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

QMD • 10 - Std Madurai District

MATHEMATICS (ENGLISH MEDIUM) • QUESTION PAPER & DETAILED SOLUTIONS

Time: 3.00 Hrs Marks: 100

PART - A • I. Choose the correct answer: 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

2. If f: A → B is a bijective function and if n(B) = 7, then n(A) is
(A) 7      (B) 49      (C) 1      (D) 14

3. If n(A) = 2 and n(B) = 3, then the total number of relations from A to B is
(A) 2³      (B) 3²      (C) 2⁵      (D) 2⁶

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. In an A.P. the first term is 1 and the common difference is 4. How many terms of the A.P. must be taken for their sum to be equal to 120?
(A) 6      (B) 7      (C) 8      (D) 9

6. The value of (1³ + 2³ + 3³ + ... + 15³) − (1 + 2 + 3 + ... + 15) is
(A) 14400      (B) 14200      (C) 14280      (D) 14520

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 square root of (256 x⁸ y⁴ z¹⁰) / (25 x⁶ y⁶ z⁶) is equal to
(A) ¹⁶/₅ |(x² y⁴) / y²|      (B) 16 |y² / (x² z⁴)|      (C) ¹⁶/₅ |y / (x z²)|      (D) ¹⁶/₅ |(x z²) / y|

9. The values of a and b if 4x⁴ − 24x³ + 76x² + ax + b is a perfect square are
(A) 100, 120      (B) 10, 12      (C) −120, 100      (D) 12, 10

10. If in triangles ABC and EDF, AB / DE = BC / FD then they will be similar, when
(A) ∠B = ∠E      (B) ∠A = ∠D      (C) ∠B = ∠D      (D) ∠A = ∠F

11. The area of triangle formed by the points (−5, 0), (0, −5) and (5, 0) is
(A) 0 sq.units      (B) 25 sq.units      (C) 5 sq.units      (D) none of these

12. If the points A(1, 1), B(2, 2) and C(3, 3) are collinear then the area of the ΔABC is
(A) 1      (B) 0      (C) 2      (D) 3

13. Which of the following is incorrect?
(A) P(A) > 1      (B) 0 ≤ P(A) ≤ 1      (C) P(∅) = 0      (D) P(A) + P(A̅) = 1

14. The probability of getting a job for a person is x / 3. If the probability of not getting the job is ²/₃, then the value of x is
(A) 2      (B) 1      (C) 3      (D) 1.5

PART - B • Answer any 10 questions (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 all first coordinates = {3, 5}
B = set of all second coordinates = {2, 4}

16. 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 t(0).
Solution:
t(C) = ⁹/₅ C + 32
Substitute C = 0: t(0) = ⁹/₅(0) + 32 = 0 + 32 = 32°F

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. 'a' and 'b' are two positive integers such that a^b × b^a = 800. Find a and b.
Solution:
800 = 2 × 400 = 2 × 2 × 200 = 2⁵ × 5²
Comparing with a^b × b^a: a = 2, b = 5 (or a = 5, b = 2)

19. Find the sum: 3 + 1 + ¹/₃ + ... ∞.
Solution:
First term a = 3, common ratio r = 1 / 3. Since |r| < 1:
S_∞ = a / (1 − r) = 3 / (1 − ¹/₃) = 3 / (²/₃) = 3 × ³/₂ = ⁹/₂

20. Find the LCM of the following: 8x⁴y², 48x²y⁴.
Solution:
LCM(8, 48) = 48; LCM(x⁴y², x²y⁴) = x⁴y⁴
LCM = 48x⁴y⁴

21. Find the excluded values, if any, of the following expression: (x² + 6x + 8) / (x² + x − 2).
Solution:
(x² + 6x + 8) / (x² + x − 2) = [(x + 4)(x + 2)] / [(x + 2)(x − 1)] = (x + 4) / (x − 1)
Setting denominator to zero: x − 1 = 0 ⇒ x = 1.
Excluded value is x = 1.

22. Determine the nature of roots for the quadratic equation: x² − x − 20 = 0.
Solution:
a = 1, b = −1, c = −20
Discriminant Δ = b² − 4ac = (−1)² − 4(1)(−20) = 1 + 80 = 81 > 0
Since Δ > 0, the roots are real and unequal.

