10th Maths Quarterly Exam 2026 Question Paper & Answer Key | Sivagangai 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. The range of the relation R = {(x, x²) | x is a prime number less than 13} is
a) {2, 3, 5, 7}      b) {2, 3, 5, 7, 11}      c) {4, 9, 25, 49, 121}      d) {1, 4, 9, 25, 49, 121}

3. A plane is flying at a speed of 500 km per hour. The function represent the distance 'd' travelled by the plane in t hours is
a) f(d) = 500t      b) f(t) = 500t      c) f(t) = 500d      d) f(d) = 500d

4. If the HCF of 65 and 117 is expressible in the form of 65m − 117, then the value of m is
a) 4      b) 2      c) 1      d) 3

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

6. 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

7. Which of the following should be added to make x⁴ + 64 a perfect square
a) 4x²      b) 16x²      c) 8x²      d) −8x²

8. Graph of a quadratic polynomial is a
a) straight line      b) circle      c) parabola      d) hyperbola

9. 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

10. If in ΔABC, DE || BC, AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm then the length of AE is
a) 1.4 cm      b) 1.8 cm      c) 1.2 cm      d) 1.05 cm

11. 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)

12. The equation of a line passing through the origin and perpendicular to the line 7x − 3y + 4 = 0 is
a) 7x − 3y + 4 = 0      b) 3x − 7y + 4 = 0      c) 3x + 7y = 0      d) 7x − 3y = 0

13. The slope of the straight line 7x − ³/₁₇ = 0 is
a) 7      b) −³/₁₇      c) 0      d) undefined

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

15. A function f is defined by f(x) = 3 − 2x. Find x such that f(x²) = (f(x))².

16. Using the functions f(x) = 3 + x and g(x) = x − 4 find f ∁ g and g ∁ f.

17. Write the following sets in roster form:
i) A = {x ∈ N | 0 < x < 4}
ii) B = {x ∈ W | x < 3}
iii) C = The set of all prime numbers less than 11.

18. 'a' and 'b' are two positive integers such that a^b × b^a = 800. Find a and b.

19. Find the sum of 1 + 3 + 5 + ... + 55.

20. Find the excluded values of the expression t / (t² − 5t + 6).

21. Simplify: [x³ / (x − y)] + [y³ / (y − x)].

22. Solve: x⁴ − 13x² + 42 = 0.

23. If ΔABC is similar to ΔDEF such that BC = 3 cm, EF = 4 cm and area of ΔABC = 54 cm². Find the area of ΔDEF.

24. Find the value of 'a' for which the given points are collinear: (2, 3), (4, a) and (6, −3).

25. Find the equation of a straight line passing through (−8, 4) and making equal intercepts on the coordinate axes.

26. Check whether the given lines are parallel or perpendicular: (x / 3) + (y / 4) + ¹/₇ = 0 and (2x / 3) + (y / 2) + ¹/₁₀ = 0.

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

28. [Compulsory] Find the sum of n terms of the series: 9 + 99 + 999 + ...

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 f: A → B be a function defined by f(x) = (x / 2) − 1 where A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}. Represent f by (i) set of ordered pairs (ii) a table (iii) an arrow diagram (iv) a graph.

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

33. Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a G.P.

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. Find the GCD of the given polynomials: 3x³ + 3x² + 3x + 3 and 6x³ + 12x² + 6x + 12.

36. Find the square root of the polynomial by division method: x⁴ − 12x³ + 42x² − 36x + 9.

37. If α, β are the roots of the equation 2x² − 3x + 1 = 0 then form the equation whose roots are 1 / α² and 1 / β².

38. State and prove angle bisector theorem.

39. In ΔABC if DE || BC, AD = x, DB = x − 2, AE = x + 2 and EC = x − 1 then find the length of the sides AB and AC.

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

41. Let A(3, −4), B(9, −4), C(5, −7) and D(7, −7). Show that ABCD is a trapezium.

42. [Compulsory] Find the equation of a straight line perpendicular to 2x − 3y + 6 = 0 and passes through the midpoint of the line segment joining (−4, −7) and (6, 5).

43. a) Construct a triangle similar to a given triangle ABC with its sides equal to ⁶/₅ of the corresponding sides of the triangle ABC (scale factor ⁶/₅ > 1).
(OR) b) Draw a triangle ABC of base BC = 8 cm, ∠A = 60° and the bisector of ∠A meets BC at D such that BD = 6 cm.

