Class 10 Mathematics - NCERT Mathematics

NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers

Chapter 1 introduces the Fundamental Theorem of Arithmetic and proofs of irrationality. Students learn to find HCF and LCM using prime factorisation and to prove that numbers like √2, √3, and √5 are irrational using proof by contradiction. This chapter carries 6 marks in the CBSE Class 10 board exam 2027.

4 Exercises17 QuestionsStep-by-Step SolutionsUpdated 2026-27
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Syllabus Update 2026-27

As per the CBSE 2026-27 syllabus, Chapter 1 Real Numbers has only 2 exercises - Exercise 1.1 (Fundamental Theorem of Arithmetic, HCF, LCM) and Exercise 1.2 (Proofs of irrationality). The older exercises on Euclid's Division Algorithm and Decimal Expansions of Rational Numbers have been removed from the CBSE syllabus and will NOT be asked in the 2027 board exam. However, solutions for the older exercises are provided below for state board students (UP Board, Bihar Board, MP Board, Rajasthan Board) who may still follow the 4-exercise format.

Unit: Number Systems
6 Marks1 MCQ (1 mark) + 1 short answer on HCF/LCM by prime factorisation (2 marks) + 1 proof of irrationality (3 marks)

Table of Contents

1.1Exercise 1.1 - Fundamental Theorem of Arithmetic

Q1Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

(i) 140:
140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7
140 = 2² × 5 × 7

(ii) 156:
156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13
156 = 2² × 3 × 13

(iii) 3825:
3825 = 3 × 1275 = 3 × 3 × 425 = 3 × 3 × 5 × 85 = 3 × 3 × 5 × 5 × 17
3825 = 3² × 5² × 17

(iv) 5005:
5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13
5005 = 5 × 7 × 11 × 13

(v) 7429:
7429 = 17 × 437 = 17 × 19 × 23
7429 = 17 × 19 × 23

Q2Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers: (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

(i) 26 and 91:
26 = 2 × 13
91 = 7 × 13

HCF = 13 (common prime factor with lowest power)
LCM = 2 × 7 × 13 = 182 (all prime factors with highest power)

Verification: LCM × HCF = 182 × 13 = 2366
Product = 26 × 91 = 2366 ✓

(ii) 510 and 92:
510 = 2 × 3 × 5 × 17
92 = 2² × 23

HCF = 2
LCM = 2² × 3 × 5 × 17 × 23 = 23460

Verification: 23460 × 2 = 46920
510 × 92 = 46920 ✓

(iii) 336 and 54:
336 = 2⁴ × 3 × 7
54 = 2 × 3³

HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3024

Verification: 3024 × 6 = 18144
336 × 54 = 18144 ✓

Q3Find 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 = 2² × 3
15 = 3 × 5
21 = 3 × 7

HCF = 3 (only common factor)
LCM = 2² × 3 × 5 × 7 = 420

(ii) 17, 23, and 29:
All three are prime numbers. No common factors.

HCF = 1
LCM = 17 × 23 × 29 = 11339

(iii) 8, 9, and 25:
8 = 2³
9 = 3²
25 = 5²

No common prime factors.
HCF = 1
LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800

Q4Given that HCF(306, 657) = 9, find LCM(306, 657).

We know: HCF × LCM = Product of two numbers

9 × LCM = 306 × 657
9 × LCM = 201042
LCM = 201042 ÷ 9

LCM(306, 657) = 22338

Q5Check whether 6ⁿ can end with the digit 0 for any natural number n.

For a number to end with digit 0, it must be divisible by 10, i.e., its prime factorisation must contain both 2 and 5 as factors.

6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ

The prime factorisation of 6ⁿ contains only 2 and 3. There is no factor of 5.

Since 5 is not a prime factor of 6ⁿ for any n, 6ⁿ can never end with the digit 0.

Q6Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

First number: 7 × 11 × 13 + 13
= 13(7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
= 13 × 78

Since 13 × 78 is a product of two integers both greater than 1, it is a composite number.

Second number: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009

Since 5 × 1009 is a product of two integers both greater than 1, it is a composite number.

Q7There 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?

They will meet at the starting point after LCM(18, 12) minutes.

18 = 2 × 3²
12 = 2² × 3

LCM = 2² × 3² = 4 × 9 = 36

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

1.2Exercise 1.2 - Irrational Numbers

Q1Prove that √5 is irrational.

Proof by contradiction:

Let us assume, to the contrary, that √5 is rational.

Then √5 = p/q, where p and q are co-prime integers (HCF(p,q) = 1) and q ≠ 0.

Squaring both sides:
5 = p²/q²
p² = 5q² ... (i)

This means 5 divides p².
By the Fundamental Theorem of Arithmetic, since 5 is prime, 5 must divide p.

Let p = 5k for some integer k.

Substituting in (i):
(5k)² = 5q²
25k² = 5q²
q² = 5k² ... (ii)

This means 5 divides q² and therefore 5 divides q.

But both p and q are divisible by 5, which contradicts our assumption that p and q are co-prime (HCF should be 1, but 5 is a common factor).

This contradiction arose because of our incorrect assumption.

