10th Maths Quarterly Exam 2026 Question Paper & Answer Key | Vellore District (EM)

COMMON QUARTERLY EXAMINATION - 2026

1. If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is
a) 3      b) 2      c) 4      d) 8

2. Let n(A) = m and n(B) = n then the total number of relations that can be defined from A to B is
a) mⁿ      b) n^m      c) 2^mn − 1      d) 2^mn

3. Let f(x) = √(1 + x²), then
a) f(xy) = f(x)·f(y)      b) f(xy) ≥ f(x)·f(y)      c) f(xy) ≤ f(x)·f(y)      d) None of these

4. Using Euclid's division lemma, if the cube of any positive integer is divided by 9 the possible remainders are
a) 1, 4, 8      b) 0, 1, 8      c) 0, 1, 3      d) 1, 3, 5

5. An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P is
a) 16m      b) 62m      c) 31m      d) ³¹/₂ m

6. If the sequence t₁, t₂, t₃, ... are in A.P then the sequence t₆, t₁₂, t₁₈, ... is
a) a Geometric Progression      b) An Arithmetic progression
c) neither an Arithmetic Progression nor a Geometric Progression      d) a constant sequence

7. The solution of the system x + y − 3z = −6, −7y + 7z = 7, 3z = 9 is
a) x = 1, y = 2, z = 3      b) x = −1, y = 2, z = 3
c) x = −1, y = −2, z = 3      d) x = 1, y = −2, z = 3

8. y² + 1/y² is not equal to
a) (y⁴ + 1) / y²      b) [y + 1/y]²      c) [y − 1/y]² + 2      d) [y + 1/y]² − 2

9. The square root of (256 x⁸ y⁴ z¹⁰) / (25 x⁶ y⁶ z⁶) is equal to
a) ¹⁶/₅ |(x² z⁴) / y²|      b) 16 |y² / (x² z⁴)|      c) ¹⁶/₅ |y / (x z²)|      d) ¹⁶/₅ |(x z²) / y|

10. In ΔLMN, ∠L = 60°, ∠M = 50°. If ΔLMN ~ ΔPQR then the value of ∠R is
a) 40°      b) 70°      c) 30°      d) 110°

11. The area of triangle formed by the points (−5, 0), (0, −5) and (5, 0) is
a) 0 sq. units      b) 25 sq. units      c) 5 sq. units      d) none of these

12. If (5, 7), (3, p) and (6, 6) are collinear, then the value of p is
a) 3      b) 6      c) 9      d) 12

13. The straight line given by the equation x = 11 is
a) parallel to x axis      b) parallel to y axis
c) passing through the origin      d) passing through the point (0, 11)

14. tan θ cosec²θ − tan θ is equal to
a) sec θ      b) cot²θ      c) sin θ      d) cot θ

15. Let A = {1, 2, 3} and B = {x | x is a prime number < 10}. Find A × B and B × A.

16. Find k if f ∁ f(k) = 5 where f(k) = 2k − 1.

17. Find the least number that is divisible by the first ten natural numbers.

18. In GP 729, 243, 81, ... find t₇.

19. Find the square root of the following expression: (144 a⁸ b¹² c¹⁶) / (81 f¹² g⁴ h¹⁴).

20. Find the excluded values of the following: (7p + 2) / (8p² + 13p + 5).

21. Find LCM of 5x − 10, 5x² − 20.

22. In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD / DB = ³/₄ and AC = 15 cm, find AE.

23. Show that the points P(−1.5, 3), Q(6, −2), R(−3, 4) are collinear.

24. Find the equation of a straight line which has slope −⁵/₄ and passing through the point (−1, 2).

25. Show that the straight lines x − 2y + 3 = 0 and 6x + 3y + 8 = 0 are perpendicular.

26. Prove that tan²θ − sin²θ = tan²θ sin²θ.

27. Prove that (1 − tan²θ) / (cot²θ − 1) = tan²θ.

28. [Compulsory] If 3 + k, 18 − k, 5k + 1 are in A.P. then find k.

29. A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime numbers. Verify that (A ∩ B) × C = (A × C) ∩ (B × C).

31. Find the HCF of 396, 504, 636.

32. Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.

33. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, ..., 24 cm. How much area can be decorated with these colour papers?

34. Solve: x + y + z = 5, 2x − y + z = 9, x − 2y + 3z = 16.

35. Find the GCD of following polynomials: x⁴ + 3x³ − x − 3 and x³ + x² − 5x + 3.

36. Simplify: [(b² + 3b − 28) / (b² + 4b + 4)] ÷ [(b² − 49) / (b² − 5b − 14)].

37. State and prove Basic Proportionality Theorem.

38. Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices taken in order are (−4, −2), (−3, k), (3, −2) and (2, 3).

39. Find the equation of a straight line parallel to x-axis and passing through the point of intersection of the lines 7x − 3y = −12 and 2y = x + 3.

40. A quadrilateral has vertices at A(−4, −2), B(5, −1), C(6, 5) and D(−7, 6). Show that the mid-points of its sides form a parallelogram.

41. Prove that √[(1 + sin θ) / (1 − sin θ)] + √[(1 − sin θ) / (1 + sin θ)] = 2 sec θ.

43. a) Construct a triangle similar to a given triangle PQR with its sides equal to ⁷/₃ of the corresponding sides of the triangle PQR (Scale Factor ⁷/₃ > 1).
(OR) b) Construct a ΔPQR in which PQ = 8 cm, ∠R = 60° and the median RG from R to PQ is 5.8 cm. Find the length of the altitude from R to PQ.

1. b) 2   [Total relations = 2^n(A)×n(B) = 2^{5 × n(B)} = 1024 = 2¹⁰ ⇒ n(B) = 2]

2. d) 2^mn   [Total number of relations from A to B is 2^n(A×B) = 2^mn]

3. c) f(xy) ≤ f(x)·f(y)   [1 + x²y² ≤ (1 + x²)(1 + y²) = 1 + x² + y² + x²y² ⇒ √(1 + x²y²) ≤ √(1 + x²)√(1 + y²)]

4. b) 0, 1, 8   [The cube of any integer can be expressed in the form 9q, 9q + 1, or 9q + 8]

5. c) 31m   [S₃₁ = (31/2)(2a + 30d) = 31(a + 15d) = 31 × t₁₆ = 31m]

6. b) An Arithmetic progression   [The common difference between terms is t₁₂ − t₆ = 6d, which is constant]

7. a) x = 1, y = 2, z = 3   [3z = 9 ⇒ z = 3; −7y + 7(3) = 7 ⇒ y = 2; x + 2 − 3(3) = −6 ⇒ x = 1]

8. b) [y + 1/y]²   [Because [y + 1/y]² = y² + 1/y² + 2 ≠ y² + 1/y²]

9. d) ¹⁶/₅ |(x z²) / y|   [√[ (256/25) · (x⁸⁻⁶ z¹⁰⁻⁶) / y⁶⁻⁴ ] = ¹⁶/₅ √(x² z⁴ / y²) = ¹⁶/₅ |(x z²) / y|]

10. b) 70°   [∠N = 180° − (60° + 50°) = 70°; ΔLMN ~ ΔPQR ⇒ ∠R = ∠N = 70°]

11. b) 25 sq. units   [Area = ¹/₂ × base × height = ¹/₂ × (5 − (−5)) × 5 = ¹/₂ × 10 × 5 = 25]

12. c) 9   [Slope = (p − 7) / (3 − 5) = (6 − 7) / (6 − 5) ⇒ (p − 7) / (−2) = −1 ⇒ p − 7 = 2 ⇒ p = 9]

13. b) parallel to y axis

14. d) cot θ   [tan θ cosec²θ − tan θ = tan θ (cosec²θ − 1) = tan θ cot²θ = cot θ]

15. A = {1, 2, 3}, B = {2, 3, 5, 7}
A × B = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7)}
B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (5, 1), (5, 2), (5, 3), (7, 1), (7, 2), (7, 3)}

16. f(k) = 2k − 1
f ∁ f(k) = f(2k − 1) = 2(2k − 1) − 1 = 4k − 2 − 1 = 4k − 3
Given f ∁ f(k) = 5 ⇒ 4k − 3 = 5 ⇒ 4k = 8 ⇒ k = 2

