Important Questions for Class 10 Maths Chapter 1: Real Numbers

Real numbers are the complete set of rational and irrational numbers. They include integers, fractions, decimals, and roots: every number that exists on the number line. Class 10 Maths Chapter 1 builds this understanding through the Fundamental Theorem of Arithmetic, HCF and LCM using prime factorisation, irrationality proofs, and decimal expansion of rational numbers.

Important questions class 10 maths chapter 1 on this page are structured in the exact NCERT sequence. You get very short answer questions, short answer questions, long answer proofs, extra questions, formulas, and notes in one place.

A student who can write an irrationality proof in under four minutes and classify decimal expansions without long division will handle the highest-weightage questions in this chapter. Most marks are lost on proof-based questions instead of HCF-LCM calculations. All questions and solutions are available section by section below.

Key Takeaways

Introduction to Real Numbers Class 10

Real numbers combine all rational and irrational numbers in one set. A rational number can be written in the form p/q, where p and q are integers and q is not equal to 0. An irrational number cannot be expressed this way: its decimal expansion is non-terminating and non-repeating.

Examples of rational numbers: 22/7 (non-terminating repeating), 13/3125 (terminating), 15/1600 (terminating). Examples of irrational numbers: root 2, root 3, root 5, pi, 0.10110111011110...

Three key properties from the 2026 CBSE syllabus: the sum or difference of a rational and irrational number is irrational; the product or quotient of a non-zero rational and an irrational number is irrational; the product of two irrational numbers can be rational or irrational.

Important Questions Class 10 Maths All Chapters

Very Short Answer Questions on Real Numbers

One-mark questions appear as MCQs and objective items in CBSE 2026 board exams. These class 10 real numbers important questions test definitions, classification, and direct application of properties.

1. For some integer x, every even integer is of the form: (a) x (b) x + 1 (c) 2x (d) 2x + 1 (c) 2x. Since 2x always produces even numbers for any integer x.
2. For some integer m, every odd integer is of the form: (a) m (b) m + 1 (c) 2m (d) 2m + 1 (d) 2m + 1. Since 2m is always even, 2m + 1 is always odd.
3. Which of the following is not irrational? (a) 3 + root 7 (b) 3 - root 7 (c) (3 + root 7)(3 - root 7) (d) 3 root 7 (c) (3 + root 7)(3 - root 7) = 9 - 7 = 2, which is rational.
4. The decimal expansion of 22/7 is: (b) Non-terminating repeating. 22/7 = 3.142857142857...
5. If a = x3y2 and b = xy3, where x and y are prime numbers, then HCF(a, b) is: (b) xy2. HCF takes the smallest power of each common prime factor: x1 x y2 = xy2.
6. The addition of a rational number and an irrational number equals: Always an irrational number.
7. n2 - 1 is divisible by 8 when n is: (c) An odd integer. For odd n = 4k + 1: n2 - 1 = 8(2k2 + k), divisible by 8.
8. HCF of 26 and 91 is: 13. Since 26 = 2 x 13 and 91 = 7 x 13, HCF = 13.
9. If set A = {1, 2, 3, 4, 5, 6, 7...}, it represents: Natural numbers.
10. The product of an irrational number and a non-zero rational number is: Always irrational.

Short Answer Questions on Real Numbers

Two-mark and three-mark questions test prime factorisation, HCF-LCM calculation, and basic properties. These real numbers class 10 extra questions cover the most tested calculation formats.

