10th Maths Quarterly Exam 2026 Question Paper & Answer Key | Perambalur District (EM)
Accelerate your math revision with the official 10th Maths Quarterly Exam 2026 Question Paper set for schools across the Perambalur District. This district-level terminal evaluation serves as an invaluable diagnostic tool for Tamil Nadu Samacheer Kalvi state board students to measure their accuracy, problem-solving speed, and analytical reasoning under strict exam conditions. The paper offers balanced coverage of early high-yield chapters, including Relations and Functions, Arithmetic and Geometric Progressions, polynomial Algebra matrices, and critical Geometry theorems. Reviewing this paper alongside the comprehensive answer keys enables you to break down complex 5-mark proofs, eliminate simple algebraic errors, and perfect the structured step-by-step presentation format required to secure a perfect centum score on your final board examinations.
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COMMON QUARTERLY EXAMINATION - 2026
1. If A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then n[(A ∪ C) × B] is
a) 8 b) 12 c) 16 d) 20
2. If A and B are finite sets such that n(A) = p, n(B) = q then the total number of functions that exist from A to B is
a) 2pq b) 2pq − 1 c) qp d) pq
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. 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. 74k ≡ _______ (mod 100)
a) 1 b) 2 c) 3 d) 4
6. A = 265 and B = 264 + 263 + 262 + ... + 20. Which of the following is true?
a) B is 264 larger than A b) A and B are equal
c) B is larger than A by 1 d) A is larger than B by 1
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. The solution of (2x − 1)2 = 9 is equal to
a) −1 b) 2 c) −1, 2 d) none of these
9. Graph of a quadratic polynomial is a
a) straight line b) circle c) parabola d) hyperbola
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. A man walks near a wall such that the distance between him and the wall is 10 units. Consider the wall to be the y-axis. The path travelled by the man is
a) x = 10 b) y = 10 c) x = 0 d) y = 0
12. If the slope of PQ is 1 / √3 then the slope of the perpendicular bisector of PQ is
a) √3 b) −√3 c) 1 / √3 d) 0
13. (2, 1) is the point of intersection of two lines
a) x − y − 3 = 0 ; 3x − y − 7 = 0
b) x + y = 3 ; 3x + y = 7
c) 3x + y = 3 ; x + y = 7
d) x + 3y − 3 = 0 ; x − y − 7 = 0
14. If sin θ = cos θ, then 2 tan2θ + sin2θ − 1 is equal to
a) −3/2 b) 3/2 c) 2/3 d) −2/3
15. A relation R is given by the set {(x, y) | y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range.
16. Let f(x) = 2x + 5. If x ≠ 0 then find [f(x + 2) − f(2)] / x.
17. Find k if f ∁ f(k) = 5, where f(k) = 2k − 1.
18. Use Euclid's division Algorithm to find the Highest Common Factor (HCF) of 340 and 412.
19. If 3 + k, 18 − k, 5k + 1 are in A.P., then find k.
20. Find the sum to infinity of 9 + 3 + 1 + ...
21. Find the excluded values of (7p + 2) / (8p2 + 13p + 5).
22. Pari needs 4 hours to complete a work. His friend Yuvan needs 6 hours to complete the same work. How long will it take to complete if they work together?
23. Determine the nature of roots of x2 − x − 20 = 0.
24. 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.
25. Calculate the slope and y-intercept of the straight line 8x − 7y + 6 = 0.
26. Show that the straight lines 5x + 23y + 14 = 0 and 23x − 5y + 9 = 0 are perpendicular.
27. Prove that √[(1 + cos θ) / (1 − cos θ)] = cosec θ + cot θ.
28. [Compulsory] A car registers its positions at three different locations A(−3, −4), B(7, 2) and C(12, 5). Based on these coordinates, determine whether the car moves in a straight line.
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} and 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.
31. If f(x) = x − 1, g(x) = 3x + 1 and h(x) = x2 prove that f ∁ (g ∁ h) = (f ∁ g) ∁ h.
32. The houses of a street are numbered from 1 to 49. Senthil's house is numbered such that the sum of numbers of the houses prior to Senthil's house is equal to the sum of numbers of the houses following Senthil's house. Find Senthil's house number.
