10th Maths Quarterly Exam 2026 Question Paper & Answer Key

Accelerate your math revision with the official 10th Maths Quarterly Exam 2026 Question Paper set for English Medium schools across the scenic Tenkasi 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 English-medium 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.

Tsi10M • Tenkasi District Standard 10 Reg. No.: [__________]

Common Quarterly Examination September 2026

MATHEMATICS (ENGLISH MEDIUM)

Time: 3.00 Hrs. Date: 21-09-2026 Marks: 100

PART - I • Answer all the questions: 14 × 1 = 14

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

Answer: c) 3
Solution: n(A) = 2. Since n(A × B) = n(A) × n(B), 6 = 2 × n(B) ⇒ n(B) = 3.

2) Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. A function f : A → B given by f = {(1, 4), (2, 8), (3, 9), (4, 10)} is a
a) Many One function b) Identity function c) One to one function d) Into function

Answer: c) One to one function
Solution: Different elements in A have different images in B, making it a one-to-one function.

3) If g = {(1, 1), (2, 3), (3, 5), (4, 7)} is a function given by g(x) = αx + β then the value of α and β are
a) (−1, 2) b) (2, −1) c) (−1, −2) d) (1, 2)

Answer: b) (2, −1)
Solution: g(1) = α(1) + β = 1 ⇒ α + β = 1. g(2) = α(2) + β = 3 ⇒ 2α + β = 3. Subtracting the two gives α = 2, which gives β = −1.

4) 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

Answer: c) 7881
Solution: t_n = 1 + (n − 1)4 = 4n − 3. Adding 3 to 7881 gives 7884, which is divisible by 4 (n = 1971).

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

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

6) In Euclid's division lemma a = bq + r the remainder is always less than the
a) divisor b) dividend c) quotient d) zero

Answer: a) divisor
Solution: By Euclid's lemma, 0 ≤ r < b, where b is the divisor.

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

Answer: a) x = 1, y = 2, z = 3
Solution: 3z = 9 ⇒ z = 3. −7y + 7(3) = 7 ⇒ −7y = −14 ⇒ y = 2. x + 2 − 3(3) = −6 ⇒ x − 7 = −6 ⇒ x = 1.

8) The solution of (2x − 1)² = 9 is equal to
a) −1 b) 2 c) −1, 2 d) None of these

Answer: c) −1, 2
Solution: 2x − 1 = ±3 ⇒ 2x = 4 ⇒ x = 2, or 2x = −2 ⇒ x = −1.

9) What should be added to make x(x + 14) a perfect square?
a) 14 b) 7 c) √7 d) 49

Answer: d) 49
Solution: x(x + 14) = x² + 14x = x² + 2(x)(7). To complete the square, add 7² = 49 to obtain (x + 7)².

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

Answer: d) 5√2 cm
Solution: Since ΔABC is isosceles right-angled at C, BC = AC = 5 cm. By Pythagoras theorem, AB = √(5² + 5²) = √50 = 5√2 cm.

11) In the adjacent figure ∠BAC = 90° and AD ⊥ BC then
a) BD · CD = BC² b) AB · AC = BC²
c) BD · CD = AD² d) AB · AC = AD²

Answer: c) BD · CD = AD²
Solution: Since ΔDBA ~ ΔDAC, BD / AD = AD / CD ⇒ BD · CD = AD².

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)

Answer: b) parallel to y-axis
Solution: Any equation of the form x = c represents a vertical straight line parallel to the y-axis.

13) 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

Answer: c) 3x + 7y = 0
Solution: The perpendicular family of 7x − 3y + 4 = 0 is 3x + 7y + k = 0. Since it passes through the origin (0, 0), k = 0, giving 3x + 7y = 0.

14) When proving that a quadrilateral is a parallelogram by using slopes you must find?
a) The slopes of two sides
b) The slopes of two pairs of opposite sides
c) The length of all sides
d) Both the lengths and slopes of two sides

Answer: b) The slopes of two pairs of opposite sides
Solution: A quadrilateral is a parallelogram if both pairs of opposite sides are parallel (i.e. having equal slopes).

