Real Numbers: Important Questions and Answers for Class 10

Real numbers is one of the most important chapters in Class 10 Mathematics, as it introduce students to the properties and applications of numbers used in everyday calculations and advanced mathematical concepts. This topic plays a crucial role in developing problem-solving skills and understanding mathematical operations such as Euclid’s Division Lemma, prime factorization, irrational numbers and decimal expansions. Practising a variety of exam-oriented questions helps students become familiar with different problem patterns and solution techniques. In this guide, you will find carefully selected important questions covering all major concepts from the Class 10 Real Numbers chapter, designed according to the latest syllabus and exam pattern to support effective preparation and revision.

Our subject experts have provided detailed solutions for these problems based on the latest CBSE syllabus and the NCERT textbook. This material helps students revise the chapter easily and perform well in the final examination.

Table of Contents

• Key Concepts of Real Numbers for Class 10
• Section A: 1-Mark Questions (MCQ & VSA)
• Section B: 2-Mark Questions (Short Answer)
• Section C: 3-Mark Questions
• Section D: 5-Mark Questions (Long Answer / Proof-Type)
• Section E: HOTS & Case-Study Based Questions
• Exam Tips & Common Mistakes to Avoid

Key Concepts of Real Numbers for Class 10

• Euclid's Division Lemma

For any two positive integers a and b, there exist unique integers q (quotient) and r (remainder) such that:

a = bq + r, where 0 ≤ r < b

For example, when you divide 17 by 5, you get 17 = 5 × 3 + 2. Here, a = 17, b = 5, q = 3, r = 2. This lemma is the key to finding the HCF of large numbers without listing all their factors.

• Fundamental Theorem of Arithmetic

The fundamental theorem of arithmetic states that every composite number can be expressed as a product of prime numbers in one and only one way (ignoring the order of factors). 

For example, 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5. No matter how you factor 60, you'll always arrive at these exact primes.

• HCF × LCM Relationship

HCF(a, b) × LCM(a, b) = a × b

• Rational vs Irrational Numbers

A number is rational if it can be written as p/q where p and q are integers and q ≠ 0. Its decimal expansion either terminates (like 0.5) or is non-terminating repeating (like 0.333...). A number is irrational if its decimal expansion is non-terminating and non-repeating, like √2 = 1.41421356...

• Terminating Decimal: Quick Test

When does p/q give a terminating decimal? q = 2ⁿ × 5ᵐ   (q has no prime factor other than 2 and 5)

For example, 7/40 terminates because 40 = 2³ × 5¹. But 7/12 does not terminate because 12 = 2² × 3, and 3 is a prime factor other than 2 and 5.

 

Section A: 1-Mark Questions (MCQ & VSA)

Q1: The HCF of 96 and 404 is 4. Find their LCM.

Solution: Using HCF × LCM = Product of the two numbers 

LCM = (96 × 404) / HCF 

⇒ LCM = (96 × 404) / 4 

⇒ LCM = 38784 / 4 

⇒ LCM = 9696

Q2: Will the decimal expansion of 17/8 be terminating or non-terminating? Give a reason.

Solution: The denominator is 8 = 2³. 

Since 8 = 2³ × 5⁰, its only prime factor is 2. Therefore, 17/8 has a terminating decimal expansion.

17 ÷ 8 = 2.125  (terminates)

Q3: Write the prime factorisation of 7429.

Solution: 

1774291943723231

∴ 7429 = 17 × 19 × 23

Q4: The LCM of two numbers is 1200. Can their HCF be 500? Justify.

Solution: No. By a fundamental property, the HCF of two numbers must always divide their LCM. Here, 500 does not divide 1200 (since 1200 ÷ 500 = 2.4, which is not an integer). Hence, HCF cannot be 500.

Q5: Which of the following is an irrational number? 

(a) √4   (b) √9   (c) √7   (d) √16

Solution: Since 7 is not a perfect square and has no rational square root, √7 is irrational. The others, simplified to 2, 3 and 4 respectively, are all rational.

Answer: (c) √7. 

Section B: 2-Mark Questions (Short Answer)

Q6: Find the HCF and LCM of 510 and 92 using prime factorisation.