23. The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB.
Solution:
Perimeter(ΔABC) / Perimeter(ΔPQR) = AB / PQ
36 / 24 = AB / 10 ⇒ ³/₂ = AB / 10 ⇒ AB = (3 × 10) / 2 = 15 cm

24. Find the area of the triangle whose vertices are (−3, 5), (5, 6) and (5, −2).
Solution:
Vertices: A(−3, 5), B(5, −2), C(5, 6)
Area = ¹/₂ | [(−3(−2) + 5(6) + 5(5)) − (5(5) + (−2)(5) + 6(−3))] |
= ¹/₂ | (6 + 30 + 25) − (25 − 10 − 18) | = ¹/₂ | 61 − (−3) | = ¹/₂(64) = 32 sq. units

25. Find the value of 'a' for which the given points are collinear: (2, 3), (4, a), (6, −3).
Solution:
Slope of line 1 = (a − 3) / (4 − 2) = (a − 3) / 2
Slope of line 2 = (−3 − 3) / (6 − 2) = −6 / 4 = −³/₂
Since points are collinear: (a − 3) / 2 = −³/₂ ⇒ a − 3 = −3 ⇒ a = 0

26. What is the probability that a leap year selected at random will contain 53 Saturdays?
Solution:
A leap year has 366 days = 52 weeks + 2 days.
Sample space S = {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)} ⇒ n(S) = 7.
Favourable event A = {(Fri, Sat), (Sat, Sun)} ⇒ n(A) = 2.
P(A) = n(A) / n(S) = ²/₇

27. If P(A) = ²/₃, P(B) = ²/₅, P(A ∪ B) = ¹/₃ then find P(A ∩ B).
Solution:
P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = ²/₃ + ²/₅ − ¹/₃ = ¹/₃ + ²/₅ = (5 + 6) / 15 = ¹¹/₁₅

28. [Compulsory] Find the remainder when 7 × 13 × 19 × 23 × 29 × 31 is divided by 6.
Solution:
7 ≡ 1 (mod 6), 13 ≡ 1 (mod 6), 19 ≡ 1 (mod 6), 23 ≡ −1 (mod 6), 29 ≡ −1 (mod 6), 31 ≡ 1 (mod 6)
Product ≡ 1 × 1 × 1 × (−1) × (−1) × 1 ≡ 1 (mod 6)
Therefore, remainder is 1.

PART - C • Answer any 10 questions (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 numbers. Verify that (A ∩ B) × C = (A × C) ∩ (B × C).
Solution:
A = {1, 2, 3, 4, 5, 6, 7}, B = {2, 3, 5, 7}, C = {2}
A ∩ B = {2, 3, 5, 7}
LHS: (A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} — (1)
A × C = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}
B × C = {(2, 2), (3, 2), (5, 2), (7, 2)}
RHS: (A × C) ∩ (B × C) = {(2, 2), (3, 2), (5, 2), (7, 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) an 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: Elements of A mapped to B: 2→0, 4→1, 6→2, 10→4, 12→5
(iv) Graph: Plotting the points (2, 0), (4, 1), (6, 2), (10, 4), (12, 5) on coordinate axes.

31. If f(x) = 2x + 3, g(x) = 1 − 2x and h(x) = 3x. Prove that f ∁ (g ∁ h) = (f ∁ g) ∁ h.
Solution:
(g ∁ h)(x) = g(3x) = 1 − 6x
f ∁ (g ∁ h)(x) = f(1 − 6x) = 2(1 − 6x) + 3 = 5 − 12x — (1)
(f ∁ g)(x) = f(1 − 2x) = 2(1 − 2x) + 3 = 5 − 4x
((f ∁ g) ∁ h)(x) = 5 − 4(3x) = 5 − 12x — (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 Lemma: a = bq + r
504 = 396(1) + 108 ; 396 = 108(3) + 72 ; 108 = 72(1) + 36 ; 72 = 36(2) + 0 ⇒ HCF(396, 504) = 36
Now, HCF(36, 636): 636 = 36(17) + 24 ; 36 = 24(1) + 12 ; 24 = 12(2) + 0 ⇒ HCF = 12
Hence, HCF(396, 504, 636) = 12.

33. The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
Solution:
Let the terms be a − d, a, a + d.
Sum = 3a = 27 ⇒ a = 9
Product = (a − d)a(a + d) = 288 ⇒ 9(81 − d²) = 288 ⇒ 81 − d² = 32 ⇒ d² = 49 ⇒ d = ±7
If d = 7, the terms are 2, 9, 16; if d = −7, the terms are 16, 9, 2.

