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

Assess your mathematical skills and pace your preparation with the official 10th Maths Quarterly Exam 2026 Question Paper curated for schools across the Thiruvannamalai District. This district-level evaluation paper serves as a vital diagnostic tool for Tamil Nadu Samacheer Kalvi state board students aiming to test their problem-solving accuracy under true exam conditions. The question paper provides complete coverage of early high-yield chapters, including Relations and Functions, Numbers and Sequences (AP/GP series), core polynomial Algebra, and essential Geometry theorems. By working through this regional paper, you can easily identify your strong and weak areas, master time management during the exam, and refine the analytical skills required to score top marks on your final board examinations.


10th Maths Quarterly Exam 2026 Question Paper | Thiruvannamalai District

10 - STD

QUARTERLY EXAM - SEP 2026

MATHEMATICS
TIME: 3.00 HRS MARKS: 100
PART - 1 • I. Choose the Correct Answer: 14 × 1 = 14

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. If f: A → B is a bijective function and if n(B) = 7, then n(A) is equal to
(A) 7      (B) 49      (C) 1      (D) 14

3. If g = {(1, 1), (2, 3), (3, 5), (4, 7)} is a function given by g(x) = αx + β then the values of α and β are
(A) (−1, 2)      (B) (2, −1)      (C) (−1, −2)      (D) (1, 2)

4. In an A.P a18 − a14 = 32, then find the common difference is
(A) 8      (B) −8      (C) −4      (D) 4

5. The sum of the exponents of the prime factors in the prime factorization of 1729 is
(A) 1      (B) 2      (C) 3      (D) 4

6. The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P.
(A) 4551      (B) 10091      (C) 7881      (D) 13531

7. The next term of the sequence 3/16, 1/8, 1/12, 1/18, ... is
(A) 1/24      (B) 1/27      (C) 2/3      (D) 1/81

8. Which of the following should be added to make x4 + 64 a perfect square
(A) 4x2      (B) 16x2      (C) 8x2      (D) −8x4

9. Graph of a linear equation is a
(A) straight line      (B) circle      (C) parabola      (D) hyperbola

10. If ΔABC is an isosceles triangle with ∠C = 90° and AC = 5 cm, then AB is
(A) 2.5 cm      (B) 5 cm      (C) 10 cm      (D) 5√2 cm

11. If (5, 7), (3, p) and (6, 6) are collinear, then the value of p is
(A) 3      (B) 6      (C) 9      (D) 12

12. 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)

13. The equation of a straight line which is parallel to the line 2x + 5y − 2 = 0 is
A) 5x − 2y − 7 = 0      B) 4x + 10y − 4 = 0
C) 2x − 5y + 2 = 0      D) 5x + 2y − 2 = 0

14. If sin θ + cos θ = a and sec θ + cosec θ = b, then the value of b(a2 − 1) is equal to
(A) 2a      (B) 3a      (C) 0      (D) 2ab

PART - 2 • II. Answer any 10 of the following (Q.No. 28 is compulsory): 10 × 2 = 20

15. If A × B = {(3, 2), (3, 4), (5, 2), (5, 4)} then find A and B.

16. Show that the function defined by f(m) = m2 + m + 3 is one-one function.

17. If f(x) = x2 − 1, g(x) = x − 2 find a, if g ∁ f(a) = 1.

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

19. What is the time 100 hours after 7 a.m.?

20. If 13 + 23 + 33 + ... + k3 = 44100 then find 1 + 2 + 3 + ... + k.

21. Find the excluded values, if t / (t2 − 5t + 6).

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

23. Find the nature of the roots of the quadratic equation: 2x2 − 2x + 9 = 0.

24. D and E are respectively the points on the sides AB and AC of a ΔABC such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, show that DE || BC.

25. What is the inclination of a line whose slope is (i) 0 (ii) 1

26. Find the equation of a straight line passing through (5, −3) and (7, −4).

27. Prove: cos θ / (1 + sin θ) = sec θ − tan θ.

28. Find the slope of the following points: A(−a, b), B(3a, −b).

PART - 3 • III. Answer any 10 of the following (Q.No. 42 is compulsory): 10 × 5 = 50

29. Let 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 number. Verify that A × (B − C) = (A × B) − (A × C).

30. The function 't' which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C) = F where F = 9/5C + 32. Find:
(i) t(0)      (ii) t(28)      (iii) t(−10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Fahrenheit value.

31. If f(x) = x − 4, g(x) = x2, h(x) = 3x − 5 then prove that f ∁ (g ∁ h) = (f ∁ g) ∁ h.

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

33. In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 57/2. Find the three terms.

34. 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?