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

15) If A = {1, 3, 5} and B = {2, 3} then find A × B and B × A.

Solution:
A × B = {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}.
B × A = {(2, 1), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3)}.

16) Let f(x) = x² − 1 find f ∁ f ∁ f.

Solution:
(f ∁ f)(x) = f(x² − 1) = (x² − 1)² − 1 = x⁴ − 2x².
(f ∁ f ∁ f)(x) = f(x⁴ − 2x²) = (x⁴ − 2x²)² − 1.

17) When the positive integers a, b and c are divided by 13, the respective remainders are 9, 7 and 10. Show that a + b + c is divisible by 13.

Solution:
By Euclid's division lemma: a = 13q₁ + 9, b = 13q₂ + 7, c = 13q₃ + 10.
a + b + c = 13(q₁ + q₂ + q₃) + (9 + 7 + 10) = 13(q₁ + q₂ + q₃) + 26 = 13(q₁ + q₂ + q₃ + 2).
Since 13 is a factor, a + b + c is divisible by 13.

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

Solution:
Prime factorisation: 800 = 2⁵ × 5².
Comparing with a^b × b^a gives a = 2, b = 5 (or a = 5, b = 2).

19) Today is Tuesday. My uncle will come after 45 days. On which day will my uncle be coming?

Solution:
Let Tuesday = day 2. 45 ≡ 3 (mod 7).
Day of arrival = 2 + 3 = 5th day of the week, which is Friday.

20) Find the excluded values, if any, of the following expression: (x³ − 27) / (x³ + x² − 6x).

Solution:
Set denominator equal to zero: x³ + x² − 6x = 0 ⇒ x(x² + x − 6) = 0.
x(x + 3)(x − 2) = 0 ⇒ x = 0, −3, 2.
The excluded values are 0, −3, 2.

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

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

22) Determine the nature of roots for the following quadratic equation: 9x² − 24x + 16 = 0.

Solution:
a = 9, b = −24, c = 16.
Discriminant Δ = b² − 4ac = (−24)² − 4(9)(16) = 576 − 576 = 0.
Since Δ = 0, the roots are real and equal.

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.

Solution:
Area(ΔABC) / Area(ΔDEF) = BC² / EF² ⇒ 54 / Area(ΔDEF) = 3² / 4² = 9 / 16.
Area(ΔDEF) = (54 × 16) / 9 = 6 × 16 = 96 cm².

24) In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD / DB = ³/₄ and AC = 15 cm find AE.

Solution:
By Thales Theorem: AD / DB = AE / EC.
Let AE = x, then EC = 15 − x.
³/₄ = x / (15 − x) ⇒ 4x = 45 − 3x ⇒ 7x = 45 ⇒ AE = ⁴⁵/₇ cm ≈ 6.43 cm.

25) Show that the points P(−1.5, 3), Q(6, −2), R(−3, 4) are collinear.

Solution:
Area(ΔPQR) = ¹/₂ | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |
= ¹/₂ | −1.5(−2 − 4) + 6(4 − 3) + (−3)(3 − (−2)) | = ¹/₂ | −1.5(−6) + 6(1) − 3(5) | = ¹/₂ | 9 + 6 − 15 | = 0.
Since area = 0, the points are collinear.

26) Find the slope and y-intercept of √3 x + (1 − √3)y = 3.

Solution:
(1 − √3)y = −√3 x + 3 ⇒ y = [(−√3) / (1 − √3)] x + [3 / (1 − √3)].
Slope m = −√3 / (1 − √3) = √3 / (√3 − 1) = (3 + √3) / 2.
y-intercept c = 3 / (1 − √3) = −3(1 + √3) / 2.

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

Solution:
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) [Compulsory] If 3, x, 6.75 are in Geometric Progression, then find x and common ratio.

Solution:
Since terms are in G.P., x² = 3 × 6.75 = 20.25 ⇒ x = ±4.5.
Common ratio r = x / 3 = ±4.5 / 3 = ±1.5.

