NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1

Real Numbers include rational and irrational numbers, and Chapter 1 begins by studying important properties of positive integers.
NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1 connect Real Numbers with prime factorisation, HCF, LCM and the Fundamental Theorem of Arithmetic.

Chapter 1 Real Numbers introduces two major ideas: Euclid’s division algorithm and the Fundamental Theorem of Arithmetic. Exercise 1.1 is based mainly on prime factorisation Class 10, HCF and LCM Class 10, and applications of the Fundamental Theorem of Arithmetic Class 10. Students learn to express numbers as products of prime factors, find HCF and LCM using prime powers, verify the relation HCF × LCM = product of two numbers, and solve a circular track timing problem. The textbook states that every composite number can be expressed as a product of primes in a unique way, apart from the order of the prime factors.

Key Takeaways

  • Prime Factorisation: Every composite number can be written as a product of primes.
  • HCF: Product of the smallest powers of common prime factors.
  • LCM: Product of the greatest powers of all prime factors involved.
  • Two Numbers Rule: HCF(a, b) × LCM(a, b) = a × b.

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1 Structure 2026

Exercise No. Main Topic Question Count
Exercise 1.1 Prime factorisation 1
Exercise 1.1 HCF and LCM 3
Exercise 1.1 Composite numbers and LCM word problem 3

NCERT Class 10 Maths Chapter 1 Exercise 1.1 Solutions

Exercise 1.1 focuses on the Fundamental Theorem of Arithmetic. The chapter uses prime factorisation to find HCF and LCM and also explains why some expressions must be composite.

Class 10 Real Numbers Solutions for Prime Factorisation

Prime factorisation means writing a number as a product of prime numbers. The order of prime factors may change, but the prime factorisation remains unique.

Q1. Express each number as a product of its prime factors.

Q1(i). 140

Prime factorisation:

140 = 14 × 10

140 = 2 × 7 × 2 × 5

140 = 2² × 5 × 7

Answer:

140 = 2² × 5 × 7

Q1(ii). 156

Prime factorisation:

156 = 2 × 78

156 = 2 × 2 × 39

156 = 2² × 3 × 13

Answer:

156 = 2² × 3 × 13

Q1(iii). 3825

Prime factorisation:

3825 = 3 × 1275

3825 = 3 × 3 × 425

3825 = 3² × 5 × 85

3825 = 3² × 5 × 5 × 17

3825 = 3² × 5² × 17

Answer:

3825 = 3² × 5² × 17

Q1(iv). 5005

Prime factorisation:

5005 = 5 × 1001

1001 = 7 × 143

143 = 11 × 13

So:

5005 = 5 × 7 × 11 × 13

Answer:

5005 = 5 × 7 × 11 × 13

Q1(v). 7429

Prime factorisation:

7429 = 17 × 437

437 = 19 × 23

So:

7429 = 17 × 19 × 23

Answer:

7429 = 17 × 19 × 23

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1 for HCF and LCM

The textbook explains that HCF is found using the smallest powers of common prime factors, while LCM is found using the greatest powers of all prime factors involved.

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

Q2(i). 26 and 91

Prime factorisation:

26 = 2 × 13

91 = 7 × 13

Common prime factor:

13

So:

HCF = 13

LCM = 2 × 7 × 13

LCM = 182

Verification:

LCM × HCF = 182 × 13

LCM × HCF = 2366

Product of numbers = 26 × 91

Product of numbers = 2366

Answer:

HCF = 13

LCM = 182

Verified:

LCM × HCF = 26 × 91

Q2(ii). 510 and 92

Prime factorisation:

510 = 2 × 3 × 5 × 17

92 = 2² × 23

Common prime factor:

2

So:

HCF = 2

LCM = 2² × 3 × 5 × 17 × 23

LCM = 23460

Verification:

LCM × HCF = 23460 × 2

LCM × HCF = 46920

Product of numbers = 510 × 92

Product of numbers = 46920

Answer:

