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

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Elevate your academic preparation with the official 10th Maths Quarterly Exam 2026 Question Paper paired with verified Answer Keys for English Medium schools in the Salem District. As a key district-level terminal evaluation, this paper offers Tamil Nadu Samacheer Kalvi state board students a strategic resource to test their algebraic speed, conceptual accuracy, and logical reasoning under timed conditions. The assessment provides thorough coverage of essential early-term topics, including Relations and Functions, Numbers and Sequences (AP/GP series), polynomial equations, and foundational Geometry theorems. Practicing with this specific English-medium regional paper and utilizing the detailed step-by-step answer keys will help you identify minor calculation traps, master the distribution of step-wise marks, and perfect the structured presentation needed to achieve a perfect centum score on your final board examinations.

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10th Maths Quarterly Exam 2026 Question Paper & Answer Key | Salem District

SALEM DISTRICT 10 - STD Reg. No.: [________]

QUARTERLY EXAM - SEP 2026

MATHEMATICS (ENGLISH MEDIUM)

TIME: 3.00 HRS MARKS: 100

PART - 1 • I. Choose the Correct Answer: 14 × 1 = 14

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

Answer: (B) 2
Solution: n(A) = 5. Total number of relations = 2^{n(A) × n(B)} = 2^{5 × n(B)} = 1024 = 2¹⁰. Thus, 5 × n(B) = 10 ⇒ n(B) = 2.

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

Answer: (A) 7
Solution: A bijective function is both one-one and onto, meaning the number of elements in domain and co-domain must be equal: n(A) = n(B) = 7.

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)

Answer: (B) (2, −1)
Solution: g(1) = α(1) + β = 1 ⇒ α + β = 1. g(2) = α(2) + β = 3 ⇒ 2α + β = 3. Subtracting gives α = 2, so 2 + β = 1 ⇒ β = −1.

4. In an A.P a₁₈ − a₁₄ = 32, then the common difference is
(A) 8 (B) −8 (C) −4 (D) 4

Answer: (A) 8
Solution: a₁₈ − a₁₄ = (a + 17d) − (a + 13d) = 4d = 32. Therefore, d = 32 / 4 = 8.

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

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

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

Answer: (C) 7881
Solution: t_n = 1 + (n − 1)4 = 4n − 3. So, t_n + 3 must be a multiple of 4. Checking 7881: 7881 + 3 = 7884, which is divisible by 4 (n = 1971).

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

Answer: (B) ¹/₂₇
Solution: This is a Geometric Progression with common ratio r = (1/8) / (3/16) = ²/₃. The next term is (1/18) × (2/3) = 1/27.

8. Which of the following should be added to make x⁴ + 64 a perfect square
(A) 4x² (B) 16x² (C) 8x² (D) −8x⁴

Answer: (B) 16x²
Solution: x⁴ + 64 = (x²)² + 8². Adding 2(x²)(8) = 16x² gives (x² + 8)².

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 it is an isosceles right triangle with right angle at C, BC = AC = 5 cm. By Pythagoras theorem, AB² = 5² + 5² = 50 ⇒ AB = √50 = 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

Answer: (C) 9
Solution: Equating slopes: (p − 7) / (3 − 5) = (6 − 7) / (6 − 5) ⇒ (p − 7) / (−2) = −1 ⇒ p − 7 = 2 ⇒ p = 9.

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: An equation of the form x = c is a vertical line parallel to the Y-axis.

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

Answer: B) 4x + 10y − 4 = 0
Solution: Parallel lines have the same slope: m = −2/5. Line B has slope −4/10 = −2/5.

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

Answer: (A) 2a
Solution: a² − 1 = (sin θ + cos θ)² − 1 = 2 sin θ cos θ. b = (sin θ + cos θ) / (sin θ cos θ) = a / (sin θ cos θ). Hence, b(a² − 1) = [a / (sin θ cos θ)] × (2 sin θ cos θ) = 2a.

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

15. If A × B = {(3, 2), (3, 4), (5, 2), (5, 4)} then find A and B.

Solution:
A = set of first coordinates = {3, 5}
B = set of second coordinates = {2, 4}

16. Show that the function 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 the domain have distinct images in the co-domain, f is a one-one function.

17. If f(x) = x² − 1, g(x) = x − 2 find a, if g ∁ f(a) = 1.

Solution:
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. Find the least number that is divisible by the first ten natural numbers.

Solution:
The required number is LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10).
Prime powers: 2³ = 8, 3² = 9, 5¹ = 5, 7¹ = 7.
LCM = 2³ × 3² × 5 × 7 = 8 × 9 × 35 = 2520.

19. What is the time 100 hours after 7 a.m.?

Solution:
In a 24-hour cycle: 100 ≡ x (mod 24) ⇒ 100 = 24(4) + 4 ⇒ x = 4 hours.
Time = 7 a.m. + 4 hours = 11 a.m.

20. If 1³ + 2³ + 3³ + ... + k³ = 44100 then find 1 + 2 + 3 + ... + k.

Solution:
Identity: 1³ + 2³ + ... + k³ = (1 + 2 + ... + k)².
(1 + 2 + ... + k)² = 44100 ⇒ 1 + 2 + ... + k = √44100 = 210.

21. Find the excluded values, if t / (t² − 5t + 6).

Solution:
The expression is undefined when denominator is zero: t² − 5t + 6 = 0.
(t − 3)(t − 2) = 0 ⇒ t = 3, 2.

22. If the difference between a number and its reciprocal is ²⁴/₅, find the number.

Solution:
Let the number be x. Then x − (1/x) = ²⁴/₅ ⇒ (x² − 1) / x = ²⁴/₅.
5x² − 24x − 5 = 0 ⇒ (x − 5)(5x + 1) = 0.
Hence, x = 5 or x = −¹/₅.