17. The numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Prime factorisation: 2 = 2, 3 = 3, 4 = 2², 5 = 5, 6 = 2 × 3, 7 = 7, 8 = 2³, 9 = 3², 10 = 2 × 5.
LCM = 2³ × 3² × 5 × 7 = 8 × 9 × 35 = 2520

18. GP: 729, 243, 81, ...
First term a = 729, common ratio r = 243 / 729 = ¹/₃.
t₇ = a · r⁷⁻¹ = 729 × (¹/₃)⁶ = 729 × (1 / 729) = 1

19. √[ (144 a⁸ b¹² c¹⁶) / (81 f¹² g⁴ h¹⁴) ] = ⁴/₃ |(a⁴ b⁶ c⁸) / (f⁶ g² h⁷)|

20. (7p + 2) / (8p² + 13p + 5)
Setting denominator to zero: 8p² + 13p + 5 = (p + 1)(8p + 5) = 0.
Excluded values are −1, −⁵/₈

21. 5x − 10 = 5(x − 2)
5x² − 20 = 5(x² − 4) = 5(x + 2)(x − 2)
LCM = 5(x + 2)(x − 2) = 5(x² − 4)

22. In ΔABC, DE || BC. By Thales theorem, AD / DB = AE / EC.
Let AE = x, then EC = AC − AE = 15 − x.
³/₄ = x / (15 − x) ⇒ 4x = 45 − 3x ⇒ 7x = 45 ⇒ AE = ⁴⁵/₇ ≈ 6.43 cm

23. Area(ΔPQR) = ¹/₂ | [x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)] |
= ¹/₂ | [−1.5(−2 − 4) + 6(4 − 3) + (−3)(3 − (−2))] |
= ¹/₂ | [−1.5(−6) + 6(1) − 3(5)] | = ¹/₂ |9 + 6 − 15| = 0.
Since the area is 0, points P, Q, R are collinear.

24. y − y₁ = m(x − x₁) ⇒ y − 2 = −⁵/₄ (x − (−1))
4(y − 2) = −5(x + 1) ⇒ 4y − 8 = −5x − 5 ⇒ 5x + 4y − 3 = 0

25. For x − 2y + 3 = 0, slope m₁ = −1 / (−2) = ¹/₂.
For 6x + 3y + 8 = 0, slope m₂ = −6 / 3 = −2.
m₁ × m₂ = (¹/₂) × (−2) = −1.
Since the product of slopes is −1, the lines are perpendicular.

26. LHS = tan²θ − sin²θ = (sin²θ / cos²θ) − sin²θ
= sin²θ [ (1 / cos²θ) − 1 ] = sin²θ [ (1 − cos²θ) / cos²θ ]
= sin²θ × (sin²θ / cos²θ) = tan²θ sin²θ = RHS. Hence proved.

27. LHS = (1 − tan²θ) / (cot²θ − 1)
= [ 1 − (sin²θ / cos²θ) ] / [ (cos²θ / sin²θ) − 1 ]
= [ (cos²θ − sin²θ) / cos²θ ] / [ (cos²θ − sin²θ) / sin²θ ] = sin²θ / cos²θ = tan²θ = RHS. Hence proved.

28. In an A.P., t₂ − t₁ = t₃ − t₂
(18 − k) − (3 + k) = (5k + 1) − (18 − k)
15 − 2k = 6k − 17 ⇒ 8k = 32 ⇒ k = 4

29. A = {1, 2, 3, 4, 5, 6, 7}, B = {2, 3, 5, 7}, C = {2}
A ∩ B = {2, 3, 5, 7}
LHS = (A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} — (1)
A × C = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}
B × C = {(2, 2), (3, 2), (5, 2), (7, 2)}
RHS = (A × C) ∩ (B × C) = {(2, 2), (3, 2), (5, 2), (7, 2)} — (2)
From (1) and (2), (A ∩ B) × C = (A × C) ∩ (B × C) is verified.

