 Real Numbers

Unit 1

Real Numbers 

Exercise 1.1 

Q1 :

Use Euclid’s division algorithm to find the HCF of:

TyFZb4FK6tjanmoat8EXBri80

Answer :

(i) 135 and 225

Since 225 > 135., 

We apply the division lemma to 225 and 135 

2UqTFSZNNtjxiJOMD8EJMBUgaNnFMznCQkYq6XpyMxUI5Ij4SU8ArylQdUV9IAfHVm KH Ao93mazWQ Uyn NkFyGYWUPL0wfNKY4JSPWb6gqj3nG5dz6 3rvGPwHX5la5irfb8

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain

Aku waUBx5PHCT7rcpm2XvWCadi7sOET9ly

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain

Since the remainder is zero, the process stops. Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

Since 38220 > 196, 

we apply the division lemma to 38220 and 196 

eJh8 AeYnwAevoqqUv5z U nUWZfHYrApb JJ1AfCrCBCnDtEkHb3IIUoNKAvD8kRLFBAr8gxGWvJzg3oMQCRqfGZ qWLQWolTyTk9OU4F 3d3XmTya20YPg1MWNqdyYv61TqHM

Since the remainder is zero, the process stops. Since the divisor at this stage is 196, Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

Since 867 > 255, 

We apply the division lemma to 867 and 255 

ckXQsVdEGUHaQSAeyupEmYD XqV8oFjo3DQ nyTyixz Nh khxWrg

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain

Since the remainder is zero, the process stops.

Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.

Q2 :

Show that any positive odd integer is of the form   where q is some integer.

Answer:

Let a be any positive integer and b = 6. 

Then, by Euclid’s algorithm,

a = 6q + r 

For some integer q and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6. 

Therefore,

Also,

where is a positive integer

where is an integer 

where is an integer

Clearly, 

are of the form 2k + 1, where k is an integer.

Therefore, are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form

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 :

HCF (616, 32) will give the maximum number of columns in which they can march. We can use Euclid’s algorithm to find the HCF.

lQYiftz7iUdTIq bo05weo1SMPVBcGLCudyhLTt7Ted9Y4uZaxU5bWEjyC BP VYYWEXE7sgKnqyxCiIBWOSp4FWXUvE4rQporhi QDzslYg4YVVUa oRcgaPH8 VAOiDR5COk
Qz7SRS1ZZF x3 ONuIICmgqZstuDyMqB4iJuax eTEvFHrStnGCzDc4A7K4U7Aqm9KSBbV06iVdFCpdwWDREtDj1w2wzvZiY47 VAODRjjiFrOYmglw4YoOmJHB6bJdaW R1 v0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

Q4 :

Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

Answer :

Let a be any positive integer and b = 3. Then a = 3q + r for some integer q ≥ 0 and r = 0, 1, 2 because

Or,
psebHhComOZ5AOdxL3R xiPNiZa2B XXjOCMA1LunylqcXTkWKtIRsQOyLykHP4zy6gk2FoXOZhGMRXwee2gbo9b9hDw EgH3QV3rBoqlksv6aXf5 omMtp3NP5qwPrh ntzpE
Where are some positive integers 

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

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 a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

tKcO 7vy8owYBLbFeDwoi S9FPmjTjrTwIucVIjvubuBwiwfgJOJ basOgCx1Yth 74i3jqSmNHqZTUdJgeuooz7TAZ Jw1kYRTSTYgYilcJNHDN879h STQTs KBjzKZsnPQKk

Therefore, every number can be represented as these three forms. 

There are three cases.

Case 1: When a = 3q,

IA mjhw YLJRskpiGCFghQ9x Uz xkc6Vb ke8A5gq87tINcwxBSXQ bX1QE6h1TjjUXiDf4cBq2ye57ahqUfCue2 NB3YXyX9Vh1VzwyUzTKvXaVW2sKTp TH5cw9R7 kBDvmQ

Where m is an integer such that m

Case 2: When a = 3q + 1,

s N4P9MGvN0oJp2uBFnv8AU0Ux8MhdJarnbv07j 248d UqawgCLkAsPXWsfRQcVqfrT8lL7DjN4NdSV4SjhVDG7 ZszaBXDH0i6n7sRbEPbY42afZgleBdisGqo 5UD8El9v8

