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CBSE Class 10 · Maths · 2026-27

CBSE Class 10 Maths Sample Paper 2026-27

This is a full Class 10 Maths sample paper built to the CBSE board blueprint — 38 questions in Sections A to E, 80 marks, with MCQs, Assertion-Reason, short and long answers, and case-based questions. A complete answer key with solutions is included below.

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Annual Examination 2026-27

Mathematics Sample Question Paper

Class 10 · Mathematics

Time: 3 HoursMax Marks: 80

General Instructions

  • This question paper contains 38 questions. All questions are compulsory.
  • Section Section A: Questions 1–20 — Multiple Choice Questions & Assertion–Reason Questions (20 marks).
  • Section Section B: Questions 21–25 — Short Answer Questions (10 marks).
  • Section Section C: Questions 26–31 — Short Answer Questions (18 marks).
  • Section Section D: Questions 32–35 — Long Answer Questions (20 marks).
  • Section Section E: Questions 36–38 — Case Study Based Questions (12 marks).
  • Marks for each question are indicated against it.

Section A — Multiple Choice Questions & Assertion–Reason Questions

(20 marks)

Choose the correct option. Each question has an Assertion (A) and a Reason (R). Choose the correct option.

  1. Q1. If p is a prime number, then √p is:

    [1]
    (A) always rational(B) always irrational(C) rational only if p=2(D) sometimes rational

  2. Q2. Two numbers are in the ratio 3:4 and their LCM is 72. The sum of the numbers is:

    [1]
    (A) 42(B) 36(C) 84(D) 21

  3. Q3. For the rational number m/140 to have a terminating decimal expansion, how many integer values can m take from 1 to 140?

    [1]
    (A) 20(B) 14(C) 10(D) 21

  4. Q4. For what value of k will the polynomial x^2 - 7x + k have zeroes 3 and 4?

    [1]
    (A) 7(B) 10(C) 12(D) 14

  5. Q5. If α and β are the zeroes of the quadratic polynomial 2x^2 - 5x - 3, then the value of α^2β + αβ^2 is:

    [1]
    (A) -15/4(B) 15/4(C) -15/2(D) 15/2

  6. Q6. If the polynomial p(x) = x^3 + ax^2 + bx + c has (x+1) as a factor, leaves remainder 4 when divided by (x-2), and p(1) = 0, then the value of a+b+c is:

    [1]
    (A) -2(B) -1(C) 0(D) 1

  7. Q7. If a pair of linear equations in two variables is given by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, and a1/a2 ≠ b1/b2, then which of the following is true about the pair?

    [1]
    (A) It has no solution.(B) It has a unique solution.(C) It has infinitely many solutions.(D) It has two solutions.

  8. Q8. A fraction becomes 1/2 when 2 is subtracted from the numerator and it becomes 1/2 when 1 is added to its denominator. What is the fraction?

    [1]
    (A) 3/5(B) 4/7(C) 5/9(D) No such fraction exists

  9. Q9. For what values of a and b does the following pair of linear equations have infinitely many solutions? 2x + 3y = 7; (a − b)x + (a + b)y = 3a + b − 2

    [1]
    (A) a = 5, b = 1(B) a = 1, b = 5(C) a = 5, b = 0(D) a = 0, b = 1

  10. Q10. The roots of the equation x^2 - 5x + 6 = 0 are:

    [1]
    (A) 2 and 3(B) -2 and -3(C) 2 and -3(D) -2 and 3

  11. Q11. For what value of p does the equation x^2 - 4x + p = 0 have real and distinct roots?

    [1]
    (A) p < 4(B) p > 4(C) p = 4(D) p ≥ 4

  12. Q12. For what value of k (k ≠ 1) does the quadratic equation (k - 1)x^2 + 2(k - 1)x + 2 = 0 have real and equal roots?

    [1]
    (A) 3(B) 1(C) 2(D) -3

  13. Q13. Which term of the arithmetic progression 21, 18, 15, ... becomes zero?

    [1]
    (A) 7th(B) 8th(C) 9th(D) 10th

  14. Q14. How many terms of the arithmetic progression 3, 6, 9, ... must be taken so that the sum is 165?

    [1]
    (A) 9(B) 10(C) 11(D) 12

  15. Q15. In an arithmetic progression with 60 terms, the sum of the first 20 terms is 210 and the sum of the first 30 terms is 465. What is the sum of the last 10 terms?

