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

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Boost your exam preparation with the official 10th Standard Maths Quarterly Exam 2026 Question Paper and Answer Key specifically for Virudhunagar District (English Medium). This comprehensive resource includes the actual exam paper alongside verified solutions designed to help you analyze your performance. Practising with this district-level paper is an excellent way for English medium students to master core mathematical concepts, improve time management, and build confidence for the upcoming board exams.

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

VNR10M • Virudhunagar District Standard 10 Reg. No.: [_________]

Common Quarterly Exam, September - 2026

MATHEMATICS (ENGLISH MEDIUM)

Time: 3.00 Hrs. Marks: 100

Part - 1 • I. Choose the best answer: 14 × 1 = 14

1) If the ordered pairs (a + 2, 4) and (5, 2a + b) are equal then (a, b) is
a) (2, −2) b) (5, 1) c) (2, 3) d) (3, −2)

Answer: d) (3, −2)
Solution: Equating corresponding coordinates: a + 2 = 5 ⇒ a = 3. 2a + b = 4 ⇒ 2(3) + b = 4 ⇒ 6 + b = 4 ⇒ b = −2. Hence (a, b) = (3, −2).

2) If n(A) = p, n(B) = q, then the total number of relations that exist from A to B is
a) 2pq b) 2^pq − 1 c) p^q d) 2^pq

Answer: d) 2^pq
Solution: The number of elements in A × B is pq. Every relation is a subset of A × B, so total relations = 2^{n(A × B)} = 2^pq.

3) f(x) = (x + 1)³ − (x − 1)³ represents a function which is
a) linear b) cubic c) reciprocal d) quadratic

Answer: d) quadratic
Solution: (x³ + 3x² + 3x + 1) − (x³ − 3x² + 3x − 1) = 6x² + 2. Since the degree is 2, it is a quadratic function.

4) The sum of the exponents of the prime factors in the prime factorisation of 1729 is
a) 1 b) 2 c) 3 d) 4

Answer: c) 3
Solution: Prime factorization: 1729 = 7¹ × 13¹ × 19¹. Sum of exponents = 1 + 1 + 1 = 3.

5) 7^4k ≡ _______ (mod 100)
a) 1 b) 2 c) 3 d) 4

Answer: a) 1
Solution: 7⁴ = 2401 ≡ 1 (mod 100). Therefore, (7⁴)^k ≡ 1^k ≡ 1 (mod 100).

6) If 1 + 2 + 3 + ... + n = 15 then find the value of 1³ + 2³ + ... + n³.
a) 125 b) 625 c) 225 d) 3375

Answer: c) 225
Solution: Identity: 1³ + 2³ + ... + n³ = (1 + 2 + ... + n)² = 15² = 225.

7) The square root of (256 x⁸ y⁴ z¹⁰) / (25 x⁶ y⁶ z⁶) is equal to
a) ¹⁶/₅ |(x² z⁴) / y²| b) 16 |y² / (x² z²)| c) ¹⁶/₅ |(x z²) / y| d) ¹⁶/₅ |y / (x z²)|

Answer: c) ¹⁶/₅ |(x z²) / y|
Solution: Simplifying inside square root: (256/25) · (x⁸⁻⁶ z¹⁰⁻⁶ / y⁶⁻⁴) = (256/25)(x² z⁴ / y²). Taking square root gives ¹⁶/₅ |(x z²) / y|.

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) Graph of a linear equation is a
a) straight line b) circle c) parabola d) hyperbola

Answer: a) straight line
Solution: Every linear equation of first degree in two variables represents a straight line.

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 an isosceles right-angled triangle at C, BC = AC = 5 cm. By Pythagoras theorem, AB = √(5² + 5²) = √50 = 5√2 cm.

11) 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

Answer: a) 1.4 cm
Solution: By Thales theorem: AD / AB = AE / AC ⇒ 2.1 / 3.6 = AE / 2.4 ⇒ AE = (2.1 × 2.4) / 3.6 = 1.4 cm.

12) The slope of the line which is perpendicular to a line joining the points (1, 1) and (0, 5) is
a) −4 b) −¹/₄ c) ¹/₄ d) 4

Answer: c) ¹/₄
Solution: Slope of line joining (1, 1) and (0, 5) is m₁ = (5 − 1) / (0 − 1) = −4. Slope of perpendicular line m₂ = −1 / m₁ = −1 / (−4) = ¹/₄.

13) If (5, 7), (3, P) and (6, 6) are collinear, then the value of P is
a) 3 b) 6 c) 9 d) 12

Answer: c) 9
Solution: Slope equality: (P − 7) / (3 − 5) = (6 − 7) / (6 − 5) ⇒ (P − 7) / (−2) = −1 ⇒ P − 7 = 2 ⇒ P = 9.

