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

COMMON QUARTERLY EXAMINATION - 2026 - 27

1. If n(A × B) = 6 and A = {1, 3} then n(B) is
a) 1      b) 2      c) 3      d) 6

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

3. Given f(x) = (−1)^x is a function from N to Z, then the range of f is
a) {1}      b) N      c) {1, −1}      d) Z

4. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
a) 2025      b) 5220      c) 5025      d) 2520

5. 7^4k ≡ _______ (mod 100)
a) 1      b) 2      c) 3      d) 4

6. The value of (1³ + 2³ + 3³ + ... + 15³) − (1 + 2 + 3 + ... + 15) is
a) 14400      b) 14200      c) 14280      d) 14520

7. If (x − 6) is the HCF of x² − 2x − 24 and x² − kx − 6 then the value of k is
a) 3      b) 5      c) 6      d) 8

8. Graph of a linear equation is a
a) Straight line      b) Circle      c) Parabola      d) Hyperbola

9. If a³ / (a − b) is added with b³ / (b − a) then the new expression?
a) a² + ab + b²      b) a² − ab + b²      c) a³ + b³      d) a³ − b³

10. The perimeters of two similar triangles ΔABC and ΔPQR are 36 cm and 24 cm respectively. If PQ = 10 cm then the length of AB is
a) 6 ²/₃ cm      b) (10√6) / 3 cm      c) 66 ²/₃ cm      d) 15 cm

11. In a ΔABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm, and DC = 3 cm, the length of the side AC is
a) 6 cm      b) 4 cm      c) 3 cm      d) 8 cm

12. The point of intersection of 3x − y = 4 and x + y = 8 is
a) (5, 3)      b) (2, 4)      c) (3, 5)      d) (4, 4)

13. If slope of the line PQ is 1 / √3 then slope of perpendicular bisector of PQ is
a) √3      b) −√3      c) 1 / √3      d) 0

14. If x = a tan θ and y = b sec θ then
a) (y² / b²) − (x² / a²) = 1      b) (x² / a²) − (y² / b²) = 1
c) (x² / a²) + (y² / b²) = 1      d) (x² / a²) − (y² / b²) = 0

15. Let A = {1, 2, 3, 4, ..., 45} and R be the relation defined as "square of a number" on A. Write R as a subset of A × A. Also, find the domain and range of R.

16. If f(x) = 3x − 2, g(x) = 2x + k and if f ∁ g = g ∁ f then find the value of k.

17. a and b are two positive integers, such that a^b × b^a = 800. Find a and b.

18. Find the sum of 1 + 4 + 9 + 16 + ... + 225.

19. Find the excluded values of: (x² + 6x + 8) / (x² + x − 2).

20. If the difference between a number and its reciprocal is ²⁴/₅, find the number.

21. If α, β are the roots of the equation 3x² + 7x − 2 = 0, find the values of (α / β) + (β / α).

22. If ΔABC ~ ΔDEF such that area of ΔABC is 9 cm² and the area of ΔDEF is 16 cm². If BC = 2.1 cm, find the length of EF.

23. In a ΔABC, AD is the bisector of ∠BAC. If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

24. Find the slope of a line joining the points (5, √5) and Origin.

25. Find the equation of line passing through the point (3, −4) and having slope −⁵/₇.

26. Find the intercepts made by the line 4x + 3y + 12 = 0 on the coordinate axes.

27. Prove that √[(1 + cos θ) / (1 − cos θ)] = cosec θ + cot θ.

28. [Compulsory] The ratio of three numbers is 2:5:7. If 7 is subtracted from the second number, the resulting three numbers are in A.P. Find the numbers.

29. Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4} and C = {3, 5}. Verify that A × (B ∪ C) = (A × B) ∪ (A × C).

30. Let A = {1, 2, 3, 4} and B = {2, 5, 8, 11, 14} be two sets. Let f: A → B be a function given by f(x) = 3x − 1. Represent this function (i) by arrow diagram (ii) in a table form (iii) as a set of ordered pairs (iv) a graph.