1. Find the LCM and HCF of 26 and 91. Verify that LCM x HCF = product of the two numbers. 26 = 2 x 13; 91 = 7 x 13. HCF(26, 91) = 13. LCM(26, 91) = 2 x 7 x 13 = 182. Verification: 182 x 13 = 2366 and 26 x 91 = 2366.
2. Find the LCM and HCF of 510 and 92. 510 = 2 x 3 x 5 x 17; 92 = 22 x 23. HCF(510, 92) = 2. LCM(510, 92) = 22 x 3 x 5 x 17 x 23 = 23460. Verification: 23460 x 2 = 46920 and 510 x 92 = 46920.
3. Express each number as a product of its prime factors: (i) 140 = 22 x 5 x 7. (ii) 156 = 22 x 3 x 13. (iii) 3825 = 32 x 52 x 17. (iv) 5005 = 5 x 7 x 11 x 13. (v) 7429 = 17 x 19 x 23.
4. Given that HCF(306, 657) = 9, find LCM(306, 657). HCF x LCM = Product of the two numbers. 9 x LCM = 306 x 657. LCM = 201042 / 9 = 22338.
5. Find the HCF of 96 and 404 and verify that HCF x LCM = product of the two numbers. 96 = 25 x 3; 404 = 22 x 101. HCF = 22 = 4. LCM = (96 x 404) / 4 = 9696. Verification: 4 x 9696 = 38784 and 96 x 404 = 38784.
6. If HCF of 65 and 117 is expressible in the form 65m - 117, find the value of m. 117 = 65 x 1 + 52; 65 = 52 x 1 + 13; 52 = 13 x 4 + 0. HCF(65, 117) = 13. 65m - 117 = 13 → 65m = 130 → m = 2.
7. The largest number that divides 70 and 125, leaving remainders 5 and 8 respectively. 70 - 5 = 65; 125 - 8 = 117. HCF(65, 117) = 13.
8. What is the HCF of the smallest prime number and the smallest composite number? Smallest prime = 2; smallest composite = 4. HCF(2, 4) = 2.

Long Answer Questions on Real Numbers

Four-mark and five-mark questions test deeper application: form-based proofs, Euclid-type problems, and LCM word problems. These real numbers class 10 important questions with solutions cover the highest-probability long answer topics.

1. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5. Let a be any positive integer and b = 6. By Euclid's division algorithm: a = 6q + r, where 0 is less than or equal to r less than 6. So r = 0, 1, 2, 3, 4, 5, giving a = 6q, 6q+1, 6q+2, 6q+3, 6q+4, 6q+5. Since 6q, 6q+2, 6q+4 are even, any positive odd integer takes the form 6q+1, 6q+3, or 6q+5. Hence proved.
2. If n is an odd integer, show that n2 - 1 is divisible by 8. Any odd positive integer n = 4q + 1 or 4q + 3. Case 1: n = 4q + 1 → n2 - 1 = (4q+1)2 - 1 = 16q2 + 8q = 8q(2q + 1), divisible by 8. Case 2: n = 4q + 3 → n2 - 1 = (4q+3)2 - 1 = 16q2 + 24q + 8 = 8(2q2 + 3q + 1), divisible by 8. Hence proved.
3. Show that the square of any positive integer is either of the form 3m or 3m + 1. Let x be any positive integer and y = 3. By Euclid's algorithm: x = 3q + r, where r = 0, 1, 2. Case 1: x = 3q → x2 = 9q2 = 3(3q2) = 3m. Case 2: x = 3q + 1 → x2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1. Case 3: x = 3q + 2 → x2 = 9q2 + 12q + 4 = 3(3q2 + 4q + 1) + 1 = 3m + 1. Hence proved.
4. Show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8. Let x be any positive integer and y = 3. By Euclid's algorithm: x = 3q, 3q+1, or 3q+2. Case 1: x = 3q → x3 = 27q3 = 9(3q3) = 9m. Case 2: x = 3q + 1 → x3 = 27q3 + 27q2 + 9q + 1 = 9(3q3 + 3q2 + q) + 1 = 9m + 1. Case 3: x = 3q + 2 → x3 = 27q3 + 54q2 + 36q + 8 = 9(3q3 + 6q2 + 4q) + 8 = 9m + 8. Hence proved.
5. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round, Ravi takes 12 minutes. After how many minutes will they meet again at the starting point? LCM(18, 12) = 36. They will meet again at the starting point after 36 minutes.

Important Questions on the Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorisation is unique except for the order of factors. This theorem is the backbone of HCF-LCM calculations and irrationality proofs in Class 10.