Where m is an integer such that

Case 3: When a = 3q + 2,

kzRhCR60ZVXVOrx2tMbw5xegwvTkO8sNE m Ht70JqNqQQXIh0jSWiDonk0q9qKQcIEYcgfLh7y9e8sqKZpy75Wvf43gVz62bUTHhTR2FI3Hdt6xSK2gk6lq 8GQZekskli9EbU

Where m is an integer such that Therefore, the cube of any positive integer is of the form

Exercise 1.2

Q1:

Express each number as product of its prime factors:

2i5OwNxkUXE

Answer :

Tfq0xOaIbKpmHMNUVavypM6dJkZgFagUZ3lGcpFow44vfyc 094mn8xMvgZWF757GLygzqw4vH2mnTG7qOh3QBfIWKj9ksuhb 9V H9P7SpgOowBH4SmzRYBbSdPmbt zCOMvjU

Q2 :

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

rAR OjF2H 2sUXCVSGRu9ONPIzEh0IKcvkGURKOjGGdmMxMTU8BhBgJqQZ o9kclvJpivDw3E2Hl5BcnrUKb88ZE4Z5VV5TqYlmP0P7tpoLj0fZurOlrwzdh8bJQI j1YlQYihA

Answer :

3smT Puy E7VtKNmm5JWw7OgH4Q0 TN4NPH5GJrtGtGSKd1XOlysUUohvvI6ZPFFlJvTEpT6z85pWcUFeVEHNBNgTX9fNeaN0YiQZsBobBDpxYLMUMmCpxl

Hence, product of two numbers = HCF × LCM

E4m053EpgzxSLdgZpKjJqgT WGPGK oMyg1bX huSvlmGynq8znEqftHmCSs5MuwQYTlcvIdr2gAux BL vRGXTqqeO RIGzkGh3GoOslCVf6nK4nfupfjOCuPWK9QXAE7zEov8

Hence, product of two numbers = HCF × LCM

l7jIcGn

Hence, product of two numbers = HCF × LCM

Q3 :

Find the LCM and HCF of the following integers by applying the prime factorization method.

oQV5iVaJyZ7 ziz53 h8jt2XS FTGLuyI9Z7PcyX4rFyLatzx5y0K42yQVOYdoiCZ1x0V7uJkYcJjyWglUj5BfUCvPJac W4cu761kSSWQ1BrJof44uKD m65nSwPCNbYGDN9YM

Answer :

SJgWjxcqeuAXw BUli rkajMUeq1OCBC7dUe
Bv5Vd78jDA32wz2LHJ95qEymU9fmRYH 9vb4 FdmHFyMi7wKCMxQi3suN2k7Y79nkWSZVwb

*Q4 :

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

Answer :

Q5 :

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

Answer :

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as

Prime factorization of

It can be observed that 5 is not in the prime factorization of 6n. Hence, for any value of n, 6n will not be divisible by 5.

Therefore, 6n cannot end with the digit 0 for any natural number n.

Q6 :

Explain why are composite numbers.

Answer :

Numbers are of two types – prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

KUZa2EmTRXHpXZOdltiCEkfldIca6qgskyjn0q4SRDZFEg9JRoNLt83hBsUPiA11yVIZOiDvgF nGXEiGc467ZiW4btUt2zG60sbgSmuOYbYsIQAvfNCy T2UG3XlL6EggYU3X4
32BbzU1BtDYzUto5cLbNpjCvgBSK9rn2Wl J 2O zYz4hDM4XVXmx8FMyHZh6q74geUYQ 9K3svtiYZOCk Y5JqDQQIKu5NU9HiafMC

The given expression has 6 and 13 as its factors. Therefore, it is a composite number. 

tag E2AxlQyh2ijDUfCFaswtnIuDORYD3e8YvHj7DY3egirW3 NVn6VlrDcsUWjCSaPhU0mKR7X7A3zmTDbGHB7 gie PkDaNfdZDdmuZ9mHODLv4mIO YTrRzuQZpGsrum uak
0VytvVwa i2Mlqorb6gYLiWvr mBBVsq62sI nILQxGGNsOHofftiKXZXPiERD2l4ycPmt96Xuy6i 4w4ZRd0Eyqh PguTQoJZEeeN D4CHK HR6m WZ7TP8 jUaLF5NgeEDfhc

1009 cannot be factorized further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a 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 :

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.

kHsL7FVKUW2bG0svNPgoFuw6ZvKCkBqCYqLbMU 0aj9XvyysrogVJdVQYrF6lcAQp0PR9KS HVqhp9QiReBLvIVH4PK4qeFPkpch5hzHWYsxG2v Yzne01uGeGoiBVwrLTcNNnE

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Exercise 1.3

Q1 :

Prove that   is irrational. 