33. Find the sum to n terms of the series 3 + 33 + 333 + ...
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 polynomials 3x4 + 6x3 − 12x2 − 24x and 4x4 + 14x3 + 8x2 − 8x.
36. If 36x4 − 60x3 + 61x2 − mx + n is a perfect square, find the values of m and n.
37. State and prove the Basic Proportionality Theorem.
38. Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are taken in the order (−4, −2), (−3, k), (3, −2) and (2, 3).
39. A line makes positive intercepts on coordinate axes whose sum is 7 and it passes through (−3, 8). Find its equation.
40. Find the equation of the perpendicular bisector of the line joining the points A(−4, 2) and B(6, −4).
41. If cos α / cos β = m and cos α / sin β = n, then prove that (m2 + n2)cos2β = n2.
42. [Compulsory] If α, β are the roots of the equation 3x2 + 7x − 2 = 0, find the values of:
i) (α / β) + (β / α) ii) (α2 / β) + (β2 / α)
43. a) Construct a triangle similar to a given triangle LMN with its sides equal to 4/5 of the corresponding sides of the triangle LMN (scale factor 4/5 < 1).
(OR)
b) Construct a ΔPQR with the base PQ = 4.5 cm, ∠R = 35° and the median RG from R to PQ is 6 cm.
i) The constant of variation
ii) How far will it travel in 90 minutes?
iii) The required time to cover a distance of 300 km from the graph.
(OR) b) Draw the graph of xy = 24, x, y > 0. Using the graph find:
i) y when x = 3 and
ii) find x when y = 6.
1. b) 12 [A ∪ C = {a, b, p, q, r, s} ⇒ n(A ∪ C) = 6; n(B) = 2 ⇒ 6 × 2 = 12]
2. c) qp [Number of functions from set of size p to set of size q is n(B)n(A) = qp]
3. b) (2, −1) [α(1) + β = 1, α(2) + β = 3 ⇒ α = 2, β = −1]
4. b) 2 [HCF(65, 117) = 13; 65m − 117 = 13 ⇒ 65m = 130 ⇒ m = 2]
5. a) 1 [74 = 2401 ≡ 1 (mod 100) ⇒ (74)k ≡ 1k ≡ 1 (mod 100)]
6. d) A is larger than B by 1 [B is sum of G.P: 20 + 21 + ... + 264 = 265 − 1 ⇒ A − B = 1]
7. a) x = 1, y = 2, z = 3 [3z = 9 ⇒ z = 3; −7y + 21 = 7 ⇒ y = 2; x + 2 − 9 = −6 ⇒ x = 1]
8. c) −1, 2 [2x − 1 = ±3 ⇒ 2x = 4 ⇒ x = 2 or 2x = −2 ⇒ x = −1]
9. c) parabola
10. b) 70° [∠N = 180° − (60° + 50°) = 70° ⇒ ∠R = ∠N = 70°]
11. a) x = 10 [Distance from y-axis is the x-coordinate: x = 10]
12. b) −√3 [m1 · m2 = −1 ⇒ m2 = −1 / (1/√3) = −√3]
13. b) x + y = 3 ; 3x + y = 7 [2 + 1 = 3 and 3(2) + 1 = 7]
14. b) 3/2 [sin θ = cos θ ⇒ θ = 45°; 2 tan2(45°) + sin2(45°) − 1 = 2(1) + 1/2 − 1 = 3/2]
15. y = x + 3 for x ∈ {0, 1, 2, 3, 4, 5}.
Pairs: (0, 3), (1, 4), (2, 5), (3, 6), (4, 7), (5, 8).
