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

QUARTERLY EXAM - SEP 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. 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

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).

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).

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).

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)	2	4	6	8	10
Amount for each participant in ₹ (y)	180	90	60	45	36

(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.

1. (B) 2   [2^{5 × n(B)} = 1024 = 2¹⁰ ⇒ 5 × n(B) = 10 ⇒ n(B) = 2]

2. (A) 7   [Bijective function ⇒ n(A) = n(B) = 7]

3. (B) (2, −1)   [α(1) + β = 1, α(2) + β = 3 ⇒ α = 2, β = −1]

4. (A) 8   [a₁₈ − a₁₄ = 4d = 32 ⇒ d = 8]

5. (C) 3   [1729 = 7¹ × 13¹ × 19¹ ⇒ Sum of exponents = 1 + 1 + 1 = 3]

6. (C) 7881   [t_n = 1 + (n − 1)4 = 4n − 3; (7881 + 3)/4 = 1971]

7. (B) ¹/₂₇   [Common ratio r = ²/₃; ¹/₁₈ × ²/₃ = ¹/₂₇]

8. (B) 16x²   [(x²)² + 8² + 2(x²)(8) = x⁴ + 64 + 16x² = (x² + 8)²]

9. (A) straight line

10. (D) 5√2 cm   [AB² = 5² + 5² = 50 ⇒ AB = 5√2 cm]

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

12. (B) parallel to Y axis   [Line x = c is vertical, hence parallel to Y-axis]

13. (B) 4x + 10y − 4 = 0   [Dividing by 2 gives 2x + 5y − 2 = 0, which has identical slope −²/₅]

14. (A) 2a   [a² − 1 = 2 sin θ cos θ; b = (sin θ + cos θ)/(sin θ cos θ) = a/(sin θ cos θ); b(a² − 1) = 2a]

15. A × B = {(3, 2), (3, 4), (5, 2), (5, 4)}
A = Set of first coordinates = {3, 5}
B = Set of second coordinates = {2, 4}

16. f(m) = m² + m + 3 for m ∈ N
f(1) = 1² + 1 + 3 = 5
f(2) = 2² + 2 + 3 = 9
f(3) = 3² + 3 + 3 = 15
Since distinct elements in the domain N have distinct images in the co-domain N, f is a one-one function.

17. f(x) = x² − 1,   g(x) = x − 2
g ∁ f(a) = g(f(a)) = g(a² − 1) = a² − 1 − 2 = a² − 3
Given g ∁ f(a) = 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)
Prime factorisations: 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

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

20. 1³ + 2³ + 3³ + ... + k³ = 44100
[ {k(k + 1)} / 2 ]² = 44100
1 + 2 + 3 + ... + k = {k(k + 1)} / 2 = √44100 = 210

21. Expression: t / (t² − 5t + 6)
Denominator: t² − 5t + 6 = (t − 3)(t − 2) = 0
Excluded values are 3, 2

22. Let the number be x and its reciprocal be 1/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
Since Δ < 0, the roots are not real (imaginary).

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 formula: (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 and divide by (1 + sin θ):
= [ (1 − sin θ)(1 + sin θ) ] / [ cos θ(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. Points: A(−a, b), B(3a, −b)
Slope m = (y₂ − y₁) / (x₂ − x₁) = (−b − b) / [3a − (−a)] = −2b / 4a = −b / (2a)

29. A = {1, 2, 3, 4, 5, 6, 7}, B = {2, 3, 5, 7}, C = {2}
B − C = {3, 5, 7}
LHS: A × (B − C) = {(1, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7), ..., (7, 3), (7, 5), (7, 7)} (21 ordered pairs) — (1)
A × B contains 28 pairs; A × C = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}
RHS: (A × B) − (A × C) leaves the exact same 21 pairs — (2)
From (1) and (2), A × (B − C) = (A × B) − (A × C) is verified.

