CBSE Class 10 Maths Chapter 1 Real Numbers Important Questions

Chapter 1 of Class 10 Mathematics, Real Numbers, is one of the most important and foundational chapters in the CBSE syllabus. It introduces students to concepts that not only appear in board exams but also form the basis of higher mathematical learning. To prepare effectively, students must practice a wide variety of class 10 maths chapter 1 extra questions. These questions help in revising all concepts thoroughly and in understanding the different ways problems can be asked in the examination.

The chapter covers topics such as Euclid’s Division Lemma, Fundamental Theorem of Arithmetic, proofs of irrational numbers, and decimal expansions of rational numbers. Each of these concepts is essential for problem-solving. By working on real numbers class 10 extra questions with answers, students strengthen their logical thinking and problem-solving skills.

Practicing extra questions for class 10 maths chapter 1 is highly beneficial because it improves accuracy, builds confidence, and ensures better time management during exams. Questions are usually asked in different formats — from simple one-mark objective questions to detailed five-mark proofs. That is why this article provides a wide range of class 10 real numbers extra questions and solutions covering all difficulty levels.

The class 10th maths chapter 1 extra questions presented here are designed to match the exam pattern and NCERT guidelines. These ch 1 maths class 10 extra questions include short answer questions, long answer questions, and higher-order thinking problems. Solving them will ensure complete preparation and help students avoid common mistakes.

By the end of this practice, students will have covered all important aspects of the chapter. These class 10 ch 1 maths extra questions will not only enhance preparation for the board exam but also build a strong foundation for advanced mathematics. Regular practice of these problems ensures that concepts remain clear, and students can approach the exam with confidence.

Do Check: CBSE Class 10 Maths Important Questions

What is Real Number?

Real numbers is a combination of rational and irrational numbers. They are the numbers upon which we easily perform mathematical operations. All the numbers  are not imaginary are real numbers. For example, 33, -11, 7.99, 3/2, π(3.14), √2, etc.

Real Numbers Class 10 – Chapter Overview

Before attempting questions, let’s revise the key concepts of this chapter:

1. Euclid’s Division Lemma

For any two positive integers a and b, there exist unique integers q and r such that:
a = bq + r, where 0 ≤ r < b.
This lemma is widely used to find the HCF of two numbers.

2. Fundamental Theorem of Arithmetic

Every composite number can be expressed (factorised) as a product of prime numbers, and this representation is unique apart from the order of primes.
Example: 210 = 2 × 3 × 5 × 7.

3. Irrational Numbers

Numbers like √2, √3, √5 are irrational. Proofs often use the contradiction method.

4. Decimal Expansions of Rational Numbers

If a rational number is in the form p/q (q ≠ 0):

• If q has only 2 and/or 5 as its prime factors → terminating decimal.
• If q has any other prime factors → non-terminating repeating decimal.

Real numbers Class 10 Important Questions with Solutions (1 Mark)

Ques. Find the HCF of 65 and 117 using Euclid’s Division Lemma.

Answer:

65 and 117

117 = 65 × 1 + 52

65 = 52 × 1 + 13

52 = 13 × 4 + 0

Hence, HCF = 13. 

Ques. Write the decimal expansion of 13/2.

Answer:

13 ÷ 2 = 6.5

Therefore, 13/2 = 6.5 (terminating). 

Ques. State whether 7/8 has a terminating or non-terminating decimal expansion.

Answer:

Denominator = 8 = 2³ (contains only 2).

Hence, 7/8 has a terminating decimal expansion. 

Do Check: NCERT Solutions for Class 10 Maths

Class 10 Real numbers Important Questions (2 Marks)

Ques. Show that √3 is irrational.

Answer:

Assume √3 = p/q (p, q are coprime, q ≠ 0).

⇒ 3 = p²/q² ⇒ p² = 3q².

So p² divisible by 3 ⇒ p divisible by 3 ⇒ p = 3k.