Therefore, √5 is irrational. ∎

Q2Prove that 3 + 2√5 is irrational.

Proof by contradiction:

Let us assume, to the contrary, that 3 + 2√5 is rational.

Then 3 + 2√5 = a/b, where a, b are integers and b ≠ 0.

Rearranging:
2√5 = a/b − 3 = (a − 3b)/b
√5 = (a − 3b)/(2b)

Since a, b are integers, (a − 3b)/(2b) is a rational number.
This means √5 is rational.

But we have already proved in Q1 that √5 is irrational. This is a contradiction.

This contradiction arose because of our incorrect assumption that 3 + 2√5 is rational.

Therefore, 3 + 2√5 is irrational. ∎

Q3Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2

(i) Prove 1/√2 is irrational:

Assume 1/√2 is rational, i.e., 1/√2 = a/b (co-prime).
Then √2 = b/a, which is rational.
But √2 is irrational (can be proved like Q1). Contradiction.
Therefore, 1/√2 is irrational. ∎

(ii) Prove 7√5 is irrational:

Assume 7√5 is rational, i.e., 7√5 = a/b.
Then √5 = a/(7b), which is rational.
But √5 is irrational (proved in Q1). Contradiction.
Therefore, 7√5 is irrational. ∎

(iii) Prove 6 + √2 is irrational:

Assume 6 + √2 is rational, i.e., 6 + √2 = a/b.
Then √2 = a/b − 6 = (a − 6b)/b, which is rational.
But √2 is irrational. Contradiction.
Therefore, 6 + √2 is irrational. ∎

old-1.1Old Exercise 1.1 - Euclid's Division Algorithm (For State Boards)

This exercise has been removed from the CBSE 2026-27 syllabus. It is provided here for UP Board, Bihar Board, MP Board, and other state board students who still follow the older textbook.

Q1Use Euclid's division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

(i) HCF of 135 and 225:

225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0

Remainder is 0. HCF(135, 225) = 45

(ii) HCF of 196 and 38220:

38220 = 196 × 195 + 0

Remainder is 0 in the first step. HCF(196, 38220) = 196

(iii) HCF of 867 and 255:

867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0

HCF(867, 255) = 51

Q2Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Let a be any positive integer and b = 6.

By Euclid's division lemma: a = 6q + r, where 0 ≤ r < 6.
So r can be 0, 1, 2, 3, 4, or 5.

Possible forms: a = 6q, 6q+1, 6q+2, 6q+3, 6q+4, 6q+5.

Checking which are odd:
• 6q = 2(3q) → even
• 6q + 1 = 2(3q) + 1 → odd ✓
• 6q + 2 = 2(3q + 1) → even
• 6q + 3 = 2(3q + 1) + 1 → odd ✓
• 6q + 4 = 2(3q + 2) → even
• 6q + 5 = 2(3q + 2) + 1 → odd ✓

Therefore, any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5. ∎

Q3An 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?

Maximum number of columns = HCF(616, 32).

616 = 32 × 19 + 8
32 = 8 × 4 + 0

Maximum number of columns = 8

Q4Use 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.

Let a be any positive integer. By Euclid's division lemma with b = 3:
a = 3q + r, where r = 0, 1, or 2.

Case 1: a = 3q
a² = 9q² = 3(3q²) = 3m, where m = 3q²

Case 2: a = 3q + 1
a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1, where m = 3q² + 2q

Case 3: a = 3q + 2
a² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 4q + 1) + 1 = 3m + 1

In every case, a² is of the form 3m or 3m + 1. ∎

old-1.4Old Exercise 1.4 - Decimal Expansions (For State Boards)

This exercise has been removed from the CBSE 2026-27 syllabus. Provided for state board students.

Q1Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: (i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/(2³×5²) (vii) 129/(2²×5⁷×7⁵) (viii) 6/15 (ix) 35/50 (x) 77/210

A rational number p/q (in lowest terms) has a terminating decimal if q = 2ⁿ × 5ᵐ.

(i) 13/3125: 3125 = 5⁵ → Terminating
(ii) 17/8: 8 = 2³ → Terminating
(iii) 64/455: 455 = 5 × 7 × 13, contains 7 → Non-terminating repeating
(iv) 15/1600: 1600 = 2⁶ × 5² → Terminating
(v) 29/343: 343 = 7³ → Non-terminating repeating
(vi) 23/(2³×5²): Already 2ⁿ×5ᵐ form → Terminating
(vii) 129/(2²×5⁷×7⁵): Contains 7⁵ → Non-terminating repeating
(viii) 6/15 = 2/5: 5 = 5¹ → Terminating
(ix) 35/50 = 7/10: 10 = 2×5 → Terminating
(x) 77/210 = 11/30: 30 = 2×3×5, contains 3 → Non-terminating repeating

Q2Write down the decimal expansions of those rational numbers in Question 1 which have terminating decimal expansions.

(i) 13/3125 = 13×2⁵/10⁵ = 416/100000 = 0.00416
(ii) 17/8 = 17×5³/10³ = 2125/1000 = 2.125
(iv) 15/1600 = 15×5⁴/10⁶ = 9375/1000000 = 0.009375
(vi) 23/(2³×5²) = 23×5/10³ = 115/1000 = 0.115
(viii) 6/15 = 2/5 = 4/10 = 0.4
(ix) 35/50 = 7/10 = 0.7

Q3The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or irrational: (i) 43.123456789 (ii) 0.120120012000120000... (iii) 43.123456789123456789...