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

Answer: d) cot θ
Solution: tan θ (cosec²θ − 1) = tan θ · cot²θ = (1 / cot θ) · cot²θ = cot θ.

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

15) A plane is flying at a speed of 500 km per hour. Express the distance 'd' travelled by the plane as function of time t in hours.

Solution:
Distance = Speed × Time.
Speed = 500 km/hr, Time = t hours.
Function: d(t) = 500t.

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

Solution:
Domain N = {1, 2, 3, ...}.
f(1) = 1² + 1 + 3 = 5;
f(2) = 2² + 2 + 3 = 9;
f(3) = 3² + 3 + 3 = 15.
Since distinct elements in domain N have distinct images in co-domain N, f is a one-one function.

17) Find K if f ∁ f(K) = 5 where f(K) = 2K − 1.

Solution:
f(f(K)) = 5 ⇒ f(2K − 1) = 5 ⇒ 2(2K − 1) − 1 = 5.
4K − 2 − 1 = 5 ⇒ 4K − 3 = 5 ⇒ 4K = 8 ⇒ K = 2.

18) Compute x, such that 10⁴ ≡ x (mod 19).

Solution:
10² = 100 = 19 × 5 + 5 ⇒ 10² ≡ 5 (mod 19).
10⁴ = (10²)² ≡ 5² ≡ 25 ≡ 6 (mod 19).
Hence, x = 6.

19) If p² × q¹ × r⁴ × s³ = 3,15,000 then find p, q, r and s.

Solution:
Prime factorisation of 315000:
315000 = 315 × 1000 = (5 × 7 × 9) × (2³ × 5³) = 2³ × 3² × 5⁴ × 7¹.
Comparing powers with p² × q¹ × r⁴ × s³:
Base of power 2: p = 3.
Base of power 1: q = 7.
Base of power 4: r = 5.
Base of power 3: s = 2.

20) Find the sum of infinity of 9 + 3 + 1 + ...

Solution:
First term a = 9, common ratio r = 3 / 9 = ¹/₃.
S_∞ = a / (1 − r) = 9 / (1 − ¹/₃) = 9 / (²/₃) = ²⁷/₂ = 13.5.

21) Reduce the rational expression to its lowest form: (x² − 16) / (x² + 8x + 16).

Solution:
(x² − 16) / (x² + 8x + 16) = [(x − 4)(x + 4)] / (x + 4)² = (x − 4) / (x + 4).

22) Solve the following quadratic equation by factorization method: √[a(a − 7)] = 3√2.

Solution:
Squaring both sides: a(a − 7) = (3√2)² ⇒ a² − 7a = 18.
a² − 7a − 18 = 0 ⇒ (a − 9)(a + 2) = 0 ⇒ a = 9 or a = −2.

23) In the figure AD is the bisector of ∠A. If BD = 4 cm, DC = 3 cm and AB = 6 cm, find AC.
[Figure: ΔABC with angle bisector AD on BC. AB = 6 cm, BD = 4 cm, DC = 3 cm]

Solution:
By Angle Bisector Theorem: AB / AC = BD / DC.
6 / AC = 4 / 3 ⇒ 4 AC = 18 ⇒ AC = 18 / 4 = 4.5 cm.

24) The line through the points (−2, a) and (9, 3) has slope −¹/₂. Find the value of a.

Solution:
Slope m = (y₂ − y₁) / (x₂ − x₁) ⇒ (3 − a) / [9 − (−2)] = −¹/₂.
(3 − a) / 11 = −¹/₂ ⇒ 2(3 − a) = −11 ⇒ 6 − 2a = −11 ⇒ 2a = 17 ⇒ a = ¹⁷/₂ = 8.5.

25) Find the equation of a line which passes through (5, 7) and makes intercepts on the axes equal in magnitude but opposite in sign.

Solution:
Let intercepts be a and −a. The intercept equation is (x / a) + (y / −a) = 1 ⇒ x − y = a.
Passing through (5, 7): 5 − 7 = a ⇒ a = −2.
Equation: x − y = −2 ⇒ x − y + 2 = 0.

26) Find the equation of a straight line perpendicular to the line 3x − 7y − 6 = 0 and passing through the point (2, −3).

Solution:
Perpendicular line family to 3x − 7y − 6 = 0 is 7x + 3y + k = 0.
Passing through (2, −3): 7(2) + 3(−3) + k = 0 ⇒ 14 − 9 + k = 0 ⇒ 5 + k = 0 ⇒ k = −5.
Required equation: 7x + 3y − 5 = 0.

27) Prove the following identity: √[(1 + sin θ) / (1 − sin θ)] = sec θ + tan θ.