These extra questions on real numbers class 10 directly test this theorem. Every HCF-LCM calculation and every irrationality proof in the chapter depends on the uniqueness guaranteed by this theorem.

1. Check whether 6n can end with the digit 0 for any natural number n. If 6n ends in 0, it must be divisible by 5. The prime factorisation of 6n = (2 x 3)n = 2n x 3n. This contains no factor of 5. By the uniqueness of the Fundamental Theorem of Arithmetic, 6n cannot end with 0 for any natural number n.
2. Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers. 7 x 11 x 13 + 13 = 13(7 x 11 + 1) = 13 x 78 = 13 x 2 x 3 x 13. It has factors other than 1 and itself. Composite. 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5(7 x 6 x 4 x 3 x 2 x 1 + 1) = 5 x 1009. Composite.
3. Check whether there is any value of n for which 4n ends with the digit zero. 4n = (22)n = 22n. The only prime factor is 2. For a number to end in 0, it must have 5 as a prime factor. Since 5 is not a factor of 4n, there is no natural number n for which 4n ends with 0.
4. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify. No. Numbers of the form 4q + 2 are always even but never divisible by 4. Not every positive integer takes this form.
5. Show that one and only one out of n, n+1, n+2 is divisible by 3 for any positive integer n. n can be of the form 3q, 3q+1, or 3q+2. If n = 3q: n is divisible by 3; n+1 and n+2 are not. If n = 3q+1: n+2 = 3(q+1), divisible by 3; n and n+1 are not. If n = 3q+2: n+1 = 3(q+1), divisible by 3; n and n+2 are not. Hence proved.

Important Questions on HCF and LCM

HCF and LCM questions are among the most frequently asked class 10 real numbers extra questions in CBSE board exams. The key relationship: HCF x LCM = Product of two numbers. This is tested in both direct calculation and word problems.

1. Find the HCF and LCM of 6, 72, and 120 using prime factorisation. 6 = 2 x 3; 72 = 23 x 32; 120 = 23 x 3 x 5. HCF = 21 x 31 = 6. LCM = 23 x 32 x 51 = 360. Note: For three numbers, HCF x LCM does not equal the product of all three numbers.
2. Find the HCF and LCM of 306 and 657. Verify that LCM x HCF = Product of the two numbers. 306 = 2 x 32 x 17; 657 = 32 x 73. HCF = 32 = 9. LCM = 2 x 32 x 17 x 73 = 22338. LHS: 22338 x 9 = 201042. RHS: 306 x 657 = 201042. Verified.
3. If two positive integers p and q are expressed as p = ab2 and q = a3b, where a and b are primes, then LCM(p, q) is: LCM takes the greatest power of each prime: a3 x b2 = a3b2.
4. An army contingent of 616 members marches behind an army band of 32 members. Find the maximum number of columns. HCF(616, 32): 616 = 32 x 19 + 8; 32 = 8 x 4 + 0. HCF = 8 columns.
5. Ravi, Seema, and Raju complete one card in 20, 16, and 10 minutes respectively. After what time will they start a new card together? 20 = 22 x 5; 16 = 24; 10 = 2 x 5. LCM(20, 16, 10) = 24 x 5 = 80 minutes.
6. If x = p2q3 and y = p3q, what is the LCM and HCF? HCF = p2q (smallest powers of common factors). LCM = p3q3 (greatest powers of all factors).

Important Questions on Irrational Numbers

Irrationality proofs are the most important long-answer questions from Chapter 1. All proofs use contradiction and the Fundamental Theorem of Arithmetic. These real numbers class 10 extra questions with answers appear regularly as 3-mark problems in CBSE board exams.