Answer :

Let  is a rational number.

Therefore, we can find two integers a, b (b ≠ 0) such that

Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and bare co-prime.

sekJnLuQBjUASeQs 75PtIbry4Qrq92vO4SrGSP3OsV0TnZF95 66Um7rWrZa FWnaTWZZiRnCwWpmJkZkedu6quhUh7ZkPEtWJBzpuc1 dPwhRVHW5v03blLEt4IaHo 0gzf3E

Therefore, a2is divisible by 5  and it can be said that a is divisible by . Let a = 5k, where k is an integer

BW034ZJCSrM4GB8vTaEZ5ltShjbV6wetRX1t2fnT7fffN2hs8bgXb36OMb1KSa6 2S1Y3Ld33RD96pk8ACnXT0puL4

This means that b2is divisible by 5 and hence, b is divisible by . This implies that a and b have 5 as a common factor.

And this is a contradiction to the fact that a and b are co-prime. Hence,  cannot be 

expressed as  or it can be said that is irrational.

Q2 :

Prove that  is irrational. 

Answer :

Let  is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

fycYyWeF2cAWjanHwR2HFId1fgfR4MpC1OE v8HRCPGxv0feCs2wWyTvDOb3 ZgJgiYytoHb hfd8kGUPAmLHkQjX8al7OXT8KoFXbpPYcLdGMCUce3ckv3wSt5Dl1kOaYLeXck

Since a and b are integers,  will also be rational and therefore, is rational.

This contradicts the fact that   is irrational. Hence, our assumption that is rational is false. Therefore,    is irrational.

Q3 :

Prove that the following are irrationals:

Answer :

ji6vXpCeBqH7MX NI0KvtSqzd6jH0OP lA2jeMJPoiBIK

Let  is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

4l7C04bgcgeuMLQ 83Zl62hNN28 p0AJ33WGrBjXqqM57QsOm TP 56eHNNRw ZCDqiYCoSs0Fo94xOQzNiFYS2NhVniUQkMLcMW0dB47zCUbcXMbVqwBzrNoYswzKdFWI1T bQ

is rational as a and b are integers.

Therefore,  is rational which contradicts to the fact that dM 6B q s Ynrzh7l7BKf YL3IdXNH 6phVii1L84jCnh1V9  is irrational.


Hence, our assumption is false and is irrational.

OJhaGD8gpy5J4NyqcTPjEFjvui8JyecHifldXiOIKJqJPAqDD3i7AEbYn6haZ8c5f KiJx0dzt4rq5DBu4bxuqtqu3OKqHG9 uvdwInD1zu8hb DvTiwtSsfzgiuPUoSZlR7feQ

Let   is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that for some integers a &b

oqe2nr eJV1 cYyy48Bp TcSXZIFbqUX3D 729DCCdPYG UnlnbAhAXs5EtzFB5RKhIDnuqhyE3gkW99j9 83B10tACalolYGHQQUq0fAUeshDuyJiFlfWnePdGZ

is rational as a and b are integers. Therefore,   should be rational.

This contradicts the fact that is irrational. 

Therefore, our assumption that is rational is false. Hence,  is irrational.

WG4UjgNgKNBy4Vd8 ku4M35XQkUzU3Hp1wYfkWUEu2A9kkGlymRUS0Qvp SCwn9jAzpfCMG 2Zd0R0H8l1SR2ADFh1JJBbw8xdWl6hq3o9c0xpI1m8mbetMe7

Let    be rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

5JHLP3KgecXxuDdOmcy2mAoVSJHKgGuSOReHumxcvJabR6QBi93 Id7qsstEEIvsbybC2H6ximKUosWuX9wMCjP4nCnkmOJ0ktC5Q7 4f 4AWAIelOg1uf84lz J CMvgKppAGU

Since a and b are integers,  is alsorational and hence,should be rational. This contradicts the fact that is irrational. Therefore, our assumption is false and hence,


Exercise 1.4

Q1 :

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:

knk3298OEaFXgF cYxR1NyF OYUU6G2yu

Answer :

(i) rB8t ygFICQAcmuiLYnsAXDHTcuv NEIWMnzIihXvBMaheO2TeKIiFSpJMdKk62j1ehgIVFVyHj R3ljvUF9pHN93xS XKL8Xjni FMwGKeqTJCoLr2IbLQSIkSjY20ye 3eCWBTa wNO2iXGGbptaSb7rAI93G8yILGJYya9oJg8CJxpP8 1 X4GLPtEwXyD6E

The denominator is of the form.