Substitute: (3k)² = 3q² ⇒ 9k² = 3q² ⇒ q² = 3k² ⇒ q divisible by 3.

Thus both p and q divisible by 3 → contradiction.

Hence, √3 is irrational. 

Ques. Express 360 as a product of its prime factors.

Answer:

360 = 2 × 2 × 2 × 3 × 3 × 5

= 2³ × 3² × 5. 

Ques. Without actual division, check whether 441 will have a terminating decimal expansion.

Answer:

441 = 3² × 7² (denominator factorisation has primes other than 2 or 5).

Hence, decimal expansion is non-terminating repeating. 

Ques. Find HCF and LCM of 13 and 17 by prime factorisation method. (2013)

Solution:

13 = 1 × 13; 17 = 1 × 17

HCF = 1 and LCM = 13 × 17 = 221

Ques. Find LCM of numbers whose prime factorisation are expressible as 3 × 5² and 3² × 7². (2014)

Solution: LCM (3 × 5², 3² × 7²) = 3² × 5² × 7² = 9 × 25 × 49 = 11025

Ques. Find the largest number which divides 70 and 125 leaving remainder 5 and 8 respectively. (2015)

Solution: It is given that on dividing 70 by the required number, there is a remainder 5.

This means that 70 – 5 = 65 is exactly divisible by the required number.

Similarly, 125 – 8 = 117 is also exactly divisible by the required number.

65 = 5 × 13

117 = 3² × 13

HCF = 13

Required number = 13

Ques. Show that 3√7 is an irrational number. (2016)

Solution: Let us assume, to the contrary, that 3√7 is rational.

That is, we can find coprime a and b (b ≠ 0) such that 3√7 = \(\frac { a }{ b }\)

Rearranging, we get √7 = \(\frac { a }{ 3b }\)

Since 3, a and b are integers, \(\frac { a }{ 3b }\) is rational, and so √7 is rational.

But this contradicts the fact that √7 is irrational.

So, we conclude that 3√7 is irrational.

Do Check: CBSE Class 10 Maths Sample Papers

Ques. Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number? (2015)

Solution:

17 × 5 × 11 × 3 × 2 + 2 × 11 …(i)

= 2 × 11 × (17 × 5 × 3 + 1)

= 2 × 11 × (255 + 1)

= 2 × 11 × 256

Number (i) is divisible by 2, 11 and 256, it has more than 2 prime factors. Therefore (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number.

Ques. Check whether 4n can end with the digit 0 for any natural number n. (2015)

Solution: 4ⁿ = (2²)ⁿ = 2²ⁿ

The only prime in the factorization of 4ⁿ is 2.

There is no other prime in the factorization of 4ⁿ = 2²ⁿ

(By uniqueness of the Fundamental Theorem of Arithmetic).

5 does not occur in the prime factorization of 4ⁿ for any n.

Therefore, 4ⁿ does not end with the digit zero for any natural number n.

Ques. Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons. (2017 OD)

Solution: No, LCM = Product of the highest power of each factor involved in the numbers.

HCF = Product of the smallest power of each common factor.

We can conclude that LCM is always a multiple of HCF, i.e., LCM = k × HCF

We are given that,

LCM = 175 and HCF = 15

175 = k × 15

⇒ 11.67 = k

But in this case, LCM ≠ k × HCF

Therefore, two numbers cannot have LCM as 175 and HCF as 15.

Class 10 Chapter 1 Real Numbers Important Questions (3 Marks)

Ques. Find the HCF and LCM of 96 and 404 using prime factorisation.

Answer: 96 = 2⁵ × 3

404 = 2² × 101

HCF = 2² = 4

LCM = (2⁵ × 3 × 101) = 9696

Check: HCF × LCM = 4 × 9696 = 38784 = 96 × 404 

Ques. Use Euclid’s Division Lemma to show that the square of any positive integer is either of the form 3m or 3m + 1.