(i) 43.123456789 - Terminating decimal (finite digits). Rational.

(ii) 0.120120012000120000... - Non-terminating and non-repeating (the gap of zeros keeps growing). Irrational.

(iii) 43.123456789123456789... - Non-terminating but repeating (the block 123456789 repeats). Rational.

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Previous Year Board Exam Questions

CBSE 20263 marks

Prove that 4 − 2√5 is an irrational number, given that √5 is irrational.

Assume 4 − 2√5 is rational, i.e., 4 − 2√5 = a/b.
Then 2√5 = 4 − a/b = (4b − a)/b.
√5 = (4b − a)/(2b), which is rational.
But √5 is given to be irrational. Contradiction.
Therefore, 4 − 2√5 is irrational. ∎

CBSE 20243 marks

Find the HCF and LCM of 72 and 120 using the prime factorisation method. Also verify that HCF x LCM = product of two numbers.

72 = 2^3 x 3^2
120 = 2^3 x 3 x 5

HCF = 2^3 x 3 = 24 (lowest powers of common primes)
LCM = 2^3 x 3^2 x 5 = 360 (highest powers of all primes)

Verification: HCF x LCM = 24 x 360 = 8640
Product = 72 x 120 = 8640
Since 8640 = 8640, HCF x LCM = product of two numbers is verified.

CBSE 20253 marks

Prove that √2 is an irrational number.

Assume √2 = p/q (co-prime). Then p² = 2q², so 2|p. Let p = 2k. Then 4k² = 2q², so q² = 2k², meaning 2|q. Both p,q divisible by 2 contradicts co-prime. Hence √2 is irrational. ∎

CBSE 20232 marks

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

6ⁿ = (2×3)ⁿ = 2ⁿ × 3ⁿ. For a number to end in 0, it needs both 2 and 5 as factors. Since 5 is not a factor of 6ⁿ, it cannot end with 0.

CBSE 20232 marks

Find the HCF and LCM of 72 and 120.

72 = 2³ × 3². 120 = 2³ × 3 × 5.
HCF = 2³ × 3 = 24.
LCM = 2³ × 3² × 5 = 360.

CBSE 20232 marks

Using prime factorisation, find HCF and LCM of 96 and 120.

96 = 2⁵ × 3. 120 = 2³ × 3 × 5.
HCF = 2³ × 3 = 24.
LCM = 2⁵ × 3 × 5 = 480.

CBSE 20233 marks

Find the greatest 3-digit number which is divisible by 18, 24 and 36.

LCM(18, 24, 36): 18 = 2×3², 24 = 2³×3, 36 = 2²×3².
LCM = 2³ × 3² = 72.
Greatest 3-digit number = 999.
999 ÷ 72 = 13.875. So 72 × 13 = 936.
Greatest 3-digit number divisible by 18, 24, and 36 is 936.

CBSE 2023 Compartment3 marks

If √2 is irrational, prove that (5 − 2√2) is also an irrational number.

Assume 5 − 2√2 = a/b (rational).
2√2 = 5 − a/b = (5b − a)/b.
√2 = (5b − a)/(2b), which is rational.
But √2 is irrational. Contradiction.
Hence 5 − 2√2 is irrational. ∎

CBSE 20203 marks

Prove that √3 is an irrational number.

Assume √3 = p/q (co-prime). Then p² = 3q², so 3|p. Let p = 3k. Then 9k² = 3q², q² = 3k², so 3|q. Both divisible by 3 contradicts co-prime. Hence √3 is irrational. ∎

CBSE 20203 marks

The traffic lights at three different road crossings change after every 48s, 72s, and 108s. If they change simultaneously at 7 am, at what time will they change together next?

LCM(48, 72, 108):
48 = 2⁴ × 3, 72 = 2³ × 3², 108 = 2² × 3³.
LCM = 2⁴ × 3³ = 16 × 27 = 432 seconds = 7 minutes 12 seconds.
They will change together at 7:07:12 am.

CBSE 20202 marks

Find the HCF and LCM of 26, 65, 117 using prime factorisation.

26 = 2 × 13. 65 = 5 × 13. 117 = 3² × 13.
HCF = 13. LCM = 2 × 3² × 5 × 13 = 1170.

CBSE 20193 marks

Prove that 2 + 5√3 is an irrational number, given that √3 is irrational.

Assume 2 + 5√3 = a/b.
5√3 = (a − 2b)/b.
√3 = (a − 2b)/(5b), which is rational. Contradicts √3 being irrational.
Hence 2 + 5√3 is irrational. ∎

CBSE 20191 mark

Write the smallest number which is divisible by both 306 and 657.

The smallest number divisible by both 306 and 657 is their LCM.

Given: HCF(306, 657) = 9
Using HCF x LCM = product of two numbers:
9 x LCM = 306 x 657 = 201042
LCM = 201042 / 9

Smallest number = LCM(306, 657) = 22338

CBSE 20191 mark

If a = x³y² and b = xy³, where x and y are prime numbers, find LCM(a, b).