    [1]
    (A) 555(B) 500(C) 510(D) 525

  16. Q16. The areas of two similar triangles are in the ratio 1:9. What is the ratio of their corresponding sides?

    [1]
    (A) 1:3(B) 1:9(C) 1:81(D) 3:1

  17. Q17. Assertion (A): The HCF of 26 and 91 is 13. Reason (R): The HCF of two numbers is the product of the smallest powers of common prime factors.

    [1]
    (A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.

  18. Q18. Assertion (A): The number 5 - √3 is irrational. Reason (R): The difference of a rational number and an irrational number is an irrational number.

    [1]
    (A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.

  19. Q19. Assertion (A): For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b. Reason (R): The HCF of two numbers is always a factor of their LCM.

    [1]
    (A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.

  20. Q20. Assertion (A): The sum of the zeroes of the polynomial 3x^2 - 5x - 2 is 5/3. Reason (R): For a quadratic polynomial ax^2 + bx + c, the sum of zeroes is -b/a.

    [1]
    (A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.

Section B — Short Answer Questions

(10 marks)

Answer the following questions briefly.

  1. Q21. State the Fundamental Theorem of Arithmetic.

    [2]

  2. Q22. Prove that √5 is irrational.

    [2]

  3. Q23. Prove that the square of any positive integer is of the form 5q, 5q+1, or 5q+4, for some integer q.

    [2]

  4. Q24. Find the sum and product of the zeros of the polynomial $5x^2 - 3x + 2$.

    [2]

  5. Q25. The zeros of the cubic polynomial $x^3 - 12x^2 + 39x - 28$ are in arithmetic progression. Find the zeros.

    [2]

Section C — Short Answer Questions

(18 marks)

Answer the following questions.

  1. Q26. Find the LCM and HCF of 12, 15 and 21 by the prime factorization method.

    [3]

  2. Q27. Prove that √3 is irrational.

    [3]

  3. Q28. Prove that one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer.

    [3]

  4. Q29. Find the quadratic polynomial whose zeroes are \(3 + 2\sqrt{2}\) and \(3 - 2\sqrt{2}\).

    [3]

  5. Q30. If \(\alpha\) and \(\beta\) are zeroes of \(2x^2 + 5x + k\) and \(\alpha^2 + \beta^2 = \frac{21}{4}\), find the value of \(k\).

    [3]

  6. Q31. If \((x+1)\) and \((x-2)\) are factors of \(x^3 + ax^2 + bx - 6\), find \(a\) and \(b\).

    [3]

Section D — Long Answer Questions

(20 marks)

Answer the following questions in detail.

  1. Q32. Prove that √3 is an irrational number.

    [5]

  2. Q33. Find the HCF and LCM of 96 and 404 by using prime factorization method. Hence, verify that HCF × LCM = product of the two numbers.

    [5]

  3. Q34. Show that the square of any positive integer cannot be expressed in the form 5q + 2 or 5q + 3, where q is an integer.

    [5]

  4. Q35. Find the zeroes of the quadratic polynomial 3x^2 – x – 4 and verify the relationship between the zeroes and its coefficients.

    [5]

Section E — Case Study Based Questions

(12 marks)

Read the passage and answer the questions that follow.

  1. Q36. Sara is checking whether certain rational numbers have terminating decimal expansions. She knows that a reduced fraction a/b has a terminating decimal if and only if b has no prime factors other than 2 and 5. She considers the number 27/400.

    [4]

  2. Q37. Three traffic signals at a square change after 18 seconds, 24 seconds, and 32 seconds respectively. They changed together at 9:00 am.

    [4]

  3. Q38. Let p and q be two distinct prime numbers. Consider the number √p + √q. We aim to prove that this number is irrational by contradiction.

    [4]

Answer Keyincluded with every paper

  1. Q1. always irrational

    By contradiction, the square root of any prime number is irrational; this is a standard result.

  2. Q2. 42

    Let numbers be 3x and 4x; LCM = 12x = 72 ⇒ x = 6; numbers are 18 and 24; sum = 42.

  3. Q3. 20

    140 = 2² × 5 × 7. For termination, the simplified denominator must have only 2 and 5, so the factor 7 must cancel. Thus m must be a multiple of 7. From 1 to 140, multiples of 7: 7,14,…,140 are 20 in number.

  4. Q4. 12

    If zeroes are 3 and 4, the polynomial is (x-3)(x-4) = x^2 - 7x + 12. Comparing with x^2 - 7x + k gives k=12.