HCF = 2

LCM = 23460

Verified:

LCM × HCF = 510 × 92

Q2(iii). 336 and 54

Prime factorisation:

336 = 2⁴ × 3 × 7

54 = 2 × 3³

Common prime factors:

2 and 3

Smallest powers:

2¹ and 3¹

So:

HCF = 2 × 3

HCF = 6

Greatest powers:

2⁴, 3³ and 7

LCM = 2⁴ × 3³ × 7

LCM = 16 × 27 × 7

LCM = 3024

Verification:

LCM × HCF = 3024 × 6

LCM × HCF = 18144

Product of numbers = 336 × 54

Product of numbers = 18144

Answer:

HCF = 6

LCM = 3024

Verified:

LCM × HCF = 336 × 54

NCERT Class 10 Maths Chapter 1 Exercise 1.1 Solutions for Three Integers

For three integers, find HCF using the smallest powers of common prime factors and LCM using the greatest powers of all prime factors.

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

Q3(i). 12, 15 and 21

Prime factorisation:

12 = 2² × 3

15 = 3 × 5

21 = 3 × 7

Common prime factor:

3

So:

HCF = 3

LCM = 2² × 3 × 5 × 7

LCM = 4 × 3 × 5 × 7

LCM = 420

Answer:

HCF = 3

LCM = 420

Q3(ii). 17, 23 and 29

17, 23 and 29 are prime numbers.

They have no common prime factor except 1.

So:

HCF = 1

LCM = 17 × 23 × 29

LCM = 11339

Answer:

HCF = 1

LCM = 11339

Q3(iii). 8, 9 and 25

Prime factorisation:

8 = 2³

9 = 3²

25 = 5²

There is no common prime factor.

So:

HCF = 1

LCM = 2³ × 3² × 5²

LCM = 8 × 9 × 25

LCM = 1800

Answer:

HCF = 1

LCM = 1800

Class 10 Maths Real Numbers Solutions for HCF and LCM Relation

For two positive integers:

HCF × LCM = product of the two numbers

This relation is used when HCF is already given.

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

Given:

HCF(306, 657) = 9

Use:

HCF × LCM = product of the two numbers

So:

9 × LCM = 306 × 657

LCM = (306 × 657)/9

First divide:

306 ÷ 9 = 34

So:

LCM = 34 × 657

LCM = 22338

Answer:

LCM(306, 657) = 22338

Fundamental Theorem of Arithmetic Class 10: Application Questions

The Fundamental Theorem of Arithmetic says that prime factorisation is unique. This helps decide whether a number can end with 0 and whether a given expression is composite.

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

A number ending in 0 must be divisible by 10.

So, its prime factorisation must contain:

10 = 2 × 5

Now:

6ⁿ = (2 × 3)ⁿ

6ⁿ = 2ⁿ × 3ⁿ

The prime factorisation of 6ⁿ contains only 2 and 3.

It does not contain 5.

Therefore, 6ⁿ cannot be divisible by 10.

Answer:

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

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

First expression:

7 × 11 × 13 + 13

Take 13 common:

7 × 11 × 13 + 13 = 13(7 × 11 + 1)

Now:

7 × 11 = 77

So:

13(77 + 1) = 13 × 78

Since it is a product of numbers greater than 1, it is composite.

Answer:

7 × 11 × 13 + 13 is composite.

Second expression:

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Take 5 common:

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

Now:

7 × 6 × 4 × 3 × 2 × 1 = 1008

So:

5(1008 + 1) = 5 × 1009

Since it is a product of numbers greater than 1, it is composite.

Answer:

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is composite.

HCF and LCM Class 10 Word Problem

When two people complete rounds in different times and start together, they meet again at the starting point after the LCM of their individual times.

Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round, while Ravi takes 12 minutes for the same. If they start at the same point, at the same time and go in the same direction, after how many minutes will they meet again at the starting point?