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

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

32. Euclid's division algorithm:
504 = 396(1) + 108
396 = 108(3) + 72
108 = 72(1) + 36
72 = 36(2) + 0 ⇒ HCF(396, 504) = 36
Now, HCF(636, 36):
636 = 36(17) + 24
36 = 24(1) + 12
24 = 12(2) + 0 ⇒ HCF(636, 36) = 12
Therefore, HCF(396, 504, 636) = 12

33. Let three consecutive terms in G.P. be a/r, a, ar.
Product = (a/r)(a)(ar) = 27 ⇒ a³ = 27 ⇒ a = 3
Sum of products in pairs: (a/r)(a) + a(ar) + (a/r)(ar) = ⁵⁷/₂
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 + 12² + ... + 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)
Interchanging first two: 10x + 5y + 20z = 125 ⇒ 2x + y + 4z = 25 — (3)
Solving (1), (2), and (3):
(2) − (1) ⇒ y + 3z = 9 — (4)
(3) − 2(1) ⇒ −y + 2z = 1 — (5)
(4) + (5) ⇒ 5z = 10 ⇒ z = 2 (twenty-rupee notes)
From (4): y + 6 = 9 ⇒ y = 3 (ten-rupee notes)
From (1): x + 3 + 2 = 12 ⇒ x = 7 (five-rupee notes)

36. x⁴ − 12x³ + 42x² − 36x + 9 by long division:
Quotient: x² − 6x + 3
Required square root = |x² − 6x + 3|

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

38. Thales Theorem:
Statement: A straight line drawn parallel to a side of a triangle intersecting the other two sides divides the sides in the same ratio.
Proof: In ΔABC, DE || BC. In ΔADE and ΔABC, ∠ADE = ∠B (corresponding angles), ∠AED = ∠C (corresponding angles), ∠A is common. By AAA similarity, ΔADE ~ ΔABC ⇒ AB / AD = AC / AE ⇒ (AD + DB) / AD = (AE + EC) / AE ⇒ 1 + DB / AD = 1 + EC / AE ⇒ AD / DB = AE / EC. Hence proved.

39. Points taken in counter-clockwise order: (8, 6), (5, 11), (−5, 12), (−4, 3)
Area = ¹/₂ [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) − (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]
= ¹/₂ [(88 + 60 − 15 − 24) − (30 − 55 − 48 + 24)] = ¹/₂ [109 − (−49)] = ¹/₂(158) = 79 sq. units

40. Points: A(−4, 2), B(6, −4)
Midpoint of AB = [(−4 + 6)/2, (2 − 4)/2] = (1, −1)
Slope of AB = (−4 − 2) / (6 − (−4)) = −6 / 10 = −³/₅
Slope of perpendicular bisector m = ⁵/₃
Equation: y − (−1) = ⁵/₃(x − 1) ⇒ 3(y + 1) = 5(x − 1)
3y + 3 = 5x − 5 ⇒ 5x − 3y − 8 = 0

41. cosec θ + cot θ = p — (1)
Since cosec²θ − cot²θ = 1, cosec θ − cot θ = 1/p — (2)
Adding (1) and (2): 2 cosec θ = p + (1/p) = (p² + 1) / p
Subtracting (2) from (1): 2 cot θ = p − (1/p) = (p² − 1) / p
Dividing (2 cot θ) by (2 cosec θ):
cot θ / cosec θ = [ (p² − 1)/p ] / [ (p² + 1)/p ] ⇒ cos θ = (p² − 1) / (p² + 1). Hence proved.

42. Fixed charge = ₹100, Charge per unit = ₹6, Units consumed = x, Total bill = y
i. Equation of electricity bill: y = 6x + 100
ii. Bill for 50 units (x = 50): y = 6(50) + 100 = 300 + 100 = ₹400
iii. If bill is ₹400: 400 = 6x + 100 ⇒ 6x = 300 ⇒ x = 50 units
iv. Comparing with slope-intercept form y = mx + c, y-intercept = 100

43. Similar triangle construction with scale factor ⁷/₃ > 1 • (OR) ΔPQR construction with base QR = 5 cm, ∠P = 30°, and altitude of length 4.2 cm.