Hence, the decimal expansion of is terminating. 

(ii)

cYKAzM44jV3ZXuntVWtJYUNWTkBghm1obhLcZNgye 2eYPjNC8

The denominator is of the form

Hence, the decimal expansion of is terminating. 

(iii)

M6 0 YbFXAUEkoHfK7 6FxdIFCCoKhl082ecp8hR1YvJdgpS AD2lRB8y6i2WPkt9LDAMwofArrD

Since the denominator is not in the form and it also contains 7 and 13 as its factors, its decimal expansion will be non-terminating repeating.

(iv)   

Z52k5IPnf MEaj SG9b6KhW7pSI3663s6 jx2jxvsXh8CnqXpp4KGubaNIbEE1ELGcMbume r2 6c7wFwXgclzDS10nZbgkmA0vfyfpoPAddNrIBQAtdRynqoW3wGFZXc3cy6UY

The denominator is of the form

Hence, the decimal expansion of  is terminating.

 (v)  

PsP7kzXgHeotZK4rJqnzCEY yFakPMoXSwIRo23MbYO0WXFmJ3Fsry1RePieMy55Yh3zOOK7 u09zFkKYBEFJVkrhlk0AG1eMPSJqUqp4BqPGfpdokzmcSRToILCVvH tpMTMm4

Since the denominator is not in the form and it has 7 as its factor, the decimal expansion of is non-terminating repeating.

(vi)

The denominator is of the form

Hence, the decimal expansion of is terminating. 

(vii)

Since the denominator is not of the form , and it also has 7 as its factor, the decimal expansion of    is non-terminating repeating.

(viii)

The denominator is of the form 5n.

Hence, the decimal expansion of  is terminating. 

(ix)

cPDvpOzLYFRyWpQ3oktqnUqX 0B4Mlfx7dYqRbJc4Xn94S6Ky Ea97er7PQoPUG8oApvPQq8dfc9Yx6 sNZK1ZV7Ddvf5QPlhxFWW3sfNXW w9YkKOUy0IVB0dsMHXT5yyzfJg

The denominator is of the form .

Q2 :

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

Answer :

(i)

(ii)

3ki0sndAYNN5m30Q0ZSMrcNQpz23vYwRtD
wy1x 0CZUPv8hHASQqBucgFSW1btL2uGEsI84RR9K2ZYA1j8ZejIJbT24AcbJ8u1m


(viii)

JnOxrE9hBQ0OPUN73cU8Sh 6ZAysmm5CO1jCd4zbi23 L9OgvvYDZz4Q2G V M4awUtHLqC637ADFJcEtlJdcySm7XfzksFlea6ykdYck4eCW0hU6CupF6yOR UDRTapSwtls18
xQu8IKHHKTz7OYLP6IibM053M5MZfAIMdjFmDGP W PulY63lBlxVjyPoCWUz7DM xwtzC6Qg7LQWhCrVawDN605rEDIQoaaqT9OYgjZs7YYGGQnF oRhXkbJdmpPnsAbXYYth4
S46Jxbme6QCgtajQcKRBVQP5c 91KTVzbdeS7dcwVYpX49Z hkJG LDBow3 XsPRLJaKTqY7hxgtDIwAP8VAKCmZuQDpaQ3IHs9 QSi6aqaSCnO0HEM9Uod1Jb5L1VFoBSIPho

*Q3:

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 what can you say about the prime factor of q?

(i) 43.123456789 (ii) 0.120120012000120000… (iii) h7DhkEvQ8ufxU8MPB7zVxN4vVoEAjPjB47RDPCxJhfe4ynsJqUzCII8R yR 8xRR4G7VCldH8k roe6nYzoSay hqala1hcJFQzlEvPZif hbLEZ p99olo9IfSJBzujELw1FPo

Answer :

(i) 43.123456789

Since this number has a terminating decimal expansion, it is a rational number of the form  and q is of the form

i.e., the prime factors of q will be either 2 or 5 or both. 

(ii) 0.120120012000120000 …

The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number. 

(iii) h7DhkEvQ8ufxU8MPB7zVxN4vVoEAjPjB47RDPCxJhfe4ynsJqUzCII8R yR 8xRR4G7VCldH8k roe6nYzoSay hqala1hcJFQzlEvPZif hbLEZ p99olo9IfSJBzujELw1FPo

Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form

and q is not of the form   i.e., the prime factors of q will also have a factor other than 2 or 5.

NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers

Real Number is one of the important topics in Maths as it has a weightage of 6 marks in class 10 (Number system – Real Numbers) Maths board exams. The average number of questions asked from this chapter is usually 3. Three questions were asked from this chapter in the previous year board examination (2018).

  1. One out of three questions in part A (1 marks).
  2. One out of three questions in part B (2 marks).
  3. One out of three questions in part C (3 marks).

This chapter talks about

  • Euclid’s Division Algorithm
  • The Fundamental Theorem of Arithmetic
  • Revisiting Rational & Irrational Numbers
  • Decimal Expansions

List of Exercises in class 10 Maths Chapter 1:

Exercise 1.1 Solutions 5 Question ( 4 long, 1 short)
Exercise 1.2 Solutions 7 Question ( 4 long, 3 short)
Exercise 1.3 Solutions 3 Question ( 3 short)
Exercise 1.4 Solutions 3 Question ( 3 short)

Real Numbers is introduced in Class 9 and this is discussed more in details in Class 10. NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers provides the answers for the questions present in this chapter. The chapter discusses the real numbers and their applications. The divisibility of integers using Euclid’s division algorithm says that any positive integer a can be divided by another positive integer b such that the remainder will be smaller than b. On the other hand, The Fundamental Theorem of Arithmetic works on multiplication of positive integers.

The chapter starts with the introduction of real numbers in section 1.1 followed by two very important topics in section 1.2 and 1.3

  • Euclid’s Division Algorithm – It includes 5 questions based on Theorem 1.1 – Euclid’s Division Lemma.
  • The Fundamental Theorem of Arithmetic – Explore the applications of this topic which talks about the multiplication of positive integers, through solutions of the 7 problems in Exercise 1.2.

Next, it discusses the following topics which were introduced in class 9.

  • Revisiting Rational & Irrational Numbers – In this, the solutions for 3 problems in Exercise 1.3 are given which also use the topic in the last Exercise 1.2.
  • Decimal Expansions – It explores when the decimal expansion of a rational number is terminating and when it is recurring. It includes a total of 3 problems with sub-parts in Exercise 1.4

Key Features of NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers

  • These NCERT Solutions let you solve and revise the whole syllabus of class 10.
  • After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
  • It follows NCERT guidelines which help in preparing the students accordingly.
  • It contains all the important questions from the examination point of view.
  • It helps in scoring well in maths in exams.

Conclusion

The NCERT Solutions for Class Xth Maths Chapter 1 – Real Numbers provide a comprehensive understanding of the fundamental concepts of numbers, including rational and irrational numbers, prime numbers, and composite numbers. The chapter covers important topics such as Euclid’s division algorithm, the fundamental theorem of arithmetic, and the properties of real numbers.
The solutions provided by the NCERT are well-structured and easy to understand, with step-by-step explanations for all the exercises. The solutions are designed to help students develop their problem-solving skills and gain a deeper understanding of the concepts covered in the chapter.
Overall, the NCERT Solutions for Class Xth Maths Chapter 1 – Real Numbers are an excellent resource for students who want to excel in the subject. The solutions will not only help them score well in their exams but also prepare them for more advanced topics in mathematics. It is crucial to have a clear understanding of the concepts covered in this chapter as it forms the foundation for all mathematical concepts.

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

The main topics covered in the NCERT Solutions for Class 10 Maths Real Numbers are Euclid’s division algorithm, the fundamental theorem of arithmetic, revisiting rational & irrational numbers and decimal expansions. Based on these concepts, our experts created the solutions effectively. By giving importance to these concepts students can score good marks in class tests as well as in Class 10 board exams.

How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 1?

There are four exercises in the NCERT Solutions for Class 10 Maths Real Numbers. These Solutions of NCERT Maths Class 10 Chapter 1 help the students in solving the problems adroitly and efficiently. They also focus on cracking the Solutions of Maths in such a way that it is easy for the students to understand.

Are NCERT Solutions for Class 10 Maths Chapter 1 important from the exam point of view?

Yes, the first chapter Real Numbers of NCERT Solutions for Class 10 is important from an exam perspective. It contains short answer questions as well as long answer questions to increase the ability of problem-solving. These solutions are created by the SWC’s expert faculty to help students in the preparation of their examinations.


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