Given: a = x^3 y^2 and b = x y^3, where x and y are prime numbers.

LCM = take the highest power of each prime factor:

  • Highest power of x: x^3 (from a)
  • Highest power of y: y^3 (from b)

LCM(a, b) = x^3 y^3

CBSE 20191 mark

If HCF(336, 54) = 6, find LCM(336, 54).

Using the property: HCF x LCM = product of two numbers

6 x LCM = 336 x 54
6 x LCM = 18144
LCM = 18144 / 6

LCM(336, 54) = 3024

CBSE 20243 marks

Prove that (2 + sqrt(3))/5 is an irrational number, given that sqrt(3) is irrational.

Assume (2 + sqrt(3))/5 is rational, i.e., (2 + sqrt(3))/5 = a/b where a, b are integers and b != 0.

Then 2 + sqrt(3) = 5a/b
sqrt(3) = 5a/b - 2 = (5a - 2b)/b

Since a, b are integers, (5a - 2b)/b is rational.
This means sqrt(3) is rational.
But sqrt(3) is given to be irrational. This is a contradiction.

Therefore, (2 + sqrt(3))/5 is irrational.

CBSE 2021 (Term 1)1 mark

The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. Find the HCF.

Let HCF = x. Then LCM = 14x.
LCM + HCF = 600
14x + x = 600
15x = 600
HCF = 40

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Practice PYQ Patterns

Assertion-Reason Questions

(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true

AR1

Assertion (A): The product of two consecutive positive integers is divisible by 2.

Reason (R): In any two consecutive integers, one is always even.

Answer: (A)

Both are true. The reason correctly explains the assertion - since one of any two consecutive integers must be even (divisible by 2), their product is always divisible by 2.

AR2

Assertion (A): (7 × 13 × 11) + 11 and (7 × 6 × 5 × 4 × 3 × 2 × 1) + 3 are composite numbers.

Reason (R): (3 × 12 × 101) + 4 is not a composite number.

Answer: (C)

Assertion is TRUE: 7×13×11+11 = 11(91+1) = 11×92 (composite). 7!+3 = 3(7×6×5×4×2×1+1) = 3×1681 (composite). Reason is FALSE: 3×12×101+4 = 3636+4 = 3640 = 2³×5×7×13 which IS composite.

AR3

Assertion (A): 5ⁿ cannot end with the digit 0, where n is a natural number.

Reason (R): Prime factorisation of 5 has only two factors: 1 and 5.

Answer: (A)

Both true. 5ⁿ = 5ⁿ (no factor of 2), so it can never be divisible by 10 and hence cannot end in 0. The reason correctly explains this - since 5 has no factor of 2.

AR4

Assertion (A): 15ⁿ ends with the digit zero, where n is a natural number.

Reason (R): Any number ends with digit zero if its prime factorisation is of the form 2ᵐ × 5ⁿ.

Answer: (D)

Assertion is FALSE: 15¹ = 15 (ends in 5, not 0). 15² = 225 (ends in 5). 15ⁿ never ends in 0 because 15ⁿ = 3ⁿ × 5ⁿ (no factor of 2). Reason is TRUE as a general statement.

AR5

Assertion (A): The HCF of two numbers is 5 and their product is 150. Then their LCM is 40.

Reason (R): For any two positive integers, HCF(a,b) × LCM(a,b) = a × b.

Answer: (D)

Reason is TRUE. But Assertion is FALSE: LCM = 150/5 = 30, not 40.

AR6

Assertion (A): The HCF of two numbers is 18 and their product is 3042. Then their LCM is 169.

Reason (R): For any two positive integers, HCF(a,b) + LCM(a,b) = a × b.

Answer: (C)

Assertion is TRUE: LCM = 3042/18 = 169. Reason is FALSE: the correct formula is HCF × LCM = a × b (multiplication, not addition).

AR7

Assertion (A): HCF of 24, 30, and 42 is 6.

Reason (R): HCF of two or more numbers = product of the smallest power of each common prime factor.

Answer: (A)

24=2³×3, 30=2×3×5, 42=2×3×7. Common primes: 2¹×3¹ = 6. Both assertion and reason are true, and the reason explains the assertion.

Case Study Questions

CS1Prime Number GameCBSE 2024

Ms. Mukta planned a prime number game for her class. She announces the number 2 and asked the first student to multiply it by a prime number and then pass it to the second student. Each student multiplied the number by a prime number and passed it forward. The last student got the number 173250.

(i) What is the least prime number used by students? (1 mark)

First find the prime factorisation: 173250 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11 = 2¹ × 3² × 5³ × 7¹ × 11¹.

The teacher announced 2 (the starting number). The students used: 3, 3, 5, 5, 5, 7, 11.

The least prime number used by students is 3.

(ii)(a) How many students are in the class? (2 marks)

Total prime factor multiplications = 2 (from 3²) + 3 (from 5³) + 1 (from 7) + 1 (from 11) = 7.

The teacher gave the first factor (2). Students did 7 multiplications.

There are 7 students in the class.