  5. Q5. -15/4

    α+β = 5/2, αβ = -3/2. Then α^2β + αβ^2 = αβ(α+β) = (-3/2)*(5/2) = -15/4.

  6. Q6. -1

    p(-1)=0 ⇒ -1 + a - b + c = 0 ⇒ a - b + c = 1. p(1)=0 ⇒ 1 + a + b + c = 0 ⇒ a + b + c = -1. Subtracting gives 2b = -2 ⇒ b = -1. Then a + c = 0. Also p(2)=4 ⇒ 8 + 4a + 2b + c = 4 ⇒ 8 + 4a -2 + c = 4 ⇒ 4a + c = -2. Solving with a + c = 0 gives a = -2/3, c = 2/3. So a+b+c = -2/3 -1 + 2/3 = -1.

  7. Q7. It has a unique solution.

    For a pair of linear equations a1x+b1y+c1=0 and a2x+b2y+c2=0, if a1/a2 ≠ b1/b2, the lines intersect at exactly one point, giving a unique solution.

  8. Q8. No such fraction exists

    Let fraction = x/y. Conditions: (x-2)/y = 1/2 → 2x - y = 4; and x/(y+1) = 1/2 → 2x - y = 1. The two equations are inconsistent, so no solution.

  9. Q9. a = 5, b = 1

    Condition: 2/(a-b) = 3/(a+b) = 7/(3a+b-2). From first equality: 2a+2b = 3a-3b → a = 5b. Using first and third: 2(3a+b-2) = 7(a-b). Substitute a=5b → 2(15b+b-2)=7(5b-b) → 32b-4=28b → 4b=4 → b=1, a=5.

  10. Q10. 2 and 3

    Factorising: x^2 - 5x + 6 = (x - 2)(x - 3) = 0, so x = 2 or x = 3.

  11. Q11. p < 4

    For real and distinct roots, discriminant > 0. So, (-4)^2 - 4 × 1 × p > 0 ⇒ 16 - 4p > 0 ⇒ p < 4.

  12. Q12. 3

    For equal roots, discriminant = 0. Here, a = k-1, b = 2(k-1), c = 2. So, [2(k-1)]^2 - 4(k-1)·2 = 0 ⇒ 4(k-1)^2 - 8(k-1) = 0 ⇒ 4(k-1)[(k-1)-2] = 0 ⇒ k = 1 or k = 3. Since k ≠ 1, k = 3.

  13. Q13. 8th

    Here a=21, d= –3. Set a_n = 0 ⇒ 21 + (n–1)(–3)=0 ⇒ –3n + 24 = 0 ⇒ n = 8. Thus the 8th term is zero.

  14. Q14. 10

    a=3, d=3. S_n = n/2[2a+(n–1)d] = n/2[6+3n–3] = n/2(3n+3) = (3n(n+1))/2. Set =165 ⇒ n(n+1)=110 ⇒ n=10 (since 10×11=110).

  15. Q15. 555

    Using S_n = n/2[2a+(n-1)d]: S₂₀ = 10(2a+19d)=210 ⇒ 2a+19d=21; S₃₀ = 15(2a+29d)=465 ⇒ 2a+29d=31. Solving gives d=1, a=1. Last term a₆₀ = 1+59=60. Sum of terms from 51 to 60 = 10×(average of a₅₁ and a₆₀). a₅₁ = 1+50=51, a₆₀=60, average=55.5 ⇒ sum=555. Alternately, S₆₀ – S₅₀ = 30×61 – 25×51 = 1830 – 1275 = 555.

  16. Q16. 1:3

    For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides. So if area ratio = 1/9, side ratio = √(1/9) = 1/3.

  17. Q17. Both A and R are true and R is the correct explanation of A.

    26 = 2 × 13, 91 = 7 × 13. The only common prime factor is 13. According to R, the HCF is the product of the smallest powers, i.e., 13^1 = 13. Thus, both A and R are true, and R correctly explains A.

  18. Q18. Both A and R are true and R is the correct explanation of A.

    5 is rational and √3 is irrational. The difference of a rational and an irrational is always irrational. Thus, A is true, R is a general property that directly explains A.

  19. Q19. Both A and R are true but R is not the correct explanation of A.

    The relation HCF × LCM = a × b is a well-known theorem derived from prime factorisation. R is also true: since LCM is a multiple of each number, it is divisible by the HCF. However, R alone does not explain or imply the product equality; it only shows a divisibility relationship, not the exact formula. Hence, both are true but R is not the correct explanation.