Sonia completes one round in:

18 minutes

Ravi completes one round in:

12 minutes

They will meet again at the starting point after:

LCM of 18 and 12

Prime factorisation:

18 = 2 × 3²

12 = 2² × 3

LCM = 2² × 3²

LCM = 4 × 9

LCM = 36

Answer:

They will meet again at the starting point after 36 minutes.

Real Numbers Class 10: Concepts Used in Exercise 1.1

Exercise 1.1 is based on the Fundamental Theorem of Arithmetic and its use in finding HCF and LCM. These concepts are important for understanding divisibility and prime factorisation.

Real Numbers Class 10

Real Numbers include rational and irrational numbers. In Chapter 1, students study positive integers through divisibility, prime factorisation, HCF and LCM.

Prime Factorisation Class 10

Prime factorisation means expressing a number as a product of prime numbers.

Copy-friendly example:

140 = 2² × 5 × 7

Fundamental Theorem of Arithmetic Class 10

Every composite number can be expressed as a product of primes, and this factorisation is unique except for the order of prime factors.

Copy-friendly statement:

Composite number = product of prime factors

HCF and LCM Class 10

HCF is found using the smallest powers of common prime factors.

LCM is found using the greatest powers of all prime factors involved.

Copy-friendly relation for two numbers:

HCF(a, b) × LCM(a, b) = a × b

Euclid Division Algorithm Class 10

Euclid division algorithm is another method used to find the HCF of two positive integers. In Exercise 1.1, the main method used is prime factorisation.

Quick Formula Table for NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1

Concept Copy-Friendly Result Used In
Prime factorisation Composite number = product of primes Q1
HCF Product of smallest powers of common prime factors Q2, Q3
LCM Product of greatest powers of all prime factors Q2, Q3, Q7
HCF-LCM relation HCF × LCM = product of two numbers Q2, Q4

Useful Links for Class 10 Maths NCERT Solutions

Section Useful Links
Class 10 Maths NCERT Solutions NCERT Solutions for Class 10 Maths
Chapter 1 NCERT Solutions for Class 10 Maths Chapter 1
Chapter 2 NCERT Solutions for Class 10 Maths Chapter 2
Chapter 3 NCERT Solutions for Class 10 Maths Chapter 3
Chapter 4 NCERT Solutions for Class 10 Maths Chapter 4
Chapter 5 NCERT Solutions for Class 10 Maths Chapter 5
Chapter 6 NCERT Solutions for Class 10 Maths Chapter 6
Chapter 7 NCERT Solutions for Class 10 Maths Chapter 7
Chapter 8 NCERT Solutions for Class 10 Maths Chapter 8
Chapter 9 NCERT Solutions for Class 10 Maths Chapter 9
Chapter 10 NCERT Solutions for Class 10 Maths Chapter 10
Chapter 11 NCERT Solutions for Class 10 Maths Chapter 11
Chapter 12 NCERT Solutions for Class 10 Maths Chapter 12
Chapter 13 NCERT Solutions for Class 10 Maths Chapter 13
Chapter 14 NCERT Solutions for Class 10 Maths Chapter 14

Q.1 Use Euclid’s division algorithm to find the HCF of:

  1. 135 and 225
  2. 196 and 38220
  3. 867 and 255

Ans.