Domain = {0, 1, 2, 3, 4, 5}
Range = {3, 4, 5, 6, 7, 8}
16. f(x) = 2x + 5
f(x + 2) = 2(x + 2) + 5 = 2x + 9
f(2) = 2(2) + 5 = 9
[f(x + 2) − f(2)] / x = [(2x + 9) − 9] / x = 2x / x = 2
17. f(k) = 2k − 1
f ∁ f(k) = f(2k − 1) = 2(2k − 1) − 1 = 4k − 3
Given f ∁ f(k) = 5 ⇒ 4k − 3 = 5 ⇒ 4k = 8 ⇒ k = 2
18. By Euclid's division algorithm:
412 = 340 × 1 + 72
340 = 72 × 4 + 52
72 = 52 × 1 + 20
52 = 20 × 2 + 12
20 = 12 × 1 + 8
12 = 8 × 1 + 4
8 = 4 × 2 + 0
Therefore, HCF(340, 412) = 4
19. Since 3 + k, 18 − k, 5k + 1 are in A.P:
(18 − k) − (3 + k) = (5k + 1) − (18 − k)
15 − 2k = 6k − 17 ⇒ 8k = 32 ⇒ k = 4
20. 9 + 3 + 1 + ... is an infinite G.P.
a = 9, r = 3 / 9 = 1/3 (−1 < r < 1)
S∞ = a / (1 − r) = 9 / (1 − 1/3) = 9 / (2/3) = 27/2 (or 13.5)
21. Setting denominator to zero: 8p2 + 13p + 5 = 0
(p + 1)(8p + 5) = 0 ⇒ p = −1 or p = −5/8
Excluded values are −1, −5/8
22. In 1 hour, Pari completes 1/4 of the work and Yuvan completes 1/6 of the work.
Together in 1 hour = 1/4 + 1/6 = (3 + 2) / 12 = 5/12 of the work.
Time taken together = 12/5 hours = 2.4 hours (2 hours 24 minutes)
23. x2 − x − 20 = 0 ⇒ a = 1, b = −1, c = −20
Δ = b2 − 4ac = (−1)2 − 4(1)(−20) = 1 + 80 = 81 > 0
Since Δ > 0 and a perfect square, roots are real and unequal (rational).
24. Area(ΔABC) / Area(ΔDEF) = BC2 / EF2
54 / Area(ΔDEF) = 32 / 42 = 9 / 16
Area(ΔDEF) = (54 × 16) / 9 = 96 cm2
25. Line: 8x − 7y + 6 = 0 ⇒ 7y = 8x + 6 ⇒ y = (8/7)x + 6/7
Comparing with y = mx + c:
Slope m = 8/7, y-intercept c = 6/7
26. Slope of line 1 (5x + 23y + 14 = 0): m1 = −5/23
Slope of line 2 (23x − 5y + 9 = 0): m2 = −23 / (−5) = 23/5
m1 × m2 = (−5/23) × (23/5) = −1
Since the product of slopes is −1, the lines are perpendicular.
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. Points: A(−3, −4), B(7, 2), C(12, 5)
Slope of AB = [2 − (−4)] / [7 − (−3)] = 6 / 10 = 3/5
Slope of BC = (5 − 2) / (12 − 7) = 3 / 5 = 3/5
Since Slope of AB = Slope of BC and point B is common, the points are collinear.
Therefore, the car moves in a straight line.
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.
f(2) = 0, f(4) = 1, f(6) = 2, f(10) = 4, f(12) = 5
(i) Set of ordered pairs: f = {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
(ii) A table:
| x | 2 | 4 | 6 | 10 | 12 |
| f(x) | 0 | 1 | 2 | 4 | 5 |
(iv) A graph: Points (2, 0), (4, 1), (6, 2), (10, 4), (12, 5) plotted on axes.
31. f(x) = x − 1, g(x) = 3x + 1, h(x) = x2
(g ∁ h)(x) = g(x2) = 3x2 + 1
f ∁ (g ∁ h)(x) = f(3x2 + 1) = (3x2 + 1) − 1 = 3x2 — (1)
(f ∁ g)(x) = f(3x + 1) = (3x + 1) − 1 = 3x
((f ∁ g) ∁ h)(x) = 3(x2) = 3x2 — (2)
From (1) and (2), f ∁ (g ∁ h) = (f ∁ g) ∁ h is proved.