Answer: Let a be any positive integer.

On dividing by 3, we get:

a = 3q, 3q + 1, or 3q + 2.

Case 1: a = 3q ⇒ a² = 9q² = 3(3q²) ⇒ form 3m.

Case 2: a = 3q + 1 ⇒ a² = (3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 ⇒ form 3m + 1.

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

Thus proved. 

Do Check: CBSE Class 10 Maths Syllabus

Ques. Find the decimal expansion of 77/210.

Answer: Simplify fraction: 77/210 = 11/30.

Denominator 30 = 2 × 3 × 5 (contains 3).

Hence, non-terminating repeating decimal.

11 ÷ 30 = 0.3666... 

Class 10 Maths Chapter 1 Real Numbers Important Questions (4–5 Marks)

Ques. Prove that there are infinitely many primes of the form 6n + 5.

Answer: Assume there are finitely many such primes: p1, p2, p3…pn.

Consider N = (6p1p2…pn) – 1.

Now, N is of the form 6n – 1 = 6m + 5.

So N is divisible by a prime of the form 6n + 5, which is not in our list.

Contradiction.

Hence, infinitely many primes of the form 6n + 5 exist. 

Ques. A merchant has 120 L of oil of one kind and 180 L of another kind. He wants to sell the oil by filling equal quantities in tins without mixing. Find the greatest measure of oil in each tin.

Answer: We need the HCF of 120 and 180.

120 = 2³ × 3 × 5

180 = 2² × 3² × 5

HCF = 2² × 3 × 5 = 60.

So the greatest measure = 60 litres per tin. 

Ques. 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.

On dividing a by 3:

a = 3q, 3q + 1, or 3q + 2.

Case 1: a = 3q ⇒ a³ = 27q³ = 9(3q³) ⇒ form 9m.

Case 2: a = 3q + 1 ⇒ a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 ⇒ form 9m + 1.

Case 3: a = 3q + 2 ⇒ a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 ⇒ form 9m + 8.

Thus proved.

Ques. If two positive integers x and y are expressible in terms of primes as x = p²q³ and y = p³q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain. (2014)

Solution: x = p²q³ and y = p³q

LCM = p³q³

HCF = p²q …..(i)

Now, LCM = p³q³

⇒ LCM = pq² (p²q)

⇒ LCM = pq² (HCF)

Yes, LCM is a multiple of HCF.

Explanation:

Let a = 12 = 2² × 3

b = 18 = 2 × 3²

HCF = 2 × 3 = 6 …(ii)

LCM = 2² × 3² = 36

LCM = 6 × 6

LCM = 6 (HCF) …[From (ii)]

Here LCM is 6 times HCF.

Do Check: CBSE Class 10 Maths Notes

Ques. Show that one and only one out of n, (n + 1) and (n + 2) is divisible by 3, where n is any positive integer. (2015)

Solution: Let n, n + 1, n + 2 be three consecutive positive integers.

We know that n is of the form 3q, 3q + 1, or 3q + 2.

Case I. When n = 3q,

In this case, n is divisible by 3,

but n + 1 and n + 2 are not divisible by 3.

Case II. When n = 3q + 1,

In this case n + 2 = (3q + 1) + 2

= 3q + 3

= 3(q + 1 ), (n + 2) is divisible by 3,

but n and n + 1 are not divisible by 3.

Case III.

When n = 3q + 2, in this case,

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

= 3q + 3 = 3 (q + 1 ), (n + 1) is divisible by 3,

but n and n + 2 are not divisible by 3.

Hence, one and only one out of n, n + 1 and n + 2 is divisible by 3.

Ques. Find the HCF and LCM of 306 and 657 and verify that LCM × HCF = Product of the two numbers. (2016 D)

Solution:

306 = 2 × 3² × 17

657 = 32 × 73

HCF = 3² = 9

LCM = 2 × 3² × 17 × 73 = 22338

L.H.S. = LCM × HCF = 22338 × 9 = 201042

R.H.S. = Product of two numbers = 306 × 657 = 201042

L.H.S. = R.H.S.