  20. Q20. Both A and R are true and R is the correct explanation of A.

    Sum of zeroes = -(-5)/3 = 5/3. The reason states the correct formula, so it directly explains the assertion.

  21. Q21. 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.

    The Fundamental Theorem of Arithmetic states that every composite number can be factorised uniquely as a product of primes, regardless of the order of the factors.

  22. Q22. √5 is irrational.

    Assume √5 = a/b, a,b coprime integers, b≠0. Then 5 = a^2/b^2 => a^2 = 5b^2. So 5 divides a^2, hence 5 divides a. Let a=5c. Then (5c)^2 = 5b^2 => 25c^2=5b^2 => 5c^2=b^2, so 5 divides b^2, hence 5 divides b. Thus a and b have common factor 5, contradiction. Therefore √5 is irrational.

  23. Q23. The square of any positive integer is of the form 5q, 5q+1, or 5q+4.

    By Euclid's division, any positive integer a = 5m, 5m+1, 5m+2, 5m+3, or 5m+4. Squaring each: (5m)^2=25m^2=5(5m^2)=5q; (5m+1)^2=25m^2+10m+1=5(5m^2+2m)+1=5q+1; (5m+2)^2=25m^2+20m+4=5(5m^2+4m)+4=5q+4; (5m+3)^2=25m^2+30m+9=5(5m^2+6m+1)+4=5q+4; (5m+4)^2=25m^2+40m+16=5(5m^2+8m+3)+1=5q+1. Thus all squares are of required forms.

  24. Q24. Sum = 3/5, Product = 2/5

    For $ax^2+bx+c=0$, sum = $-b/a$, product = $c/a$. Here $a=5, b=-3, c=2$. So sum = $-(-3)/5 = 3/5$, product = $2/5$.

  25. Q25. 1, 4, 7

    Let zeros be $a-d$, $a$, $a+d$. Sum = $3a = 12 \Rightarrow a=4$. Sum of pairwise products = $a(a-d)+a(a+d)+(a-d)(a+d) = 3a^2 - d^2 = 3(16) - d^2 = 48 - d^2 = 39 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3$. Thus zeros are $4-3=1$, $4$, $4+3=7$.

  26. Q26. LCM = 420, HCF = 3

    Prime factors: 12 = 2^2 × 3, 15 = 3 × 5, 21 = 3 × 7. HCF is the product of smallest powers of common primes: 3^1 = 3. LCM is the product of greatest powers of all primes: 2^2 × 3 × 5 × 7 = 420.

  27. Q27. √3 is irrational.

    Assume √3 is rational, i.e., √3 = a/b where a and b are co-prime and b ≠ 0. Squaring: 3 = a²/b² ⇒ a² = 3b². So 3 divides a² ⇒ 3 divides a. Let a = 3c. Then 9c² = 3b² ⇒ b² = 3c². Hence 3 divides b² ⇒ 3 divides b. Thus a and b have a common factor 3, contradicting co-primality. Therefore √3 is irrational.

  28. Q28. Exactly one of n, n+2, n+4 is divisible by 3.

    By Euclid's division lemma, any positive integer n is of the form 3q, 3q+1 or 3q+2. Case I: n = 3q, then n is divisible by 3, but n+2 = 3q+2 and n+4 = 3q+4 = 3(q+1)+1 are not. Case II: n = 3q+1, then n is not divisible by 3, n+2 = 3q+3 = 3(q+1) is divisible by 3, but n+4 = 3q+5 = 3(q+1)+2 is not. Case III: n = 3q+2, then n and n+2 = 3q+4 = 3(q+1)+1 are not divisible by 3, but n+4 = 3q+6 = 3(q+2) is. Thus exactly one is divisible by 3.

  29. Q29. \(x^2 - 6x + 1\)

    Sum of zeroes = \((3+2\sqrt{2}) + (3-2\sqrt{2}) = 6\). Product of zeroes = \((3)^2 - (2\sqrt{2})^2 = 9 - 8 = 1\). Required polynomial is \(x^2 - (\text{sum})x + \text{product} = x^2 - 6x + 1\).

  30. Q30. \(k = 1\)

    \(\alpha+\beta = -\frac{5}{2}\), \(\alpha\beta = \frac{k}{2}\). \(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \frac{25}{4} - k\). Equating to \(\frac{21}{4}\) gives \(\frac{25}{4} - k = \frac{21}{4}\), so \(k = 1\).