31. HCF of 396, 504, 636:
504 = 396(1) + 108
396 = 108(3) + 72
108 = 72(1) + 36
72 = 36(2) + 0 ⇒ HCF(396, 504) = 36
636 = 36(17) + 24
36 = 24(1) + 12
24 = 12(2) + 0 ⇒ HCF(636, 36) = 12
Therefore, HCF(396, 504, 636) = 12

32. Required number = HCF(1230 − 12, 1926 − 12) = HCF(1218, 1914)
1914 = 1218(1) + 696
1218 = 696(1) + 522
696 = 522(1) + 174
522 = 174(3) + 0
The largest number is 174.

33. Total Area = 10² + 11² + ... + 24² = Σ₁²⁴ k² − Σ₁⁹ k²
= [(24 × 25 × 49) / 6] − [(9 × 10 × 19) / 6] = 4900 − 285 = 4615 cm²

34.
x + y + z = 5 — (1)
2x − y + z = 9 — (2)
x − 2y + 3z = 16 — (3)
(1) + (2) ⇒ 3x + 2z = 14 — (4)
(2) × 2 − (3) ⇒ 3x − z = 2 — (5)
(4) − (5) ⇒ 3z = 12 ⇒ z = 4
From (5): 3x − 4 = 2 ⇒ 3x = 6 ⇒ x = 2
From (1): 2 + y + 4 = 5 ⇒ y = −1
Solution: x = 2, y = −1, z = 4

35. Let f(x) = x⁴ + 3x³ − x − 3 and g(x) = x³ + x² − 5x + 3.
Dividing f(x) by g(x) gives remainder 3x² + 6x − 9 = 3(x² + 2x − 3).
Dividing g(x) by (x² + 2x − 3) gives remainder 0.
Therefore, GCD = x² + 2x − 3

36. [(b² + 3b − 28) / (b² + 4b + 4)] ÷ [(b² − 49) / (b² − 5b − 14)]
= [ (b + 7)(b − 4) / (b + 2)² ] × [ (b − 7)(b + 2) / ((b − 7)(b + 7)) ]
= [(b − 4)(b + 7) / ((b + 2)(b + 7))] = (b − 4) / (b + 2)

37. Basic Proportionality Theorem (Thales Theorem): Statement and Proof (Refer standard textbook).

38. Vertices in order: (−4, −2), (−3, k), (3, −2), (2, 3)
Area = ¹/₂ | [(−4k + 6 + 9 − 4) − (6 + 3k − 4 − 12)] | = 28
| −4k + 11 − (3k − 10) | = 56 ⇒ |−7k + 21| = 56
−7k + 21 = 56 ⇒ −7k = 35 ⇒ k = −5

39. 7x − 3y = −12 — (1) and −x + 2y = 3 — (2)
(2) × 7 ⇒ −7x + 14y = 21. Adding to (1): 11y = 9 ⇒ y = ⁹/₁₁.
A line parallel to x-axis is of the form y = c ⇒ y = ⁹/₁₁ ⇒ 11y − 9 = 0

40. A(−4, −2), B(5, −1), C(6, 5), D(−7, 6)
Midpoints: E = (¹/₂, −³/₂), F = (¹¹/₂, 2), G = (−¹/₂, ¹¹/₂), H = (−¹¹/₂, 2).
Slope of EF = ⁷/₁₀, Slope of GH = ⁷/₁₀ ⇒ EF || GH.
Slope of FG = −⁷/₁₂, Slope of EH = −⁷/₁₂ ⇒ FG || EH.
Since both pairs of opposite sides are parallel, EFGH is a parallelogram.

41. LHS = √[ (1 + sin θ)² / cos²θ ] + √[ (1 − sin θ)² / cos²θ ]
= [(1 + sin θ) / cos θ] + [(1 − sin θ) / cos θ] = (1 + sin θ + 1 − sin θ) / cos θ
= 2 / cos θ = 2 sec θ = RHS. Hence proved.

42.
i) f(3) = 3 + 2 = 5
ii) f(0) = 2
iii) f(−1.5) = −1.5 − 1 = −2.5
iv) f(2) = 2 + 2 = 4 ; f(−2) = −2 − 1 = −3 ⇒ f(2) + f(−2) = 4 + (−3) = 1

43. a) Similar triangle construction with scale factor ⁷/₃ > 1 • b) ΔPQR construction with base PQ = 8 cm, ∠R = 60°, median RG = 5.8 cm, and altitude length measured as 3.8 cm.