(ii)(b) What is the highest prime number used by students? (1 mark)

From the factorisation 2 × 3² × 5³ × 7 × 11, the primes used by students are 3, 5, 7, 11.

The highest prime number used is 11.

(iii) Which prime number has been used the maximum number of times? (1 mark)

From 2¹ × 3² × 5³ × 7¹ × 11¹:

  • 3 was used 2 times
  • 5 was used 3 times
  • 7 and 11 were used once each

5 has been used the maximum number of times (3 times).

CS2Book Donation DriveCBSE 2023 Compartment

Some people donated books to a public library on International Book Giving Day. The number of books donated were: History = 96, Science = 240, Mathematics = 336. These books have to be arranged in minimum number of stacks such that each stack contains books of only one subject and the number of books in each stack is the same.

(i) How many books are arranged in each stack? (1 mark)

Number of books per stack = HCF(96, 240, 336).
96 = 2⁵ × 3, 240 = 2⁴ × 3 × 5, 336 = 2⁴ × 3 × 7.
HCF = 2⁴ × 3 = 48 books per stack.

(ii) How many stacks are used to arrange all the Mathematics books? (1 mark)

Mathematics books = 336. Books per stack = 48.
Stacks = 336 ÷ 48 = 7 stacks.

(iii)(a) Determine the total number of stacks. (2 marks)

History stacks = 96/48 = 2
Science stacks = 240/48 = 5
Maths stacks = 336/48 = 7

Total stacks = 2 + 5 + 7 = 14.

MCQ Practice

MCQ1. If HCF(a, b) = 4 and a × b = 9696, then LCM(a, b) is:

(A) 2424

(B) 1212

(C) 9696

(D) 4848

Answer: (A) - LCM = (a × b)/HCF = 9696/4 = 2424.

MCQ2. The decimal expansion of 13/3125 is:

(A) terminating

(B) non-terminating repeating

(C) non-terminating non-repeating

(D) none

Answer: (A) - 3125 = 5⁵. Denominator has only 5 as prime factor (form 2⁰×5⁵), so it terminates.

MCQ3. If p is a prime number, the HCF of p² and p³ is:

(A) p

(B) p²

(C) p³

(D) p⁶

Answer: (B) - HCF = lowest power of common prime = p².

MCQ4. The LCM of two numbers is 9 times their HCF. The sum of LCM and HCF is 500. The HCF is:

(A) 45

(B) 50

(C) 55

(D) 40

Answer: (B) - LCM = 9×HCF. So 9×HCF + HCF = 500 → 10×HCF = 500 → HCF = 50.

MCQ5. For some integer m, every even integer is of the form:

(A) m

(B) m + 1

(C) 2m

(D) 2m + 1

Answer: (C) - Even integers are multiples of 2, so they can be written as 2m.

MCQ6. If two positive integers p and q can be expressed as p = ab² and q = a³b, where a and b are prime numbers, then LCM(p, q) is:

(A) ab

(B) a²b²

(C) a³b²

(D) a³b³

Answer: (C) - LCM = highest powers = a³ × b² = a³b².

MCQ7. The product of a non-zero rational and an irrational number is:

(A) always irrational

(B) always rational

(C) rational or irrational

(D) one

Answer: (A) - A non-zero rational times an irrational is always irrational. If the result were rational, dividing by the rational number would make the irrational number rational - a contradiction.

MCQ8. If the HCF of 65 and 117 is expressible in the form 65m − 117, then the value of m is:

(A) 1

(B) 2

(C) 3

(D) 4

Answer: (B) - HCF(65,117) = 13. So 65m − 117 = 13 → 65m = 130 → m = 2.

MCQ9. The number 6ⁿ, for any natural number n, can never end with the digit:

(A) 2

(B) 6

(C) 0

(D) 8

Answer: (C) - 6ⁿ = (2×3)ⁿ. To end in 0, needs factor of 5. Since 6ⁿ has no factor of 5, it never ends in 0.

MCQ10. The least number that is divisible by all the numbers from 1 to 10 is:

(A) 10

(B) 100

(C) 504

(D) 2520

Answer: (D) - LCM(1,2,...,10) = 2³×3²×5×7 = 2520.

MCQ11. If n is any natural number, then 6ⁿ − 5ⁿ always ends with:

(A) 1

(B) 3

(C) 5

(D) 7

Answer: (A) - 6ⁿ always ends in 6. 5ⁿ always ends in 5. 6−5 = 1. So 6ⁿ−5ⁿ always ends in **1**.

MCQ12. The ratio of HCF to LCM of the least composite number and the least prime number is:

(A) 1:2

(B) 2:1

(C) 1:1

(D) 1:3

Answer: (A) - Least composite = 4, least prime = 2. HCF(4,2) = 2, LCM(4,2) = 4. Ratio HCF:LCM = 2:4 = 1:2.

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Important Questions - Mark-wise

1 Mark Questions

State the Fundamental Theorem of Arithmetic.

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

What is the HCF of two prime numbers?

The HCF of any two prime numbers is always 1, since primes have no common factor other than 1.

Express 140 as a product of its prime factors.

140 = 2² × 5 × 7.

Write the prime factorisation of 96.