Solution: 510 = 2 × 3 × 5 × 17 

92 = 2² × 23

HCF = product of common prime factors raised to their lowest powers

HCF(510, 92) = 2 

LCM = product of all prime factors raised to their highest powers. 

LCM(510, 92) = 2² × 3 × 5 × 17 × 23 = 23460 

Verification: HCF × LCM = 2 × 23460 = 46920 = 510 × 92

Q7: Given HCF (306, 657) = 9, find LCM (306, 657).

Solution: HCF × LCM = Product of numbers 

⇒ 9 × LCM = 306 × 657 

⇒ 9 × LCM = 201,042 

⇒ LCM = 201,042 / 9 = 22,338

Q8: Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution: For any number to end with 0, it must be divisible by 10, which means it must have both 2 and 5 as prime factors.

6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ 

Prime factors of 6ⁿ are only 2 and 3. 

5 is NOT a prime factor of 6ⁿ for any n. 

∴ 6ⁿ can NEVER end with the digit 0.

Q9: Find the largest number which divides 245 and 1029, leaving remainder 5 in each case.

Solution: Since the remainder is 5 in both cases, subtract 5 from each number to find what the divisor exactly divides:

Numbers to consider: (245 - 5) = 240 and (1029 - 5) = 1024 

240 = 2⁴ × 3 × 5 

1024 = 2¹⁰ 

HCF(240, 1024) = 2⁴ = 16 

∴ The largest number is 16.

Section C: 3-Mark Questions

Q10: Prove that √5 is irrational.

Solution: Proof by Contradiction

Assume that √5 is rational. Then we can write:

√5 = p/q where p, q are integers, q ≠ 0, and HCF(p, q) = 1 

Squaring both sides: 5 = p²/q² 

p² = 5q² ... (1) 

This means 5 divides p². 

By a theorem: if a prime divides p², it also divides p.

 ∴ 5 divides p. 

Let p = 5m for some integer m. 

Substituting in (1): (5m)² = 5q² 

25m² = 5q² 

5m² = q² 

This means 5 divides q², so 5 divides q. 

But now, 5 divides both p and q. 

This contradicts our assumption that HCF(p, q) = 1. 

∴ Our assumption was wrong. √5 is irrational. 

Q11: Prove that 3 + 2√5 is irrational.

Solution: Proof by Contradiction

Assume 3 + 2√5 is rational. 

Then:

3 + 2√5 = p/q (where HCF(p, q) = 1, q ≠ 0) 

Rearranging: 2√5 = (p/q) - 3 

2√5 = (p - 3q)/q 

√5 = (p - 3q)/(2q) 

Since p and q are integers, (p - 3q)/(2q) is rational. 

So √5 would have to be rational. 

But √5 is irrational, which is a contradiction.

 ∴ 3 + 2√5 is irrational.

Q12: Find the HCF of 867 and 255 using Euclid's Division Algorithm.

Solution: Apply the algorithm: divide the larger number by the smaller, then replace with the new remainder, and repeat until the remainder is 0.

Step 1: 867 = 255 × 3 + 102 

Step 2: 255 = 102 × 2 + 51 

Step 3: 102 = 51 × 2 + 0 

The remainder is now 0. 

The divisor at this step is the HCF. 

HCF(867, 255) = 51

Q13: Three bells toll at intervals of 9, 12, and 15 minutes respectively. If they toll together at 10:00 AM, when will they toll together again?

Solution: The bells will toll together again after the LCM of 9, 12, and 15 minutes.

9 = 3² 

12 = 2² × 3 

15 = 3 × 5 

LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes = 3 hours 

They toll together again at 10:00 AM + 3 hours = 1:00 PM

Section D: 5-Mark Questions (Long Answer / Proof-Type)

Q14: 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.

Solution: Let x be any positive integer. 

When divided by 3, the possible remainders are 0, 1, or 2. 

So by Euclid's Lemma, x can be written as one of:

x = 3q or x = 3q + 1 or x = 3q + 2 

Case 1: x = 3q 

x² = (3q)² = 9q² 

= 3(3q²) = 3m where m = 3q² 

Case 2: x = 3q + 1 

x² = (3q + 1)² = 9q² + 6q + 1 

= 3(3q² + 2q) + 1 = 3m + 1 where m = 3q² + 2q 

Case 3: x = 3q + 2 x² = (3q + 2)² 

= 9q² + 12q + 4 = 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1 = 3m + 1 where m = 3q² + 4q + 1 

∴ The square of any positive integer is of the form 3m or 3m+1.