(i)135 and 225Step 1: Since 225>135, we apply the division lemmato 225 and 135 to get225=135×1+90Step 2:Since remainder 90≠0, we apply the division lemmato 135 and 90 to get135=90×1+45Step 3:We consider the new divisor 90 and new remainder 45,and apply the division lemma to get90=2×45+0The remainder has now become zero, so the procedure stops.Since the divisor at this stage is 45,the HCF of 135 and 225 is 45.(ii)  196 and 38220Step 1:Since 38220>196, we apply the division lemmato 38220 and 196 to get38220=196×195+0Since the remainder is zero, the procedure stops.The divisor at this stage is 196.Therefore, HCF of 196 and 38220 is 196.(iii)  867 and 255Step 1:Since 867>255, we apply the division lemmato 867 and 255 to get867=255×3+102Step 2:Since remainder 102≠0,we apply the division lemma to 255 and 102 to get255=102×2+51Step 3:We consider the new divisor 102 and new remainder 51,and apply the division lemma to get102=51×2+0Since the remainder is zero, the procedure stops.The divisor at this stage is 51.Therefore, HCF of 867 and 255 is 51.

Q.2 Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Ans.

Let a is a positive integer and b=6.By Euclid’s algorithm, we geta=6q+r  for some integer q≥0,and 0≤r<6.Therefore,a=6q  or  6q+1  or  6q+2 or  6q+3 or  6q+4  or  6q+5Also, 6q+1=2×3q+1=2k1+1, where k1=3q.6q+3=(6q+2)+1=2(3q+1)+1=2k2+1,where k2=3q+1.6q+5=(6q+4)+1=2(3q+2)+1=2k3+1, where k3=3q+2.Clearly, 6q+1, 6q+3, 6q+5 are of the form 2k+ 1, where k  is an integer.Therefore, 6q+1, 6q+3, 6q+5 are not divisible by 2and so these represent odd numbers.Thus, any odd integer can be expressed in the form 6q+1, or  6q+3,  or  6q+5.

Q.3 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?

Ans.

HCF of 616 and 32 will give the maximum number of columns inwhich each group can march.By Euclid’s algorithm, we get616=32×19+8and32=8×4+0The HCF (616, 32) is 8.Therefore, each group can march in 8 columns.

Q.4 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.]

Ans.

Let a be a positive integer and b=3.By Euclid’s Algorithm,a=3q+r for some integer q≥0 and 0≤r<3.The possible remainders are 0, 1 and 2. Therefore,a can be 3q or 3q+1 or 3q+2.Thus,     a2=9q2 or (3q+1)2 or (3q+2)2           =9q2 or (9q2+6q+1) or (9q2+12q+4)           =3×(3q2) or 3(3q2+2q)+1 or 3(3q2+4q+1)+1           =3k1   or 3k2+1 or 3k3+1where k1,  k2 and  k3 are some positive integers.Hence, square of any positive integer is either of the form3m or 3m + 1 for some integer m.

Q.5 Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+ 8.

Ans.

Let a be a positive integer and b=3a=3q+r, where q≥0 and 0≤r<3Therefore,a=3q or 3q+1 or 3q+2 Therefore, an integer can be represented by these three forms. Case 1: When a = 3q, thena3=(3q)3=27q3=9(3q3)=9mwhere m is an integer and m=3q3.Case 2: When a=3q+1, then                     a3=(3q+1)3               ⇒a3=27q3+27q2+9q+1               ⇒a3=9(3q3+3q2+q)+1               ⇒a3=9m+1Where m is an integer such that m=3q3+3q2+ qCase 3: When a=3q+2, then                     a3= (3q+2)3               ⇒a3=27q3+54q2+36q+8              ⇒a3=9(3q3+6q2+4q)+ 8              ⇒a3=9m+ 8Where m is an integer such that m=3q3+6q2+4qTherefore, the cube of any positive integer is of the form9m, 9m+1,or 9m+8.

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FAQs (Frequently Asked Questions)

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1 cover prime factorisation, HCF, LCM, the HCF-LCM relation, composite numbers and an LCM-based circular path word problem.

The prime factorisation of 140 is 2² × 5 × 7.

The HCF is 13 and the LCM is 182.

6ⁿ cannot end with the digit 0 because its prime factorisation contains only 2 and 3, while a number ending in 0 must also have factor 5.

Sonia and Ravi will meet again at the starting point after 36 minutes. This is the LCM of 18 and 12.