1. b) 2

2. c) {4, 9, 25, 49, 121}

3. b) f(t) = 500t

4. b) 2

5. c) 14280

6. b) 5

7. b) 16x²

8. c) parabola

9. d) 5√2 cm

10. a) 1.4 cm

11. b) parallel to Y axis

12. c) 3x + 7y = 0

13. d) undefined (slope = −a/b = −7/0 = ∞)

14. d) cot θ

15. f(x) = 3 − 2x
f(x²) = (f(x))² ⇒ 3 − 2x² = (3 − 2x)²
3 − 2x² = 9 + 4x² − 12x
6x² − 12x + 6 = 0 ⇒ x² − 2x + 1 = 0
(x − 1)² = 0 ⇒ x = 1

16. f(x) = 3 + x, g(x) = x − 4
f ∁ g(x) = f(x − 4) = 3 + x − 4 = x − 1
g ∁ f(x) = g(3 + x) = 3 + x − 4 = x − 1
Hence, f ∁ g = g ∁ f

17. Roster form:
i) A = {x ∈ N | 0 < x < 4} = {1, 2, 3}
ii) B = {x ∈ W | x < 3} = {0, 1, 2}
iii) C = Prime numbers less than 11 = {2, 3, 5, 7}

18. a^b × b^a = 800
800 = 2⁵ × 5²
Comparing exponents and bases: a = 2, b = 5 or a = 5, b = 2

19. Sum of first n odd numbers: 1 + 3 + 5 + ... + l = [(l + 1) / 2]²
Here l = 55
Sum = [(55 + 1) / 2]² = (56 / 2)² = (28)² = 784

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

21. [x³ / (x − y)] + [y³ / (y − x)] = [x³ / (x − y)] − [y³ / (x − y)]
= (x³ − y³) / (x − y) = [(x − y)(x² + xy + y²)] / (x − y) = x² + xy + y²

22. Let x² = a
a² − 13a + 42 = 0 ⇒ (a − 7)(a − 6) = 0
a = 7 ⇒ x² = 7 ⇒ x = ±√7
a = 6 ⇒ x² = 6 ⇒ x = ±√6

23. ΔABC ~ ΔDEF
Area(ΔABC) / Area(ΔDEF) = BC² / EF²
54 / Area(ΔDEF) = 3² / 4² = 9 / 16
Area(ΔDEF) = (54 × 16) / 9 = 6 × 16 = 96 cm²

24. Points (2, 3), (4, a), (6, −3) are collinear ⇒ Area of triangle = 0
¹/₂ [(2a − 12 + 18) − (12 + 6a − 6)] = 0
(2a + 6) − (6a + 6) = 0 ⇒ −4a = 0 ⇒ a = 0

25. Equal intercepts: a = b
Intercept form: (x / a) + (y / a) = 1 ⇒ x + y = a
Passes through (−8, 4) ⇒ −8 + 4 = a ⇒ a = −4
Equation: x + y = −4 ⇒ x + y + 4 = 0

26. Line 1: (x / 3) + (y / 4) + ¹/₇ = 0 ⇒ m₁ = −a / b = −(¹/₃) / (¹/₄) = −⁴/₃
Line 2: (2x / 3) + (y / 2) + ¹/₁₀ = 0 ⇒ m₂ = −(²/₃) / (¹/₂) = −⁴/₃
Since m₁ = m₂, the given lines are parallel.

27. LHS = √[ ((1 + cos θ)(1 + cos θ)) / ((1 − cos θ)(1 + cos θ)) ] = √[ (1 + cos θ)² / (1 − cos²θ) ]
= √[ (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. S_n = 9 + 99 + 999 + ... to n terms
= (10 − 1) + (100 − 1) + (1000 − 1) + ... to n terms
= (10 + 10² + 10³ + ... + 10ⁿ) − n
= [ {10(10ⁿ − 1)} / (10 − 1) ] − n = [ {10(10ⁿ − 1)} / 9 ] − n

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.
(i) f(4) = 3(4) − 2 = 12 − 2 = 10
(ii) f(−2) = (−2)² − 2 = 4 − 2 = 2
(iii) f(1) = 1² − 2 = −1 ⇒ f(4) + 2f(1) = 10 + 2(−1) = 10 − 2 = 8