35. There are 12 pieces of five, ten and twenty rupee currencies whose total value is 105. When first 2 sorts are interchanged in their numbers its value will be increased by 20. Find the number of currencies in each sort.

36. Find the square root of the following: x4 − 12x3 + 42x2 − 36x + 9.

37. A flock of swans contained x2 members. As the clouds gathered, 10x went to a lake and one-eighth of the members flew away to a garden. The remaining three pairs played about in the water. How many swans were there in total?

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. Find the equation of the perpendicular bisector of the line joining the points A(−4, 2) and B(6, −4).

41. If cosec θ + cot θ = p then prove that cos θ = (p2 − 1) / (p2 + 1).

42. A household has a fixed electricity charge of ₹100. In addition, ₹6 charged for every unit of electricity consumed. Let x be the number of units consumed and y be the total bill.
i. Find the equation of the electricity bill.
ii. Find the bill for 50 units.
iii. If the bill is ₹400, find the number of units consumed.
iv. Find the y-intercept.
PART - 4 • IV. Answer all the questions: 2 × 8 = 16

43. 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) Construct a triangle ΔPQR such that QR = 5 cm, ∠P = 30° and the altitude from P to QR is of length 4.2 cm.

44. Graph the following linear function y = 1/2 x. Identify the constant of variation and verify it with the graph. Also (i) find y when x = 9 (ii) find x when y = 7.5.
(OR) A school announces that for a certain competitions, the cash prize will be distributed for all the participants equally as shown below:

No. of participants (x)246810
Amount for each participant in ₹ (y)18090604536
(i) Find the constant of variation.
(ii) Graph the above data and hence, find how much will each participant get if the number of participants are 12.
ANSWERS & COMPLETE SOLUTIONS
Part - 1:

1. (B) 2

2. (A) 7

3. (B) (2, −1)

4. (A) 8

5. (C) 3

6. (C) 7881

7. (B) 1/27

8. (B) 16x2

9. (A) straight line

10. (D) 5√2 cm

11. (C) 9

12. (B) parallel to Y axis

13. (B) 4x + 10y − 4 = 0

14. (A) 2a

Part - 2:

15. A × B = {(3, 2), (3, 4), (5, 2), (5, 4)}
A = {3, 5},   B = {2, 4}

16. f(m) = m2 + m + 3
f(1) = 5, f(2) = 9, f(3) = 15
Since distinct elements in the domain have distinct images in the co-domain, f is a one-one function.

17. g ∁ f(a) = 1 ⇒ g(a2 − 1) = 1 ⇒ a2 − 1 − 2 = 1 ⇒ a2 − 3 = 1
a2 = 4 ⇒ a = ±2

18. Least number divisible by first ten natural numbers = LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
= 23 × 32 × 5 × 7 = 8 × 9 × 5 × 7 = 2520

19. 100 ≡ x (mod 24) ⇒ 100 = 24(4) + 4 ⇒ x = 4
Time 100 hours after 7 a.m. = 7 a.m. + 4 hours = 11 a.m.

20. 13 + 23 + ... + k3 = 44100
[ {k(k + 1)} / 2 ]2 = 44100 ⇒ 1 + 2 + ... + k = √44100 = 210

21. t / (t2 − 5t + 6) = t / [(t − 3)(t − 2)]
Setting denominator to zero: (t − 3)(t − 2) = 0 ⇒ t = 3, 2

22. 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

23. 2x2 − 2x + 9 = 0 ⇒ a = 2, b = −2, c = 9
Discriminant Δ = b2 − 4ac = (−2)2 − 4(2)(9) = 4 − 72 = −68 < 0
Hence, the equation has no real roots (imaginary roots).

24. DB = AB − AD = 5.6 − 1.4 = 4.2 cm
EC = AC − AE = 7.2 − 1.8 = 5.4 cm
AD / DB = 1.4 / 4.2 = 1/3 ; AE / EC = 1.8 / 5.4 = 1/3
Since AD / DB = AE / EC, by converse of Thales theorem, DE || BC.

25. (i) Slope m = 0 ⇒ tan θ = 0 ⇒ θ = 0°
(ii) Slope m = 1 ⇒ tan θ = 1 ⇒ θ = 45°

26. Two-point form: (y − y1) / (y2 − y1) = (x − x1) / (x2 − x1)
(y + 3) / (−4 + 3) = (x − 5) / (7 − 5) ⇒ (y + 3) / (−1) = (x − 5) / 2
2y + 6 = −x + 5 ⇒ x + 2y + 1 = 0

27. RHS = sec θ − tan θ = (1 / cos θ) − (sin θ / cos θ) = (1 − sin θ) / cos θ
Multiply numerator and denominator by (1 + sin θ):
= (1 − sin2θ) / [cos θ(1 + sin θ)] = cos2θ / [cos θ(1 + sin θ)] = cos θ / (1 + sin θ) = LHS. Hence proved.