Solution:
LHS = √[ ((1 + sin θ)(1 + sin θ)) / ((1 − sin θ)(1 + sin θ)) ] = √[ (1 + sin θ)² / cos²θ ]
= (1 + sin θ) / cos θ = (1 / cos θ) + (sin θ / cos θ) = sec θ + tan θ = RHS. Hence proved.

28) [Compulsory] If α and β are the roots of x² + 7x + 10 = 0 find the values of (α − β).

Solution:
x² + 7x + 10 = 0 ⇒ α + β = −7, αβ = 10.
(α − β)² = (α + β)² − 4αβ = (−7)² − 4(10) = 49 − 40 = 9.
Hence, α − β = ±√9 = ±3.

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

29) Let A = {x ∈ W | x < 3}, B = {x ∈ N | 1 < x ≤ 5} and C = {3, 5, 7} then verify that A × (B ∪ C) = (A × B) ∪ (A × C).

Solution:
A = {0, 1, 2}, B = {2, 3, 4, 5}, C = {3, 5, 7}.
B ∪ C = {2, 3, 4, 5, 7}.
LHS = A × (B ∪ C) = {(0, 2), (0, 3), (0, 4), (0, 5), (0, 7), (1, 2), (1, 3), (1, 4), (1, 5), (1, 7), (2, 2), (2, 3), (2, 4), (2, 5), (2, 7)} — (1)
A × B = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5)}.
A × C = {(0, 3), (0, 5), (0, 7), (1, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7)}.
RHS = (A × B) ∪ (A × C) = Same 15 elements as in (1) — (2).
From (1) and (2), LHS = RHS. Hence verified.

30) Let A = {1, 2, 3, 4} and B = {2, 5, 8, 11, 14} be two sets. Let f: A → B be a function given by f(x) = 3x − 1. Represent this function (i) by arrow diagram (ii) in a table form (iii) as a set of ordered pairs (iv) in a graphical form.

Solution:
f(1) = 2, f(2) = 5, f(3) = 8, f(4) = 11.
(i) Arrow diagram: 1 → 2, 2 → 5, 3 → 8, 4 → 11.
(ii) Table form:

x	1	2	3	4
f(x)	2	5	8	11

(iii) Set of ordered pairs: f = {(1, 2), (2, 5), (3, 8), (4, 11)}.
(iv) Graphical form: Plotting the points (1, 2), (2, 5), (3, 8), (4, 11) on Cartesian axes.

31) Consider the functions f(x), g(x), h(x) as given below. Show that (f ∁ g) ∁ h = f ∁ (g ∁ h), where f(x) = x − 1, g(x) = 3x + 1 and h(x) = x².

Solution:
(f ∁ g)(x) = f(3x + 1) = (3x + 1) − 1 = 3x.
((f ∁ g) ∁ h)(x) = (f ∁ g)(x²) = 3(x²) = 3x² — (1)
(g ∁ h)(x) = g(x²) = 3x² + 1.
(f ∁ (g ∁ h))(x) = f(3x² + 1) = (3x² + 1) − 1 = 3x² — (2)
From (1) and (2), (f ∁ g) ∁ h = f ∁ (g ∁ h) is proved.

32) The 104th term and 4th term of an A.P. are 125 and 0. Find the sum of first 35 terms.

Solution:
t₁₀₄ = a + 103d = 125 — (1)
t₄ = a + 3d = 0 — (2) ⇒ a = −3d.
(1) − (2) ⇒ 100d = 125 ⇒ d = 125 / 100 = ⁵/₄.
a = −3(5/4) = −¹⁵/₄.
S₃₅ = (35 / 2) [2a + 34d] = (35 / 2) [2(−15/4) + 34(5/4)]
= (35 / 2) [−¹⁵/₂ + ⁸⁵/₂] = (35 / 2) [⁷⁰/₂] = (35 / 2) [35] = 1225 / 2 = 612.5.

33) Find the sum to n terms of the series 5 + 55 + 555 + ...

Solution:
S_n = 5(1 + 11 + 111 + ... to n terms)
= ⁵/₉ [9 + 99 + 999 + ... to n terms]
= ⁵/₉ [(10 − 1) + (10² − 1) + ... + (10ⁿ − 1)]
= ⁵/₉ [ {10(10ⁿ − 1) / (10 − 1)} − n ] = ⁵⁰/₈₁ (10ⁿ − 1) − ⁵ⁿ/₉.

34) Find the sum of all three digit natural numbers divisible by 9.

Solution:
Three-digit numbers divisible by 9: 108, 117, 126, ..., 999.
a = 108, d = 9, l = 999.
n = [(l − a) / d] + 1 = [(999 − 108) / 9] + 1 = (891 / 9) + 1 = 99 + 1 = 100.
S₁₀₀ = (100 / 2) [a + l] = 50 × [108 + 999] = 50 × 1107 = 55350.