PART - III • Answer any 10 questions. (Q.No. 42 is compulsory) 10 × 5 = 50

29) Let A = {x ∈ N | 1 < x < 4}, B = {x ∈ W | 0 ≤ x < 2} and C = {x ∈ N | x < 3} then verify that A × (B ∪ C) = (A × B) ∪ (A × C).

Solution:
A = {2, 3}, B = {0, 1}, C = {1, 2}.
B ∪ C = {0, 1, 2}.
LHS = A × (B ∪ C) = {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)} — (1)
A × B = {(2, 0), (2, 1), (3, 0), (3, 1)}.
A × C = {(2, 1), (2, 2), (3, 1), (3, 2)}.
RHS = (A × B) ∪ (A × C) = {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)} — (2)
From (1) and (2), LHS = RHS. Hence verified.

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) arrow diagram (iv) a graph.

Solution:
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) Table:

x	2	4	6	10	12
f(x)	0	1	2	4	5

(iii) Arrow diagram: Map elements from A to B: 2 → 0, 4 → 1, 6 → 2, 10 → 4, 12 → 5.
(iv) Graph: Plot (2, 0), (4, 1), (6, 2), (10, 4), (12, 5) on the Cartesian plane.

31) If the function f: R → R is defined by
f(x) = 2x + 7, if x < −2
f(x) = x² − 2, if −2 ≤ x < 3
f(x) = 3x − 2, if x ≥ 3
then find the values of (i) f(4) (ii) f(−2) (iii) f(4) + 2f(1) (iv) [f(1) − 3f(4)] / f(−3).

Solution:
(i) f(4) = 3(4) − 2 = 10.
(ii) f(−2) = (−2)² − 2 = 4 − 2 = 2.
(iii) f(1) = 1² − 2 = −1 ⇒ f(4) + 2f(1) = 10 + 2(−1) = 8.
(iv) f(−3) = 2(−3) + 7 = 1 ⇒ [f(1) − 3f(4)] / f(−3) = [−1 − 3(10)] / 1 = −31.

32) Find the greatest number consisting of 6 digits which is exactly divisible by 24, 15, 36?

Solution:
LCM of 24, 15, 36: 24 = 2³ × 3, 15 = 3 × 5, 36 = 2² × 3² ⇒ LCM = 2³ × 3² × 5 = 360.
Greatest 6-digit number is 999999.
999999 ÷ 360 gives quotient 2777 and remainder 279.
Required number = 999999 − 279 = 999720.

33) Find the sum of first 15 terms of the A.P. 8, 7 ¹/₄, 6 ¹/₂, 5 ³/₄, ...

Solution:
a = 8, d = 7 ¹/₄ − 8 = ²⁹/₄ − 8 = −³/₄.
S₁₅ = (15 / 2) [2(8) + (15 − 1)(−³/₄)] = (15 / 2) [16 + 14(−³/₄)]
= (15 / 2) [16 − ²¹/₂] = (15 / 2) [¹¹/₂] = ¹⁶⁵/₄ = 41 ¹/₄.

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?

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

35) Find the values of m and n if the following polynomial is a perfect square: 36x⁴ − 60x³ + 61x² − mx + n.

Solution:
Square root of 36x⁴ − 60x³ + 61x² − mx + n by division method:
Quotient is 6x² − 5x + 3.
(6x² − 5x + 3)² = 36x⁴ − 60x³ + 61x² − 30x + 9.
Comparing like terms: m = 30, n = 9.

36) A passenger train takes 1 hour more than an express train to travel a distance of 240 km from Chennai to Virudhachalam. The speed of the express train is more than that of the passenger train by 20 km per hour. Find the average speed of both the trains.

Solution:
Let the speed of the passenger train be s km/hr. Then express train speed is (s + 20) km/hr.
(240 / s) − [240 / (s + 20)] = 1 ⇒ 240 [ (s + 20 − s) / (s(s + 20)) ] = 1
4800 / (s² + 20s) = 1 ⇒ s² + 20s − 4800 = 0.
(s + 80)(s − 60) = 0 ⇒ s = 60 (since speed > 0).
Speed of passenger train = 60 km/hr.
Speed of express train = 80 km/hr.