1. Prove that root 5 is irrational. Assume root 5 is rational. Then root 5 = x/y, where x and y are coprime integers and y is not equal to 0. Squaring: 5y2 = x2. So x2 is divisible by 5, which means x is divisible by 5. Let x = 5k. Then 5y2 = 25k2, so y2 = 5k2, meaning y is also divisible by 5. This contradicts the fact that x and y are coprime. So root 5 is irrational. Hence proved.
2. Prove that root 3 is irrational. Assume root 3 is rational. Then root 3 = a/b, where a and b are coprime and b is not equal to 0. Squaring: 3b2 = a2. So a is divisible by 3. Let a = 3c. Then 3b2 = 9c2, so b2 = 3c2, meaning b is also divisible by 3. This contradicts the fact that a and b are coprime. So root 3 is irrational. Hence proved.
3. Prove that 3 + 2 root 5 is irrational. Assume 3 + 2 root 5 is rational. Then 3 + 2 root 5 = x/y, where x and y are coprime integers. Rearranging: root 5 = (x/y - 3)/2 = (x - 3y)/(2y). Since x and y are integers, the right side is rational. So root 5 would be rational. But root 5 is irrational. Contradiction. So 3 + 2 root 5 is irrational. Hence proved.
4. Prove the following are irrational: (i) 1/root 2: Assume 1/root 2 = x/y. Then root 2 = y/x, which is rational. But root 2 is irrational. Contradiction. So 1/root 2 is irrational. (ii) 7 root 5: Assume 7 root 5 = x/y. Then root 5 = x/(7y), which is rational. But root 5 is irrational. Contradiction. So 7 root 5 is irrational. (iii) 6 + root 2: Assume 6 + root 2 = x/y. Then root 2 = x/y - 6 = (x - 6y)/y, which is rational. But root 2 is irrational. Contradiction. So 6 + root 2 is irrational.

5. Show that 5 - root 3 is irrational. Assume 5 - root 3 is rational. Then root 3 = 5 - (a/b) = (5b - a)/b, which is rational. But root 3 is irrational. Contradiction. So 5 - root 3 is irrational. Hence proved.

Important Questions on Decimal Expansion of Rational Numbers

A rational number p/q has a terminating decimal expansion if and only if the denominator q has no prime factors other than 2 and 5, that is, q is of the form 2m x 5n. These class 10 maths real numbers extra questions cover both classification and calculation.

1. Without performing long division, state whether each rational number has a terminating or non-terminating repeating decimal expansion. (i) 13/3125: 3125 = 55. Only factor is 5. Terminating. (ii) 17/8: 8 = 23. Only factor is 2. Terminating. (iii) 64/455: 455 = 5 x 7 x 13. Has factor 7. Non-terminating repeating. (iv) 15/1600: 1600 = 26 x 52. Only factors 2 and 5. Terminating. (v) 29/343: 343 = 73. Has factor 7. Non-terminating repeating.

2. The decimal expansion of 14587/1250 will terminate after how many decimal places? 1250 = 2 x 54 = 21 x 54. Multiply numerator and denominator by 23 to get denominator as 104: 14587 x 8 / 10000 = 116696/10000 = 11.6696. Terminates after 4 decimal places.
3. The following real numbers have decimal expansions as given. Decide whether they are rational or irrational. (i) 43.123456789: Terminating decimal. Rational. Denominator has only 2 and 5 as prime factors. (ii) 0.120120012000120000...: Non-terminating, non-repeating. Irrational.

4. State whether 129/225775 has a terminating or non-terminating decimal expansion. 225775 has factors other than 2 and 5. Non-terminating repeating.

Real Numbers Class 10 Extra Questions

These class 10th real numbers extra questions go beyond the standard textbook exercise. They appear in school unit tests and pre-board papers for CBSE 2026.