32. Numbers divisible by 7 between 300 and 600: 301, 308, 315, ..., 595
a = 301, d = 7, l = 595
n = [(l − a) / d] + 1 = [(595 − 301) / 7] + 1 = (294 / 7) + 1 = 42 + 1 = 43
S₄₃ = (43 / 2) [a + l] = (43 / 2) [301 + 595] = (43 / 2) × 896 = 43 × 448 = 19264

33. Consecutive terms of G.P: t₂ / t₁ = t₃ / t₂
(x + 12) / (x + 6) = (x + 15) / (x + 12)
(x + 12)² = (x + 6)(x + 15)
x² + 24x + 144 = x² + 21x + 90
3x = 90 − 144 = −54 ⇒ x = −18

34. Total Area = 10² + 11² + 12² + ... + 24²
= (1² + 2² + ... + 24²) − (1² + 2² + ... + 9²)
= [(24 × 25 × 49) / 6] − [(9 × 10 × 19) / 6] = 4900 − 285 = 4615 cm²

35. f(x) = 3x³ + 3x² + 3x + 3 = 3(x³ + x² + x + 1)
g(x) = 6x³ + 12x² + 6x + 12 = 6(x³ + 2x² + x + 2)
GCD of coefficients (3, 6) = 3
Dividing (x³ + 2x² + x + 2) by (x³ + x² + x + 1) gives remainder x² + 1
Dividing (x³ + x² + x + 1) by (x² + 1) gives remainder 0
GCD = 3(x² + 1) = 3x² + 3

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

37. 2x² − 3x + 1 = 0 ⇒ α + β = ³/₂, αβ = ¹/₂
Sum of roots = (1 / α²) + (1 / β²) = (α² + β²) / (αβ)² = [(α + β)² − 2αβ] / (αβ)²
= [ (³/₂)² − 2(¹/₂) ] / (¹/₂)² = (⁹/₄ − 1) / (¹/₄) = (⁵/₄) / (¹/₄) = 5
Product of roots = (1 / α²)(1 / β²) = 1 / (αβ)² = 1 / (¹/₂)² = 4
Equation: x² − 5x + 4 = 0

38. Angle Bisector Theorem:
Statement: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.
Proof: In ΔABC, AD is the internal bisector of ∠A. Draw CE || DA meeting extended BA at E. Since ΔABD ~ ΔECD and ΔACE is isosceles (AC = AE), we get AB / AC = BD / CD. Hence proved.

39. In ΔABC, DE || BC ⇒ AD / DB = AE / EC
x / (x − 2) = (x + 2) / (x − 1) ⇒ x(x − 1) = (x − 2)(x + 2)
x² − x = x² − 4 ⇒ −x = −4 ⇒ x = 4
AB = AD + DB = x + x − 2 = 2x − 2 = 2(4) − 2 = 6
AC = AE + EC = x + 2 + x − 1 = 2x + 1 = 2(4) + 1 = 9

40. Area of quadrilateral = ¹/₂ | [(−4k + 6 + 9 − 4) − (6 + 3k − 4 − 12)] | = 28
−4k + 11 − (3k − 10) = 56 ⇒ −7k + 21 = 56 ⇒ −7k = 35 ⇒ k = −5

41. A(3, −4), B(9, −4), C(5, −7), D(7, −7)
Slope of AB = (−4 + 4) / (9 − 3) = 0
Slope of CD = (−7 + 7) / (7 − 5) = 0
Slope of BC = (−7 + 4) / (5 − 9) = ³/₄
Slope of DA = (−4 + 7) / (3 − 7) = −³/₄
Since slope of AB = slope of CD (AB || CD) and slope of BC ≠ slope of DA, ABCD is a trapezium.

42. Midpoint of (−4, −7) and (6, 5): M = [(−4 + 6) / 2, (−7 + 5) / 2] = (1, −1)
Equation of line perpendicular to 2x − 3y + 6 = 0 is 3x + 2y + k = 0
Passes through (1, −1) ⇒ 3(1) + 2(−1) + k = 0 ⇒ 3 − 2 + k = 0 ⇒ k = −1
Required equation: 3x + 2y − 1 = 0

43. a) Similar triangle construction with scale factor ⁶/₅ > 1 • b) ΔABC construction with base BC = 8 cm, vertical angle 60°, and internal bisector meeting BC at BD = 6 cm.