Q15: Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Solution: Let a be any positive integer and b = 4. 

By Euclid's Division Lemma:

a = 4q + r where r = 0, 1, 2, or 3 

So the possible forms are: 4q which is even and divisible by 4

4q + 1 which is odd

4q + 2 = (2(2q + 1)) which is even

4q + 3 which is odd

Even integers take the form 4q or 4q + 2. Odd integers must therefore be of the form 4q + 1 or 4q + 3. 

∴ Any positive odd integer is of the form 4q + 1 or 4q + 3.

Q16: Prove that √2 + √3 is irrational.

Solution: Assume √2 + √3 is rational. 

Let it equal p/q where HCF(p, q) = 1, q ≠ 0.

√2 + √3 = p/q 

Squaring both sides: (√2 + √3)² = p²/q² 

⇒ 2 + 2√6 + 3 = p²/q² 

⇒ 5 + 2√6 = p²/q² 

⇒ 2√6 = p²/q² - 5 

⇒ 2√6 = (p² - 5q²)/q² 

⇒ √6 = (p² - 5q²)/(2q²) 

The RHS is rational (p, q are integers, q ≠ 0). 

So √6 would be rational. But √6 = √2 × √3. If √6 were rational, one can show this contradicts the irrationality of √2 (or √3)  which is a known fact. This leads to contradiction 

∴ √2 + √3 is irrational.

Section E: HOTS & Case-Study Based Questions

HOTS 1: Explain why 7 × 11 × 13 + 13 is a composite number.

Solution: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) 

= 13 × (77 + 1) = 13 × 78 

= 13 × 2 × 3 × 13 = 2 × 3 × 13² 

This number has factors other than 1 and itself. 

∴ It is a composite number. 

 The maximum number of books per stack = HCF(56, 98, 21).

HOTS 2: Case Study

A school librarian is arranging 56 Maths books, 98 Science books, and 21 English books into stacks so that each stack has the same number of books and only one subject. What is the maximum number of books per stack? How many stacks are there in all?

Solution: 56 = 2³ × 7 

98 = 2 × 7² 

21 = 3 × 7 

⇒ HCF = 7 (the only common prime factor) 

Maximum books per stack = 7 

⇒ Number of Maths stacks = 56 ÷ 7 = 8 

⇒ Number of Science stacks = 98 ÷ 7 = 14 

⇒ Number of English stacks = 21 ÷ 7 = 3 

∴ Total stacks = 8 + 14 + 3 = 25 stacks

HOTS 3: There is a circular path around a sports field. Soni takes 18 minutes to complete one round and Ravi takes 12 minutes. If both start from 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?

Solution: 18 = 2 × 3² 

12 = 2² × 3 

LCM (18, 12) = 2² × 3² = 4 × 9 = 36 minutes 

They will meet at the starting point after 36 minutes.

Exam Tips & Common Mistakes to Avoid

• Write Every Step: In proof-type questions, marks are awarded for each logical step. Never skip a case, show all possibilities.

• Memorise HCF × LCM Formula: This relation is tested directly or indirectly in almost every board exam. Know it cold and verify by checking the product equals a × b.

• Don't Confuse HCF and LCM: HCF is always ≤ the smaller number. LCM is always ≥ the larger number.

• State the Contradiction Clearly: In irrationality proofs, always explicitly say "this contradicts our assumption that HCF(p,q) = 1". That line earns a mark.

• Check Terminating Decimals: Always fully simplify the fraction first, then check if the denominator has only factors of 2 and 5.

• Real-Life Application Problems: For ‘when do they meet’ or ‘max stack size’ word problems, immediately identify whether you need HCF or LCM. Same intervals: LCM. Maximum grouping: HCF.

• Assuming √(a+b) = √a + √b: This is wrong. √(4+9) = √13 ≠ 2 + 3

• Thinking all square roots are irrational: √4 = 2, √9 = 3 are rational. Only non-perfect-square roots are irrational

• Not reducing fraction before checking termination: Always simplify first; 6/15 = 2/5, which terminates, even though 15 has the factor 3.