96 = 2⁵ × 3.

What is the LCM of 8 and 9?

8 = 2³, 9 = 3². No common factor. LCM = 2³ × 3² = 72.

2-3 Mark Questions

Find the LCM and HCF of 26 and 91, and verify that LCM × HCF = product of the two numbers.

26 = 2 × 13. 91 = 7 × 13. HCF = 13. LCM = 2 × 7 × 13 = 182.
Verification: 182 × 13 = 2366. 26 × 91 = 2366. ✓

Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.

96 = 2⁵ × 3. 404 = 2² × 101.
HCF = 2² = 4.
LCM = (96 × 404)/4 = 38784/4 = 9696.

Check whether 4ⁿ can end with the digit 0 for any natural number n.

4ⁿ = (2²)ⁿ = 2²ⁿ. Contains only 2 as a prime factor, no 5. Cannot end in 0.

Two tankers contain 850 litres and 680 litres of kerosene oil. Find the maximum capacity of a container which can measure the kerosene of both the tankers exactly.

Maximum capacity = HCF(850, 680).
850 = 2 × 5² × 17. 680 = 2³ × 5 × 17.
HCF = 2 × 5 × 17 = 170 litres.

3-4 Mark Questions

Prove that √5 is an irrational number.

Assume √5 = p/q (co-prime). Then p² = 5q², so 5|p. Let p=5k. Then 25k² = 5q², q² = 5k², so 5|q. Both divisible by 5 contradicts co-prime. Hence √5 is irrational. ∎

Prove that 3 + 2√5 is irrational.

Assume 3 + 2√5 = a/b. Then √5 = (a−3b)/(2b), which is rational. But √5 is irrational. Contradiction. Hence 3 + 2√5 is irrational. ∎

Find the HCF and LCM of 6, 72 and 120 using the prime factorisation method.

6 = 2 × 3. 72 = 2³ × 3². 120 = 2³ × 3 × 5.
HCF = 2 × 3 = 6.
LCM = 2³ × 3² × 5 = 360.

Key Concepts

Fundamental Theorem of Arithmetic - every composite number has a unique prime factorisationHCF by prime factorisation - take lowest power of common primesLCM by prime factorisation - take highest power of all primesHCF × LCM = product of two numbers (only valid for exactly 2 numbers)Proof by contradiction - assume the opposite, derive a contradictionIrrational numbers - cannot be expressed as p/q; their decimal expansions are non-terminating non-repeating

Important Formulas

Euclid's Division Lemma: a = bq + r, where 0 ≤ r < b
HCF × LCM = Product of two numbers (for exactly 2 numbers only)
Terminating decimal ↔ denominator (in lowest terms) = 2ⁿ × 5ᵐ
If p is prime and p divides a², then p divides a

Study Plan - Real Numbers

1

Fundamental Theorem of Arithmetic

Read NCERT theory pages. Understand prime factorisation with examples. Solve Exercise 1.1 Q1-Q3.

2

HCF and LCM Applications

Solve Exercise 1.1 Q4-Q7. Focus on word problems (Q7 type).

3

Irrational Numbers - Proofs

Learn proof by contradiction structure. Solve Exercise 1.2 Q1-Q3. Write proofs from memory.

4

Previous Year Questions

Solve all 15 PYQs from this page. Pay attention to composite irrationality proofs (4−2√5 type).

5

MCQ + Assertion-Reason + Case Study

Practice all MCQs. Attempt assertion-reason and case study questions under timed conditions.

6

Full Revision Test

Attempt all important questions (mark-wise) as a timed test. Review common mistakes.

Common Mistakes to Avoid

  • ✗Using HCF × LCM = product for THREE numbers - this formula works only for exactly TWO numbers
  • ✗Not starting irrationality proofs with 'Let us assume, to the contrary, that √5 is rational' - examiners look for this exact phrasing
  • ✗Forgetting to write 'where p and q are co-prime integers and q ≠ 0' when assuming p/q form
  • ✗Not stating 'This contradicts our assumption that p and q are co-prime' at the end of the proof
  • ✗In composite irrationality proofs (like 4−2√5), forgetting to rearrange to isolate the irrational term
  • ✗Confusing HCF (lowest powers of COMMON primes) with LCM (highest powers of ALL primes)

Tips for Board Exam

  • ✓Irrationality proofs (√2, √3, √5) are guaranteed 3-mark questions - memorise the 6-step structure and practice writing it in under 5 minutes
  • ✓Since 2023, CBSE asks composite irrationality proofs like 'prove 4−2√5 is irrational given √5 is irrational' - these follow a simpler pattern than the basic proofs
  • ✓For HCF/LCM problems, always show the complete prime factorisation tree - examiners award 1 mark just for the factorisation step
  • ✓In case study questions, read the full context first. Sub-parts (ii), (iii) often use the answer from part (i)
  • ✓MCQ tip: for 'decimal expansion' questions, always reduce the fraction to lowest terms BEFORE checking the denominator

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About Real Numbers - Class 10 Mathematics

Chapter 1 Real Numbers is the opening chapter of CBSE Class 10 Mathematics and forms the foundation for understanding number theory. The chapter is part of the Number Systems unit, which carries 6 marks in the CBSE Class 10 board exam 2027. Real Numbers is considered one of the most scoring chapters because the question types are predictable and the proofs follow a fixed structure that can be memorised.