34. Rekha has 15 squares 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. Solve: x + y + z = 5 — (1),   2x − y + z = 9 — (2),   x − 2y + 3z = 16 — (3)
Solution:
(1) + (2) ⇒ 3x + 2z = 14 — (4)
2 × (1) + (3) ⇒ 3x + 5z = 26 — (5)
(5) − (4) ⇒ 3z = 12 ⇒ z = 4
From (4): 3x + 2(4) = 14 ⇒ 3x = 6 ⇒ x = 2
From (1): 2 + y + 4 = 5 ⇒ y = −1
Solution: x = 2, y = −1, z = 4

36. Find the square root of 64x⁴ − 16x³ + 17x² − 2x + 1 by division method.
Solution:
Dividing by long division method yields the quotient 8x² − x + 1 and remainder 0.
Square root = |8x² − x + 1|

37. A boy of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/sec. If the lamppost is 3.6 m above the ground, find the length of his shadow cast after 4 seconds.
Solution:
Distance walked in 4 s: d = 1.2 × 4 = 4.8 m. Let shadow length be x.
By similar triangles: (Height of boy) / (Height of lamp) = (Shadow) / (Total distance)
0.9 / 3.6 = x / (4.8 + x) ⇒ ¹/₄ = x / (4.8 + x) ⇒ 4x = 4.8 + x ⇒ 3x = 4.8 ⇒ x = 1.6 m

38. The floor of a hall is covered with 110 identical triangular tiles with vertices at (−3, 2), (−1, −1) and (1, 2). Find the area of the floor.
Solution:
Area of 1 tile = ¹/₂ | [(−3(−1) + (−1)(2) + 1(2)) − (2(−1) + (−1)(1) + 2(−3))] |
= ¹/₂ | (3 − 2 + 2) − (−2 − 1 − 6) | = ¹/₂ | 3 − (−9) | = ¹/₂(12) = 6 sq. units
Total area of 110 tiles = 110 × 6 = 660 sq. units

39. Find the area of the quadrilateral whose vertices are at (−9, −2), (−8, −4), (2, 2), (1, −3).
Solution:
Arranging in counter-clockwise order: A(−9, −2), B(−8, −4), C(1, −3), D(2, 2)
Area = ¹/₂ | [(36 + 24 + 2 − 4) − (16 − 4 − 6 − 18)] | = ¹/₂ | 58 − (−12) | = ¹/₂(70) = 35 sq. units

40. Two unbiased dice are rolled once (n(S) = 36). Find the probability of getting:
Solution:
(i) A doublet: A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} ⇒ P(A) = 6 / 36 = ¹/₆
(ii) Product is a prime: B = {(1, 2), (2, 1), (1, 3), (3, 1), (1, 5), (5, 1)} ⇒ P(B) = 6 / 36 = ¹/₆
(iii) Sum is a prime: n(C) = 15 ⇒ P(C) = 15 / 36 = ⁵/₁₂
(iv) Sum is 1: Impossible event ⇒ P(D) = 0

41. n(S) = 50, n(A) = 28 (NCC), n(B) = 30 (NSS), n(A ∩ B) = 18. Find the probability that:
Solution:
(i) Opted NCC but not NSS: [n(A) − n(A ∩ B)] / n(S) = (28 − 18) / 50 = 10 / 50 = ¹/₅
(ii) Opted NSS but not NCC: [n(B) − n(A ∩ B)] / n(S) = (30 − 18) / 50 = 12 / 50 = ⁶/₂₅
(iii) Exactly one of them: ¹⁰/₅₀ + ¹²/₅₀ = ²²/₅₀ = ¹¹/₂₅

42. [Compulsory] If α, β are the roots of f(x) = x² − 2x + 3, form the polynomial whose roots are: (i) α + 2, β + 2    (ii) 1/α, 1/β
Solution:
From given equation: α + β = 2, αβ = 3
(i) Sum = α + β + 4 = 6 ; Product = αβ + 2(α + β) + 4 = 3 + 4 + 4 = 11
Equation: x² − (Sum)x + Product = x² − 6x + 11 = 0
(ii) Sum = 1/α + 1/β = (α + β) / αβ = ²/₃ ; Product = 1/(αβ) = ¹/₃
Equation: x² − (²/₃)x + ¹/₃ = 0 ⇒ 3x² − 2x + 1 = 0

PART - D • Answer ALL the questions: 2 × 8 = 16

44. a) Distance - Time graph (Direct Variation: y = kx):
(i) Constant of variation: k = 50 (where y = 50x)
(ii) Distance travelled in 90 minutes (1.5 hrs): y = 50 × 1.5 = 75 km
(iii) Time required to cover 300 km: x = 300 / 50 = 6 hours
(OR)
44. b) Cash prize distribution (Inverse Variation: xy = k):
(i) Constant of variation: k = 2 × 180 = 360
(ii) When number of participants x = 12: y = 360 / 12 = Rs. 30