32. If p₁^x₁ × p₂^x₂ × p₃^x₃ × p₄^x₄ = 113400, where p₁, p₂, p₃, p₄ are primes in ascending order and x₁, x₂, x₃, x₄ are integers, find the value of p₁, p₂, p₃, p₄ and x₁, x₂, x₃, x₄.

33. Find the sum of all natural numbers between 300 and 600 which are divisible by 7.

34. Find the sum to n terms of the series 3 + 33 + 333 + ...

35. Solve: 3x − 2y + z = 2,   2x + 3y − z = 5,   x + y + z = 6.

36. If 36x⁴ − 60x³ + 61x² − mx + n is a perfect square, find the values of m and n.

37. If the roots of the equation (c² − ab)x² − 2(a² − bc)x + b² − ac = 0 are real and equal, prove that either a = 0 or a³ + b³ + c³ = 3abc.

38. State and prove Thales Theorem.

39. Find the area of the quadrilateral formed by the points (8, 6), (5, 11), (−5, 12) and (−4, 3).

40. If the vertices of a ΔABC are A(6, 2), B(−5, −1) and C(1, 9). Through the vertex A, find the equation of median.

41. If cos α / cos β = m and cos α / sin β = n, then prove that (m² + n²)cos²β = n².

42. [Compulsory] Find the equation of a line passing through the point of intersection of the lines 4x + 7y − 3 = 0 and 2x − 3y + 1 = 0 that has equal intercepts on the axes.

43. a) Construct a triangle similar to a given triangle ΔPQR with its sides equal to ³/₅ of the corresponding sides of the triangle PQR (scale factor ³/₅ < 1).
(OR) b) Construct a ΔPQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm.

1. c) 3   [n(A × B) = n(A) × n(B) ⇒ 6 = 2 × n(B) ⇒ n(B) = 3]

2. c) 2^mn − 1   [Total number of relations is 2^mn; excluding empty relation gives 2^mn − 1]

3. c) {1, −1}   [For x ∈ N, (−1)^x takes values −1 for odd numbers and 1 for even numbers]

4. d) 2520   [LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 2³ × 3² × 5 × 7 = 2520]

5. a) 1   [7⁴ = 2401 ≡ 1 (mod 100); hence (7⁴)^k ≡ 1^k ≡ 1 (mod 100)]

6. c) 14280   [(15 × 16 / 2)² − (15 × 16 / 2) = 120² − 120 = 14400 − 120 = 14280]

7. b) 5   [Since (x − 6) is a factor of x² − kx − 6, 6² − 6k − 6 = 0 ⇒ 30 = 6k ⇒ k = 5]

8. a) Straight line

9. a) a² + ab + b²   [a³/(a − b) − b³/(a − b) = (a³ − b³)/(a − b) = a² + ab + b²]

10. d) 15 cm   [Perimeter(ΔABC) / Perimeter(ΔPQR) = AB / PQ ⇒ 36 / 24 = AB / 10 ⇒ AB = 15 cm]

11. b) 4 cm   [AB / AC = BD / DC ⇒ 8 / AC = 6 / 3 = 2 ⇒ AC = 4 cm]

12. c) (3, 5)   [Adding equations: 4x = 12 ⇒ x = 3; 3 + y = 8 ⇒ y = 5]

13. b) −√3   [m₁ · m₂ = −1 ⇒ m₂ = −1 / (1/√3) = −√3]

14. a) (y² / b²) − (x² / a²) = 1   [sec²θ − tan²θ = 1 ⇒ (y/b)² − (x/a)² = 1]

15. A = {1, 2, 3, 4, ..., 45}. Relation R is "square of a number" on A.
Squares belonging to A: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36.
R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)} ⊂ A × A
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1, 4, 9, 16, 25, 36}

16. f(x) = 3x − 2,   g(x) = 2x + k
f ∁ g(x) = f(2x + k) = 3(2x + k) − 2 = 6x + 3k − 2
g ∁ f(x) = g(3x − 2) = 2(3x − 2) + k = 6x − 4 + k
f ∁ g = g ∁ f ⇒ 6x + 3k − 2 = 6x − 4 + k ⇒ 2k = −2 ⇒ k = −1