What Are Real Numbers?

Real numbers include all rational numbers (numbers that can be written as p/q where q is not zero) and all irrational numbers (numbers that cannot be written as p/q, such as the square root of 2, the square root of 3, and pi). Together, rational and irrational numbers make up the entire number line. Every point on the number line corresponds to a real number and every real number corresponds to a point on the number line.

In earlier classes, students learned about natural numbers, whole numbers, integers, and rational numbers. Class 10 Chapter 1 builds on this foundation by introducing two powerful tools: the Fundamental Theorem of Arithmetic and the concept of proof by contradiction for irrational numbers.

Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of prime numbers in a unique way, apart from the order of factors. For example, 140 can only be written as 2 squared times 5 times 7. No other combination of primes will give 140. This uniqueness is what makes the theorem so powerful.

This theorem is the backbone of Chapter 1. It is used in three major applications in this chapter. First, it is used to find the HCF (Highest Common Factor) and LCM (Lowest Common Multiple) of two or more numbers through prime factorisation. To find the HCF, we take the lowest power of each common prime factor. For LCM, we take the highest power of every prime factor that appears in any of the numbers.

A key property students must remember is that HCF multiplied by LCM equals the product of the two numbers. However, this identity holds only for exactly two numbers, not for three or more. This is one of the most common mistakes students make in board exams and is frequently tested in MCQs and assertion-reason questions.

Second, the Fundamental Theorem is used to determine whether the decimal expansion of a rational number is terminating or non-terminating repeating. A rational number p/q (in its lowest terms) has a terminating decimal expansion if and only if the prime factorisation of q is of the form 2 raised to n times 5 raised to m, where n and m are non-negative integers. If q has any prime factor other than 2 or 5, the decimal expansion will be non-terminating repeating.

Third, the theorem is used as a key step in proofs of irrationality. When we assume that the square root of a prime number is rational and arrive at a contradiction, the Fundamental Theorem guarantees that if a prime p divides a squared, then p must divide a. This logical step is essential for completing irrationality proofs.

Proofs of Irrationality

The second major topic in Chapter 1 is proving that certain numbers are irrational. Students learn to use the method of proof by contradiction, which is one of the most elegant techniques in mathematics.

The structure of an irrationality proof follows these steps. First, assume to the contrary that the number (say the square root of 5) is rational. Write it as p/q where p and q are co-prime integers and q is not zero. Second, square both sides to get p squared equals 5 times q squared. Third, conclude that 5 divides p squared, and by the Fundamental Theorem, 5 must divide p. Fourth, let p equal 5k for some integer k, substitute back, and show that 5 also divides q. Fifth, since both p and q are divisible by 5, this contradicts the assumption that they are co-prime. Sixth, conclude that the square root of 5 is irrational.

This proof structure appears in virtually every CBSE board exam as a 3-mark question. Students should memorise the six-step structure and practice writing it in under 5 minutes.

In recent years from 2023 to 2026, CBSE has evolved the irrationality question pattern. Instead of asking students to prove that the square root of 2 is irrational from scratch, they now provide that the square root of 5 or the square root of 2 or the square root of 3 is irrational and ask students to prove that a composite expression like 4 minus 2 times the square root of 5 or (3 plus the square root of 7) divided by 5 is irrational. This variation is actually simpler. It requires just 4 to 5 lines where you assume the composite expression is rational, rearrange to isolate the irrational term, show it equals a rational expression, and note the contradiction.

Euclid's Division Algorithm (For State Boards)

While Euclid's Division Algorithm has been removed from the CBSE 2026-27 syllabus, it remains part of the curriculum for state boards including UP Board, Bihar Board, MP Board, and Rajasthan Board. Euclid's Division Lemma states that for any two positive integers a and b, there exist unique integers q (quotient) and r (remainder) such that a equals b times q plus r, where r is greater than or equal to 0 and less than b.

The algorithm is a systematic way to find the HCF of two numbers by repeatedly applying the division lemma until the remainder becomes zero. The last non-zero remainder is the HCF. This method is particularly useful for large numbers where prime factorisation would be tedious.

Students studying under state boards should practice both Euclid's method and the prime factorisation method, as either can appear in their exams.

CBSE 2026-27 Syllabus Changes

For the CBSE 2026-27 academic session, Chapter 1 Real Numbers has been trimmed to contain only two exercises. Exercise 1.1 covers the Fundamental Theorem of Arithmetic with 7 questions on prime factorisation, HCF, LCM, composite numbers, and word problems. Exercise 1.2 covers proofs of irrationality with 3 questions.

The older exercises on Euclid's Division Algorithm (previously Exercise 1.1) and Decimal Expansions of Rational Numbers (previously Exercise 1.4) have been removed from the CBSE syllabus. These topics will not appear in the CBSE Class 10 board exam 2027. However, Padhaao provides solutions for these older exercises as well, since state board students in UP, Bihar, MP, Rajasthan, and Haryana may still follow the four-exercise format.