23. Find the nature of the roots of the quadratic equation: 2x² − 2x + 9 = 0.

Solution:
a = 2, b = −2, c = 9.
Discriminant Δ = b² − 4ac = (−2)² − 4(2)(9) = 4 − 72 = −68 < 0.
Since Δ < 0, the equation has no real roots (imaginary roots).

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.

Solution:
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 Basic Proportionality Theorem, DE || BC.

25. What is the inclination of a line whose slope is (i) 0 (ii) 1?

Solution:
(i) Slope m = tan θ = 0 ⇒ θ = 0°.
(ii) Slope m = tan θ = 1 ⇒ θ = 45°.

26. Find the equation of a straight line passing through (5, −3) and (7, −4).

Solution:
(y − y₁) / (y₂ − y₁) = (x − x₁) / (x₂ − x₁)
(y + 3) / (−4 + 3) = (x − 5) / (7 − 5) ⇒ (y + 3) / (−1) = (x − 5) / 2
2(y + 3) = −(x − 5) ⇒ 2y + 6 = −x + 5 ⇒ x + 2y + 1 = 0.

27. Prove: cos θ / (1 + sin θ) = sec θ − tan θ.

Solution:
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. [Compulsory] Find the slope of the line joining the points A(−a, b) and B(3a, −b).

Solution:
Slope m = (y₂ − y₁) / (x₂ − x₁)
m = (−b − b) / [3a − (−a)] = −2b / (3a + a) = −2b / 4a = −b / (2a).

PART - 3 • III. Answer any 10 of the following (Q.No. 42 is compulsory): 10 × 5 = 50

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).

Solution:
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).
RHS: A × B contains 28 pairs; A × C = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}.
(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. The function 't' which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C) = F where F = ⁹/₅ C + 32. Find: (i) t(0) (ii) t(28) (iii) t(−10) (iv) the value of C when t(C) = 212 (v) the temperature when the Celsius value is equal to the Fahrenheit value.

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

31. If f(x) = x − 4, g(x) = x², h(x) = 3x − 5 then prove that f ∁ (g ∁ h) = (f ∁ g) ∁ h.

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

32. Find the HCF of 396, 504, 636.

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

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.

Solution:
Let three terms be a/r, a, ar.
Product = (a/r)(a)(ar) = 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.
When a = 3, r = ³/₂: the terms are 2, 3, ⁹/₂.
When a = 3, r = ²/₃: the terms are ⁹/₂, 3, 2.

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

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.

Solution:
Let the number of five, ten, and twenty rupee notes be x, y, z.
x + y + z = 12 — (1)
5x + 10y + 20z = 105 ⇒ x + 2y + 4z = 21 — (2)
10x + 5y + 20z = 125 ⇒ 2x + y + 4z = 25 — (3)
(2) − (1) ⇒ y + 3z = 9 — (4)
(3) − 2(1) ⇒ −y + 2z = 1 — (5)
Adding (4) and (5): 5z = 10 ⇒ z = 2.
From (4): y + 3(2) = 9 ⇒ y = 3.
From (1): x + 3 + 2 = 12 ⇒ x = 7.
Therefore: 7 five-rupee notes, 3 ten-rupee notes, 2 twenty-rupee notes.

36. Find the square root of the following: x⁴ − 12x³ + 42x² − 36x + 9.

Solution:
By polynomial long division method for square root:
First term of quotient: x². Subtract (x²)² = x⁴.
Next divisor: 2x² − 6x. Quotient term: −6x. Subtract −12x³ + 36x² leaving 6x² − 36x + 9.
Next divisor: 2x² − 12x + 3. Quotient term: 3. Subtract 6x² − 36x + 9 leaving remainder 0.
Required square root = |x² − 6x + 3|.

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?

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

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

Solution:
Arranging 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 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)
3y + 3 = 5x − 5 ⇒ 5x − 3y − 8 = 0.

41. If cosec θ + cot θ = p then prove that cos θ = (p² − 1) / (p² + 1).

Solution:
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. [Compulsory] A household has a fixed electricity charge of Rs. 100. In addition, Rs. 6 charged for every unit of electricity consumed. Let x be the number of units consumed and y be the total bill.
i. Find the equation of the electricity bill.
ii. Find the bill for 50 units.
iii. If the bill is Rs. 400, find the number of units consumed.
iv. Find the y-intercept.

Solution:
i. Total bill equation: y = 6x + 100
ii. For x = 50 units: y = 6(50) + 100 = 300 + 100 = Rs. 400
iii. If y = 400: 400 = 6x + 100 ⇒ 6x = 300 ⇒ x = 50 units
iv. Comparing with y = mx + c: y-intercept = 100

PART - 4 • IV. Answer all the questions: 2 × 8 = 16

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

Solution Outline:
a) 1. Construct ΔPQR. 2. Draw acute ray QX below QR. Mark 7 equal segments Q₁ to Q₇. 3. Join Q₃ to R. 4. Draw line through Q₇ parallel to Q₃R meeting extended QR at R'. 5. Draw line through R' parallel to RP meeting extended QP at P'. ΔP'QR' is the required similar triangle.
b) 1. Draw base QR = 5 cm. 2. At Q, draw ∠RQE = 30° downwards. Draw QF ⊥ QE. 3. Draw perpendicular bisector of QR meeting QF at O and QR at G. 4. With centre O and radius OQ, draw circle. 5. On perpendicular bisector, mark distance 4.2 cm from G and draw parallel line to QR intersecting circle at P and P'. 6. Join PQ and PR to complete ΔPQR.

44. a) Graph the 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.
(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 = 4 × 90 = 360
(ii) When number of participants x = 12 ⇒ Amount y = 360 / 12 = Rs. 30