17. a^b × b^a = 800
800 = 32 × 25 = 2⁵ × 5²
Comparing bases and powers: a = 2, b = 5 (or a = 5, b = 2)

18. 1 + 4 + 9 + 16 + ... + 225 = 1² + 2² + 3² + ... + 15²
Sum = [n(n + 1)(2n + 1)] / 6 = [15 × 16 × 31] / 6 = 5 × 8 × 31 = 1240

19. Expression: (x² + 6x + 8) / (x² + x − 2)
Denominator: x² + x − 2 = (x + 2)(x − 1) = 0
Excluded values are −2, 1

20. Let the number be x.
x − (1/x) = ²⁴/₅ ⇒ (x² − 1)/x = ²⁴/₅ ⇒ 5x² − 24x − 5 = 0
(x − 5)(5x + 1) = 0 ⇒ x = 5 or x = −¹/₅

21. 3x² + 7x − 2 = 0 ⇒ α + β = −⁷/₃,   αβ = −²/₃
(α / β) + (β / α) = (α² + β²) / αβ = [(α + β)² − 2αβ] / αβ
= [(−⁷/₃)² − 2(−²/₃)] / (−²/₃) = (⁴⁹/₉ + ⁴/₃) / (−²/₃) = (⁶¹/₉) × (−³/₂) = −⁶¹/₆

22. Area(ΔABC) / Area(ΔDEF) = BC² / EF²
9 / 16 = (2.1)² / EF² ⇒ 3 / 4 = 2.1 / EF
EF = (2.1 × 4) / 3 = 2.8 cm

23. By Angle Bisector Theorem, BD / DC = AB / AC = 10 / 14 = 5 / 7
BC = BD + DC = 6 cm
BD = 6 × (5 / 12) = 2.5 cm
DC = 6 × (7 / 12) = 3.5 cm

24. Slope of line joining (5, √5) and (0, 0):
m = (y₂ − y₁) / (x₂ − x₁) = (√5 − 0) / (5 − 0) = √5 / 5 = 1 / √5

25. y − y₁ = m(x − x₁) ⇒ y − (−4) = −⁵/₇ (x − 3)
7(y + 4) = −5(x − 3) ⇒ 7y + 28 = −5x + 15 ⇒ 5x + 7y + 13 = 0

26. 4x + 3y = −12 ⇒ 4x / (−12) + 3y / (−12) = 1
x / (−3) + y / (−4) = 1
x-intercept = −3,   y-intercept = −4

27. LHS = √[ ((1 + cos θ)(1 + cos θ)) / ((1 − cos θ)(1 + cos θ)) ] = √[ (1 + cos θ)² / sin²θ ]
= (1 + cos θ) / sin θ = (1 / sin θ) + (cos θ / sin θ) = cosec θ + cot θ = RHS. Hence proved.

28. Let the numbers be 2k, 5k, 7k.
Given 2k, 5k − 7, 7k are in A.P.
2(5k − 7) = 2k + 7k ⇒ 10k − 14 = 9k ⇒ k = 14
Numbers are 2(14), 5(14), 7(14) = 28, 70, 98

29. A = {0, 1}, B = {2, 3, 4}, C = {3, 5}
B ∪ C = {2, 3, 4, 5}
LHS: A × (B ∪ C) = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} — (1)
A × B = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}
RHS: (A × B) ∪ (A × C) = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} — (2)
From (1) and (2), A × (B ∪ C) = (A × B) ∪ (A × C) is verified.