35) Simplify: [(x³ − y³) / (3x² + 9xy + 6y²)] × [(x² + 2xy + y²) / (x² − y²)].

Solution:
x³ − y³ = (x − y)(x² + xy + y²)
3x² + 9xy + 6y² = 3(x² + 3xy + 2y²) = 3(x + y)(x + 2y)
x² + 2xy + y² = (x + y)²
x² − y² = (x − y)(x + y)
Multiplying and cancelling common factors:
= [ (x − y)(x² + xy + y²) / (3(x + y)(x + 2y)) ] × [ (x + y)² / ((x − y)(x + y)) ]
= (x² + xy + y²) / [3(x + 2y)].

36) If 9x⁴ + 12x³ + 28x² + ax + b is a perfect square, find the values of a and b.

Solution:
Using polynomial long division for square root:
Quotient is 3x² + 2x + 4.
(3x² + 2x + 4)² = 9x⁴ + 12x³ + 28x² + 16x + 16.
Equating coefficients: a = 16, b = 16.

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

Solution:
α + β = −a / 7, αβ = ²/₇.
(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 prove: Basic Proportionality 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.
Proof: In ΔABC, let line 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.
Therefore, AB / AD = AC / AE ⇒ (AD + DB) / AD = (AE + EC) / AE.
1 + (DB / AD) = 1 + (EC / AE) ⇒ DB / AD = EC / AE.
Taking reciprocals: AD / DB = AE / EC. Hence proved.

39) 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).

Solution:
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 (or −7K + 21 = −56 ⇒ K = 11).

40) 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.
Intercept form: (x / a) + [y / (7 − a)] = 1.
Passing 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 = 4.
Equation: (x / 3) + (y / 4) = 1 ⇒ 4x + 3y − 12 = 0.

41) If cos θ / (1 + sin θ) = 1 / a, then prove that (a² − 1) / (a² + 1) = sin θ.

Solution:
a = (1 + sin θ) / cos θ = sec θ + tan θ.
a² − 1 = (sec θ + tan θ)² − 1 = sec²θ + tan²θ + 2 sec θ tan θ − 1 = 2 tan²θ + 2 sec θ tan θ = 2 tan θ(tan θ + sec θ).
a² + 1 = (sec θ + tan θ)² + 1 = sec²θ + tan²θ + 2 sec θ tan θ + 1 = 2 sec²θ + 2 sec θ tan θ = 2 sec θ(sec θ + tan θ).
(a² − 1) / (a² + 1) = [2 tan θ(sec θ + tan θ)] / [2 sec θ(sec θ + tan θ)] = tan θ / sec θ = sin θ. Hence proved.

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

Solution:
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) ⇒ 5x − 3y − 8 = 0.

Part - IV • Answer the following: 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) 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.

Construction Steps Summary:
a) 1. Construct arbitrary ΔPQR. 2. Draw acute ray QX below QR. Mark 7 equal segments Q₁ to Q₇ on QX. 3. Join Q₃ to R. 4. Draw line through Q₇ parallel to Q₃R meeting extended line segment QR at R'. 5. Draw line through R' parallel to RP meeting extended line segment QP at P'. ΔP'QR' is the required similar triangle.
b) 1. Draw base BC = 8 cm. 2. At B, draw ∠CBE = 60° downwards and draw BF ⊥ BE. 3. Draw perpendicular bisector of BC meeting BF at O. 4. Draw circle with centre O and radius OB. 5. On BC, mark point D such that BD = 6 cm. 6. Let the perpendicular bisector cut the circle arc at I. Join ID and extend it to intersect circle at A. 7. Join AB and AC to get ΔABC.

44) Draw the graph of xy = 24, x, y > 0. Using the graph find (i) y when x = 3 and (ii) x when y = 6.
(OR) A bus is travelling at a uniform speed of 50 km/hr. Draw the time distance graph and hence find: (i) the constant of variation (ii) how far will it travel in 90 minutes? (iii) the time required to cover a distance of 300 km from the graph.

Solution:
First Option:
xy = 24 is an inverse variation with points: (1, 24), (2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2), (24, 1).
(i) From graph, when x = 3 ⇒ y = 8.
(ii) From graph, when y = 6 ⇒ x = 4.
Second Option:
Direct variation: y = kx where speed is 50 km/hr.
(i) Constant of variation: k = 50 (or ⁵/₆ km/min).
(ii) Distance travelled in 90 minutes (1.5 hours): y = 50 × 1.5 = 75 km.
(iii) Time required to cover 300 km: x = 300 / 50 = 6 hours (360 minutes).