  31. Q31. \(a = 2\), \(b = -5\)

    By Factor Theorem: \(f(-1)=0 \Rightarrow (-1)^3 + a(-1)^2 + b(-1) - 6 = -1 + a - b - 6 = a - b - 7 = 0 \Rightarrow a - b = 7\). \(f(2)=0 \Rightarrow 8 + 4a + 2b - 6 = 4a + 2b + 2 = 0 \Rightarrow 2a + b + 1 = 0 \Rightarrow b = -2a - 1\). Substituting: \(a - (-2a - 1) = 7 \Rightarrow 3a + 1 = 7 \Rightarrow a = 2\). Then \(b = -2(2) - 1 = -5\).

  32. Q32. √3 is irrational.

    Assume √3 is rational. Then there exist co-prime integers a and b (b≠0) such that √3 = a/b. So 3 = a^2/b^2, or a^2 = 3b^2. Then 3 divides a^2, so 3 divides a. Let a = 3c. Then (3c)^2 = 3b^2 ⇒ 9c^2 = 3b^2 ⇒ b^2 = 3c^2. So 3 divides b^2, hence 3 divides b. Thus a and b have a common factor 3, contradicting co-primality. Hence √3 is irrational.

  33. Q33. HCF = 4, LCM = 9696. 4 × 9696 = 38784 = 96 × 404.

    Prime factorization: 96 = 2^5 × 3, 404 = 2^2 × 101. HCF = product of smallest powers of common primes = 2^2 = 4. LCM = product of greatest powers = 2^5 × 3 × 101 = 32 × 303 = 9696. Verification: HCF × LCM = 4 × 9696 = 38784; product = 96 × 404 = 38784. Hence verified.

  34. Q34. The square is always of the form 5q, 5q+1, or 5q+4, never 5q+2 or 5q+3.

    Let a be any positive integer. By Euclid’s division lemma, a = 5m, 5m+1, 5m+2, 5m+3, or 5m+4 for some integer m. Square each case: (i) (5m)^2 = 25m^2 = 5(5m^2) = 5q, remainder 0. (ii) (5m+1)^2 = 25m^2+10m+1 = 5(5m^2+2m)+1 = 5q+1. (iii) (5m+2)^2 = 25m^2+20m+4 = 5(5m^2+4m)+4 = 5q+4. (iv) (5m+3)^2 = 25m^2+30m+9 = 5(5m^2+6m+1)+4 = 5q+4. (v) (5m+4)^2 = 25m^2+40m+16 = 5(5m^2+8m+3)+1 = 5q+1. Hence, squares are only of the forms 5q, 5q+1, or 5q+4, never 5q+2 or 5q+3.

  35. Q35. Zeroes are 4/3 and –1. Sum of zeroes = 1/3; Product = –4/3, which matches –b/a and c/a respectively.

    To find zeroes, factor: 3x^2 – x – 4 = 3x^2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1). So x = 4/3, x = –1. For verification: a=3, b=-1, c=-4. Sum of zeroes = 4/3 + (–1) = 1/3 = –b/a = –(–1)/3 = 1/3. Product = (4/3)*(–1) = –4/3 = c/a = –4/3.

  36. Q36. Prime factorization: 400 = 2^4 × 5^2. The fraction 27/400 terminates because its denominator has only 2 and 5 as prime factors.

    400 ÷ 2 = 200, ÷2 = 100, ÷2 = 50, ÷2 = 25, ÷5 = 5, ÷5 = 1 → 2^4 × 5^2. Since denominator is of the form 2^n·5^m, the decimal terminates. 27/400 = 0.0675.

  37. Q37. They will next change together at 9:04:48 am. They change together 12 times between 9:00 am and 10:00 am (excluding the initial change).

    LCM of 18, 24, 32: 18 = 2×3^2, 24 = 2^3×3, 32 = 2^5 → LCM = 2^5×3^2 = 288 sec = 4 min 48 sec. Next change: 9:04:48. In one hour (3600 sec), number of intervals of 288 sec = floor(3600/288) = 12. So they change together 12 more times.

  38. Q38. Both √p + √q and √p – √q are irrational.

    Proof by contradiction: Assume sum = a/b, isolate √p, square, solve for √q giving rational expression, contradiction. For difference, similar steps yield √q rational, contradiction.

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