How This Chapter Is Tested in Board Exams

Real Numbers questions in the CBSE board exam typically follow this pattern. One MCQ worth 1 mark tests basic concepts like HCF, LCM, prime factorisation properties, or the nature of decimal expansions. One short answer question worth 2 marks asks students to find the HCF and LCM of two or three numbers using prime factorisation, or to verify that HCF times LCM equals the product of the numbers. One long answer question worth 3 marks requires proving the irrationality of a number using proof by contradiction.

Since 2023, CBSE has also introduced case study questions based on real-life scenarios involving prime factorisation. These carry 4 to 5 marks and have 3 to 4 sub-questions. Assertion-Reason questions, which test conceptual understanding without requiring calculations, also feature regularly. Students should practice both formats as they are relatively new additions to the exam pattern.

Tips for Scoring Full Marks

To score all 6 marks from this chapter, students should focus on three things. First, memorise the irrationality proof structure and practice writing proofs for the square root of 2, 3, 5, and composite forms. Second, practice finding HCF and LCM by prime factorisation quickly and accurately, always showing the complete factor tree. Third, solve at least 10 to 15 previous year board questions from this chapter to get familiar with how the examiner phrases questions and awards marks.

Common mistakes to avoid include using the HCF times LCM formula for three numbers (it only works for two), not mentioning the co-prime condition when starting an irrationality proof, confusing HCF (lowest powers of common primes) with LCM (highest powers of all primes), and not simplifying fractions to lowest terms before checking for terminating decimals.

Connection to Other Chapters

Real Numbers connects to several other chapters in Class 10 Maths. The concept of prime factorisation reappears in Chapter 2 (Polynomials) when discussing the relationship between zeros and coefficients. The proof by contradiction technique used for irrationality proofs is a general mathematical method that develops logical thinking required throughout higher mathematics. Understanding HCF and LCM is useful for solving word problems in Chapters 3 (Linear Equations) and 5 (Arithmetic Progressions).

For students preparing for competitive exams like NTSE, Olympiads, or JEE Foundation, the number theory concepts in this chapter form an essential base. Euclid's Algorithm, in particular, has applications in computer science (encryption algorithms) and is worth understanding even if it is not in the CBSE syllabus.

Frequently Asked Questions

How many exercises are in NCERT Class 10 Maths Chapter 1 for CBSE 2026-27?

For CBSE 2026-27, Chapter 1 Real Numbers has only 2 exercises: Exercise 1.1 (7 questions on Fundamental Theorem of Arithmetic, HCF, LCM) and Exercise 1.2 (3 questions on proving irrationality). The older exercises on Euclid's Division Algorithm and decimal expansions have been removed from the CBSE syllabus.

What is the Fundamental Theorem of Arithmetic?

The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorised) as a product of prime numbers, and this factorisation is unique apart from the order in which the prime factors appear. For example, 140 = 2² × 5 × 7 is the only way to express 140 as a product of primes.

How many marks does Real Numbers carry in CBSE Class 10 board exam 2027?

Real Numbers is part of the Number Systems unit which carries 6 marks in the CBSE Class 10 board exam. Typically: 1 MCQ (1 mark) + 1 short answer on HCF/LCM (2 marks) + 1 irrationality proof (3 marks).

How to prove √2 is irrational for the CBSE board exam?

Use proof by contradiction: (1) Assume √2 = p/q where p, q are co-prime and q≠0. (2) Square both sides: p² = 2q². (3) Since 2 divides p², 2 must divide p. Let p = 2k. (4) Substituting: 4k² = 2q², so q² = 2k², meaning 2 divides q. (5) Both p and q divisible by 2 contradicts them being co-prime. (6) Therefore √2 is irrational.

What is the difference between HCF and LCM in prime factorisation?

HCF = product of the LOWEST powers of COMMON prime factors only. LCM = product of the HIGHEST powers of ALL prime factors from all numbers. Example: for 12 = 2² × 3 and 18 = 2 × 3², HCF = 2¹ × 3¹ = 6, LCM = 2² × 3² = 36.

Is Chapter 1 Real Numbers deleted from the CBSE 2026-27 syllabus?

No, Chapter 1 is NOT deleted. However, it has been trimmed. Only the Fundamental Theorem of Arithmetic and Irrational Numbers sections remain (Exercises 1.1 and 1.2). Euclid's Division Algorithm and Decimal Expansions sections have been removed from CBSE but may still be in state board syllabi.

What are assertion-reason questions in Real Numbers?

Assertion-Reason questions give two statements - an Assertion (A) and a Reason (R). You must determine: (A) Both true, R explains A, (B) Both true, R doesn't explain A, (C) A true but R false, (D) A false but R true. These test conceptual understanding of properties like HCF×LCM = product of numbers.

What type of questions come from Chapter 1 in CBSE board exam?

Three types: (1) MCQs on HCF/LCM, prime factorisation, and properties - 1 mark each. (2) Short answer: find HCF and LCM of given numbers, or show a number is composite - 2 marks. (3) Long answer: prove irrationality of √2, √3, √5, or composite forms like 4−2√5 - 3 marks. Case study questions on prime factorisation were introduced in CBSE 2024.

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