31. t(C) = ⁹/₅ C + 32
(i) t(0) = ⁹/₅(0) + 32 = 32°F
(ii) t(28) = ⁹/₅(28) + 32 = 50.4 + 32 = 82.4°F
(iii) t(−10) = ⁹/₅(−10) + 32 = −18 + 32 = 14°F
(iv) ⁹/₅ C + 32 = 212 ⇒ ⁹/₅ C = 180 ⇒ C = 100°C
(v) C = ⁹/₅ C + 32 ⇒ −⁴/₅ C = 32 ⇒ C = −40°

32. 113400 = 2³ × 3⁴ × 5² × 7¹
Primes: p₁ = 2, p₂ = 3, p₃ = 5, p₄ = 7
Integers: x₁ = 3, x₂ = 4, x₃ = 2, x₄ = 1

33. Multiples of 7 between 300 and 600: 301, 308, 315, ..., 595
a = 301, d = 7, l = 595
n = [(l − a) / d] + 1 = [(595 − 301) / 7] + 1 = 42 + 1 = 43
S₄₃ = (43 / 2) [301 + 595] = (43 / 2) × 896 = 43 × 448 = 19264

34. S_n = 3 + 33 + 333 + ... to n terms
= 3(1 + 11 + 111 + ... to n terms) = ³/₉ (9 + 99 + 999 + ... to n terms)
= ¹/₃ [(10 − 1) + (10² − 1) + ... + (10ⁿ − 1)] = ¹⁰/₂₇ (10ⁿ − 1) − ⁿ/₃

35.
3x − 2y + z = 2 — (1)
2x + 3y − z = 5 — (2)
x + y + z = 6 — (3)
(1) + (2) ⇒ 5x + y = 7 — (4)
(2) + (3) ⇒ 3x + 4y = 11 — (5)
4 × (4) − (5) ⇒ 20x − 3x = 28 − 11 ⇒ 17x = 17 ⇒ x = 1
From (4): 5(1) + y = 7 ⇒ y = 2
From (3): 1 + 2 + z = 6 ⇒ z = 3
Solution: x = 1, y = 2, z = 3

36. 36x⁴ − 60x³ + 61x² − mx + n
Square root by division gives quotient: 6x² − 5x + 3
(6x² − 5x + 3)² = 36x⁴ − 60x³ + 61x² − 30x + 9
Comparing terms: m = 30,   n = 9

37. Roots are real and equal ⇒ Δ = B² − 4AC = 0
[−2(a² − bc)]² − 4(c² − ab)(b² − ac) = 0
4[(a⁴ − 2a²bc + b²c²) − (b²c² − ac³ − ab³ + a²bc)] = 0
a⁴ − 3a²bc + ab³ + ac³ = 0 ⇒ a(a³ + b³ + c³ − 3abc) = 0
Hence, either a = 0 or a³ + b³ + c³ = 3abc. Hence proved.

38. Thales Theorem: State and proof (Refer standard textbook).

39. Points in counter-clockwise order: (8, 6), (5, 11), (−5, 12), (−4, 3)
Area = ¹/₂ [(88 + 60 − 15 − 24) − (30 − 55 − 48 + 24)] = ¹/₂ [109 − (−49)] = ¹/₂(158) = 79 sq. units

40. Midpoint of BC where B(−5, −1) and C(1, 9): D = [(−5 + 1)/2, (−1 + 9)/2] = (−2, 4)
Equation of line through A(6, 2) and D(−2, 4):
(y − 2) / (4 − 2) = (x − 6) / (−2 − 6) ⇒ (y − 2) / 2 = (x − 6) / (−8)
−4(y − 2) = x − 6 ⇒ −4y + 8 = x − 6 ⇒ x + 4y − 14 = 0

41. m = cos α / cos β,   n = cos α / sin β
LHS = (m² + n²)cos²β = [ (cos²α / cos²β) + (cos²α / sin²β) ] cos²β
= cos²α [ (sin²β + cos²β) / (cos²β sin²β) ] cos²β = cos²α / sin²β = n² = RHS. Hence proved.

42. Solving 4x + 7y − 3 = 0 and 2x − 3y + 1 = 0 gives point of intersection (¹/₁₃, ⁵/₁₃).
Line with equal intercepts: x + y = a
a = ¹/₁₃ + ⁵/₁₃ = ⁶/₁₃
Equation: x + y = ⁶/₁₃ ⇒ 13x + 13y − 6 = 0

43. a) Construction of similar triangle with scale factor ³/₅ < 1 • b) Construction of ΔPQR with QR = 6.5 cm, ∠P = 60°, and altitude 4.5 cm.