32. Let Senthil's house number be x.
1 + 2 + ... + (x − 1) = (x + 1) + (x + 2) + ... + 49
Sx−1 = S49 − Sx
[(x − 1)x] / 2 = [49 × 50] / 2 − [x(x + 1)] / 2
x2 − x + x2 + x = 2450 ⇒ 2x2 = 2450 ⇒ x2 = 1225 ⇒ x = 35
33. Sn = 3 + 33 + 333 + ... to n terms
= 3/9 (9 + 99 + 999 + ... to n terms) = 1/3 [(10 − 1) + (102 − 1) + ... + (10n − 1)]
= 1/3 [ {10(10n − 1) / 9} − n ] = 10/27 (10n − 1) − n/3
34. Area = 102 + 112 + ... + 242 = Σ124 k2 − Σ19 k2
= [(24 × 25 × 49) / 6] − [(9 × 10 × 19) / 6] = 4900 − 285 = 4615 cm2
35. Let f(x) = 3x(x3 + 2x2 − 4x − 8) and g(x) = 2x(2x3 + 7x2 + 4x − 4).
GCD of leading factors (3x, 2x) = x.
Dividing (2x3 + 7x2 + 4x − 4) by (x3 + 2x2 − 4x − 8) gives remainder 3x2 + 12x + 12 = 3(x2 + 4x + 4).
Dividing (x3 + 2x2 − 4x − 8) by (x2 + 4x + 4) gives quotient (x − 2) and remainder 0.
Therefore, GCD = x(x2 + 4x + 4) = x(x + 2)2
36. Square root of 36x4 − 60x3 + 61x2 − mx + n by division method:
Quotient: 6x2 − 5x + 3
(6x2 − 5x + 3)2 = 36x4 − 60x3 + 61x2 − 30x + 9
Comparing terms: m = 30, n = 9
37. Basic Proportionality Theorem (Thales Theorem): Statement and Proof (Refer standard textbook).
38. Vertices: (−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. Intercepts: a + b = 7 ⇒ b = 7 − a
(x / a) + [y / (7 − a)] = 1, passing through (−3, 8)
−3(7 − a) + 8a = a(7 − a) ⇒ a2 + 4a − 21 = 0 ⇒ (a + 7)(a − 3) = 0
Since intercepts are positive, a = 3 ⇒ b = 4
Equation: (x / 3) + (y / 4) = 1 ⇒ 4x + 3y − 12 = 0
40. Points: A(−4, 2) and B(6, −4)
Midpoint = [(−4 + 6)/2, (2 − 4)/2] = (1, −1)
Slope of AB = (−4 − 2) / (6 + 4) = −3/5 ⇒ Perpendicular slope = 5/3
Equation: y + 1 = 5/3(x − 1) ⇒ 5x − 3y − 8 = 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 = (cos α / sin β)2 = n2 = RHS. Hence proved.
42. 3x2 + 7x − 2 = 0 ⇒ α + β = −7/3, αβ = −2/3
i) (α / β) + (β / α) = [(α + β)2 − 2αβ] / αβ = [(−7/3)2 − 2(−2/3)] / (−2/3) = −61/6
ii) (α2 / β) + (β2 / α) = [(α + β)3 − 3αβ(α + β)] / αβ = [(−7/3)3 − 3(−2/3)(−7/3)] / (−2/3) = 469/18
43. a) Similar triangle construction with scale factor 4/5 < 1 • b) ΔPQR construction with base PQ = 4.5 cm, vertical angle 35°, and median RG = 6 cm.
i) Constant of variation: k = y / x = 50 / 60 = 5/6
ii) Distance in 90 minutes: y = (5/6) × 90 = 75 km
iii) Time required to cover 300 km: x = 300 × (6/5) = 360 minutes (6 hours)
(OR) 44. b) Graph of xy = 24 (Inverse variation):
i) When x = 3 ⇒ y = 24 / 3 = 8
ii) When y = 6 ⇒ x = 24 / 6 = 4
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