Ques. Show that any positive odd integer is of the form 41 + 1 or 4q + 3 where q is a positive integer. (2016 OD)

Solution: Let a be a positive odd integer

By Euclid’s Division algorithm:

a = 4q + r …[where q, r are positive integers and 0 ≤ r < 4]

a = 4q

or 4q + 1

or 4q + 2

or 4q + 3

But 4q and 4q + 2 are both even

a is of the form 4q + 1 or 4q + 3.

Ques. Dudhnath has two vessels containing 720 ml and 405 ml of milk respectively. Milk from these containers is poured into glasses of equal capacity to their brim. Find the minimum number of glasses that can be filled. (2014)

Solution: 1st vessel = 720 ml; 2nd vessel = 405 ml

We find the HCF of 720 and 405 to find the maximum quantity of milk to be filled in one glass.

405 = 3⁴ × 5

720 = 2⁴ × 3² × 5

HCF = 3² × 5 = 45 ml = Capacity of glass

No. of glasses filled from 1st vessel = \(\frac { 720 }{ 45 }\) = 16

No. of glasses filled from 2nd vessel = \(\frac { 405 }{ 45 }\) = 9

Total number of glasses = 25

Ques. Amita, Sneha, and Raghav start preparing cards for all persons of an old age home. In order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together? (2013)

Solution: To find the earliest (least) time, they will start preparing a new card together, we find the LCM of 10, 16 and 20.

10 = 2 × 5

16 = 2⁴

20 = 2² × 5

LCM = 2⁴ × 5 = 16 × 5 = 80 minutes

They will start preparing a new card together after 80 minutes.

Ques. Prove that √5 is irrational and hence show that 3 + √5 is also irrational. (2012)

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

So, we can find integers p and q (q ≠ 0), such that

√5 = \(\frac { p }{ q }\), where p and q are coprime.

Squaring both sides, we get

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

⇒ 5q² = p² …(i)

⇒ 5 divides p²

5 divides p

So, let p = 5r

Putting the value of p in (i), we get

5q² = (5r)²

⇒ 5q² = 25r²

⇒ q² = 5r²

⇒ 5 divides q²

5 divides q

So, p and q have atleast 5 as a common factor.

But this contradicts the fact that p and q have no common factor.

So, our assumption is wrong, is irrational.

√5 is irrational, 3 is a rational number.

So, we conclude that 3 + √5 is irrational.

Extra Solved Questions – Real Numbers

Ques. Prove that √2 is irrational.

(similar to √3 proof, use contradiction method).

Ques. Show that 0.285714285714… is a rational number.

Let x = 0.285714285714… = 0.285714̅ (repeating block of 6 digits).

Multiply by 10⁶:

1000000x = 285714.285714…

Subtract: 1000000x – x = 285714 ⇒ 999999x = 285714.

x = 285714 / 999999 = 2/7.

Practice Exercise – Real Numbers Class 10 Extra Questions (Unsolved)

1. Show that 15/2⁴ × 5² has a terminating decimal expansion.
2. Find the HCF of 867 and 255 using Euclid’s Lemma.
3. Express 5005 as a product of prime factors.
4. Prove that there are infinitely many prime numbers of the form 4n + 3.
5. Show that √6 is irrational.
6. Write the decimal expansion of 23/2³ × 5².
7. If HCF of two numbers is 16 and their LCM is 160, find the product of numbers.

Tips to Solve Real Numbers Extra Questions Effectively

• Always simplify fractions before checking decimal expansion.
• For irrational proofs, contradiction method is key.
• Learn prime factorisation for quick HCF/LCM calculation.
• Solve previous year questions regularly.
• Revise NCERT exemplar problems for advanced preparation.