1. A positive integer is of the form 3q + 1, where q is a natural number. Can its square take any form other than 3m + 1? No. (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1. The square always takes the form 3m + 1.
2. Show that the square of any odd integer is of the form 4q + 1, for some integer q. Let a be any odd integer and b = 4. By Euclid's division: a = 4m + 1 or 4m + 3. (4m + 1)2 = 16m2 + 8m + 1 = 4(4m2 + 2m) + 1 = 4q + 1. (4m + 3)2 = 16m2 + 24m + 9 = 4(4m2 + 6m + 2) + 1 = 4q + 1. Hence proved.
3. Dudhnath has two vessels containing 720 ml and 405 ml of milk. Find the minimum number of glasses needed. HCF(720, 405): 720 = 24 x 32 x 5; 405 = 34 x 5. HCF = 32 x 5 = 45 ml. Glasses from first vessel: 720/45 = 16. Glasses from second vessel: 405/45 = 9. Total = 25 glasses.
4. Find HCF of 134791, 6341, and 6339. 6341 = 6339 x 1 + 2; 6339 = 2 x 3169 + 1; 2 = 1 x 2 + 0. HCF(6341, 6339) = 1. HCF(134791, 1) = 1. So HCF of all three = 1.
5. Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4. Let x = 2r + 1 and y = 2s + 1. x2 + y2 = 4r2 + 4r + 1 + 4s2 + 4s + 1 = 4(r2 + s2 + r + s) + 2. This is even but leaves remainder 2 when divided by 4. Not divisible by 4. Hence proved.
6. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5. Using Euclid's division with b = 6: a = 6q + r, where r = 0, 1, 2, 3, 4, 5. Calculating a2 for each r gives forms: 6m, 6m+1, 6m+4, 6m+3, 6m+4, 6m+1. The square of a positive integer can only be 6m, 6m+1, 6m+3, or 6m+4. It cannot be 6m+2 or 6m+5. Hence proved.

Real Numbers Class 10 Important Questions with Solutions: Proof-Based

Proof-based questions carry the highest weightage in this chapter. These extra questions from real numbers class 10 are fully solved using the contradiction method as taught in the CBSE 2026 syllabus.

1. Use Euclid's division lemma to show that the cube of any positive integer is of the form 4m, 4m + 1, or 4m + 3. Let x be any positive integer and y = 4. Then x = 4q + r, where r = 0, 1, 2, 3. r = 0: x3 = (4q)3 = 64q3 = 4(16q3) = 4m. r = 1: x3 = (4q+1)3 = 64q3 + 48q2 + 12q + 1 = 4(16q3 + 12q2 + 3q) + 1 = 4m + 1. r = 2: x3 = (4q+2)3 = 64q3 + 96q2 + 48q + 8 = 4(16q3 + 24q2 + 12q + 2) = 4m. r = 3: x3 = (4q+3)3 = 64q3 + 144q2 + 108q + 27 = 4(16q3 + 36q2 + 27q + 6) + 3 = 4m + 3. Hence proved.
2. Prove that root 2 is irrational. Assume root 2 is rational. Then root 2 = a/b, where a and b are coprime integers, b is not equal to 0. Squaring: 2b2 = a2. So 2 divides a2, and by Theorem 1.2, 2 divides a. Let a = 2c. Then 2b2 = 4c2, so b2 = 2c2. So 2 divides b. But then a and b have 2 as a common factor, contradicting that they are coprime. So root 2 is irrational. Hence proved.

Real Numbers Formulas for Class 10

Students searching for real numbers formulas for class 10 and real numbers class 10 formulas before board exams need these key identities from Chapter 1.

Real Numbers Class 10 Notes

These real numbers class 10 notes cover the essential points from the CBSE 2026 syllabus. Use them for last-minute revision before tests and board exams.

Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes. This factorisation is unique except for the order. This uniqueness is the key tool behind all HCF-LCM work and irrationality proofs.

HCF and LCM: HCF takes the smallest power of each prime factor common to both numbers. LCM takes the greatest power of every prime factor present in either number.

Irrationality proofs: All proofs use contradiction. Assume the number is rational (p/q form with coprime p and q). Show this leads to a common factor, contradicting the coprime assumption. Conclude the number is irrational.

Decimal expansion rule: A rational number p/q in lowest terms has a terminating decimal if and only if q has no prime factors other than 2 and 5. If q has any other prime factor, the expansion is non-terminating repeating.

Key results from the book: root 2, root 3, and root 5 are all irrational; the sum of a rational and irrational is always irrational; the product of a non-zero rational and irrational is always irrational; the product of two irrationals can be rational (root 2 x root 2 = 2) or irrational.