37) If α, β are the roots of 7x² + ax + 2 = 0 and if β − α = −¹³/₇, find the values of a.

Solution:
α + β = −a / 7, αβ = ²/₇.
Identity: (β − α)² = (α + β)² − 4αβ.
(−¹³/₇)² = (−a / 7)² − 4(²/₇) ⇒ ¹⁶⁹/₄₉ = (a² / 49) − ⁸/₇ = (a² − 56) / 49.
a² − 56 = 169 ⇒ a² = 225 ⇒ a = ±15.

38) State and demonstrate Thales theorem.

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

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

Solution:
Vertices 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) Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x − 4y = 1 and parallel to the line 13x + 5y + 12 = 0.

Solution:
Solving 7x + 3y = 10 — (1) and 5x − 4y = 1 — (2):
4 × (1) + 3 × (2) ⇒ 28x + 12y + 15x − 12y = 40 + 3 ⇒ 43x = 43 ⇒ x = 1.
From (1): 7(1) + 3y = 10 ⇒ 3y = 3 ⇒ y = 1. Intersection point is (1, 1).
Line parallel to 13x + 5y + 12 = 0 is of the form 13x + 5y + k = 0.
Passing through (1, 1): 13(1) + 5(1) + k = 0 ⇒ k = −18.
Required equation: 13x + 5y − 18 = 0.

41) Prove that [sin A / (1 + cos A)] + [sin A / (1 − cos A)] = 2 cosec A.

Solution:
LHS = sin A [ (1 / (1 + cos A)) + (1 / (1 − cos A)) ]
= sin A [ (1 − cos A + 1 + cos A) / (1 − cos²A) ]
= sin A [ 2 / sin²A ] = 2 / sin A = 2 cosec A = RHS. Hence proved.

42) [Compulsory] A line makes positive intercepts on coordinate axes whose sum is 7 and it passes through (−3, 8). Find its equation.

Solution:
Let the intercepts be a and b. Given a + b = 7 ⇒ b = 7 − a.
Equation: (x / a) + [y / (7 − a)] = 1.
Since it passes through (−3, 8): (−3 / a) + [8 / (7 − a)] = 1 ⇒ −3(7 − a) + 8a = a(7 − a).
−21 + 3a + 8a = 7a − a² ⇒ a² + 4a − 21 = 0 ⇒ (a + 7)(a − 3) = 0.
Since intercepts are positive, a = 3 ⇒ b = 7 − 3 = 4.
Equation: (x / 3) + (y / 4) = 1 ⇒ 4x + 3y − 12 = 0.

PART - IV • Answer all the questions: 2 × 8 = 16

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

Solution Outline:
a) 1. Draw base QR = 5 cm. 2. At Q, draw ray QE making ∠RQE = 30° downwards. Draw ray QF ⊥ QE. 3. Draw perpendicular bisector of QR meeting QF at centre O and QR at G. 4. Draw circle with centre O and radius OQ. 5. On perpendicular bisector, mark 4.2 cm from G and draw parallel line to QR intersecting circle at P. 6. Join PQ and PR to complete ΔPQR.
b) 1. Draw ΔABC. 2. Draw acute ray BX downwards from BC. Mark 6 equal segments B₁ to B₆ on BX. 3. Join B₅ to C. 4. Through B₆, draw line parallel to B₅C meeting extended BC at C'. 5. Through C', draw line parallel to CA meeting extended BA at A'. ΔA'BC' is the required triangle.

44) a) 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) b) A school announces that for a certain competition, 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 Rs. (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.

Solution:
a) Direct variation: y = kx. Constant of variation: k = ¹/₂ = 0.5.
(i) When x = 9: y = ¹/₂(9) = 4.5.
(ii) When y = 7.5: x = 2(7.5) = 15.
b) As x increases, y decreases. Inverse variation: xy = k.
i) Constant of variation: k = 2 × 180 = 360.
ii) When x = 12 participants: y = 360 / 12 = Rs. 30.