28. Slope m = (y2 − y1) / (x2 − x1) = (−b − b) / [3a − (−a)] = −2b / 4a = −b / (2a)

Part - 3:

29. A = {1, 2, 3, 4, 5, 6, 7}, B = {2, 3, 5, 7}, C = {2}
B − C = {3, 5, 7}
A × (B − C) = {(1, 3), (1, 5), (1, 7), ..., (7, 3), (7, 5), (7, 7)} (21 ordered pairs) — (1)
(A × B) − (A × C) leaves the same 21 ordered pairs — (2)
From (1) and (2), A × (B − C) = (A × B) − (A × C) is verified.

30. t(C) = 9/5C + 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) 9/5C + 32 = 212 ⇒ 9/5C = 180 ⇒ C = 180 × (5/9) = 100°C
(v) C = 9/5C + 32 ⇒ −4/5C = 32 ⇒ C = −40°

31. f(x) = x − 4, g(x) = x2, h(x) = 3x − 5
(f ∁ g)(x) = x2 − 4 ⇒ ((f ∁ g) ∁ h)(x) = (3x − 5)2 − 4 — (1)
(g ∁ h)(x) = (3x − 5)2 ⇒ (f ∁ (g ∁ h))(x) = (3x − 5)2 − 4 — (2)
From (1) and (2), (f ∁ g) ∁ h = f ∁ (g ∁ h).

32. By Euclid's division algorithm:
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
Hence, HCF(396, 504, 636) = 12

33. Let the terms be a/r, a, ar.
Product: (a/r)(a)(ar) = 27 ⇒ a3 = 27 ⇒ a = 3
Sum of products in pairs: (a2/r) + a2r + a2 = 57/2 ⇒ 9[(1 + r2 + r)/r] = 57/2
6r2 − 13r + 6 = 0 ⇒ (3r − 2)(2r − 3) = 0 ⇒ r = 2/3 or 3/2
The terms are 2, 3, 9/2 or 9/2, 3, 2

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

35. Let five, ten, and twenty rupee notes be x, y, z.
x + y + z = 12 — (1)
5x + 10y + 20z = 105 ⇒ x + 2y + 4z = 21 — (2)
10x + 5y + 20z = 125 ⇒ 2x + y + 4z = 25 — (3)
Solving (1), (2), (3) gives: x = 7 (₹5 notes), y = 3 (₹10 notes), z = 2 (₹20 notes)

36. Square root of x4 − 12x3 + 42x2 − 36x + 9 by long division gives quotient x2 − 6x + 3.
Square root = |x2 − 6x + 3|

37. Total swans = x2.
x2 − 10x − 1/8x2 = 6 ⇒ 7x2 − 80x − 48 = 0
Factoring: (x − 12)(7x + 4) = 0 ⇒ x = 12 (since x > 0)
Total number of swans = x2 = 122 = 144

38. Thales Theorem (Basic Proportionality Theorem):
State and prove: A line drawn parallel to one side of a triangle intersecting the other two sides divides them in the same ratio: AD / DB = AE / EC.

39. Vertices in 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 A(−4, 2) and B(6, −4): M = (1, −1)
Slope of AB = (−4 − 2) / (6 + 4) = −3/5 ⇒ Perpendicular slope m = 5/3
Equation: y + 1 = 5/3(x − 1) ⇒ 5x − 3y − 8 = 0

41. cosec θ + cot θ = p ⇒ cosec θ − cot θ = 1/p
Adding gives 2 cosec θ = (p2 + 1)/p ; Subtracting gives 2 cot θ = (p2 − 1)/p
Dividing cot θ by cosec θ gives: cos θ = (p2 − 1) / (p2 + 1). Hence proved.

42.
(i) Equation of electricity bill: y = 6x + 100
(ii) For x = 50 units: y = 6(50) + 100 = ₹400
(iii) If bill is ₹400: 400 = 6x + 100 ⇒ 6x = 300 ⇒ x = 50 units
(iv) Comparing with y = mx + c, y-intercept = 100

Part - 4:

43. Similar triangle construction with scale factor 7/3 > 1 • (OR) ΔPQR construction with base 5 cm, vertical angle 30°, and altitude 4.2 cm.

44. Graph of y = 1/2 x: Constant of variation k = 1/2 (Direct variation).
(i) When x = 9 ⇒ y = 4.5
(ii) When y = 7.5 ⇒ x = 15
(OR)
Prize money distribution: xy = 360 (Inverse variation), k = 360.
When participants x = 12 ⇒ Amount y = 360 / 12 = ₹30.

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