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

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Master your board exam preparation with the official 10th Standard Maths Quarterly Exam 2026 Question Paper and Answer Key released for Ranipet District (English Medium). This dedicated resource provides students with the actual regional examination paper alongside comprehensive, step-by-step solutions verified by expert educators. Self-evaluating with this real-time question paper allows English medium students to understand key question patterns, evaluate their speed and accuracy, and target specific areas for improvement before the final board exams.

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

RANIPET DISTRICT Standard X Reg. No.: [___________]

COMMON QUARTERLY EXAMINATION 2026

MATHEMATICS (ENGLISH MEDIUM)

Time: 3.00 hrs Marks: 100

Part - 1 • 1. 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 relations = 2^{n(A) × n(B)} = 2^{5 × n(B)} = 1024 = 2¹⁰. Hence, 5 × n(B) = 10 ⇒ n(B) = 2.

2. Let n(A) = m and n(B) = n then the total number of relations that can be defined from A to B is
a) mⁿ b) n^m c) 2^mn − 1 d) 2^mn

Answer: d) 2^mn
Solution: Every relation from A to B is a subset of A × B. Since n(A × B) = mn, the total number of subsets is 2^mn.

3. Let f(x) = √(1 + x²), then
a) f(xy) = f(x)·f(y) b) f(xy) ≥ f(x)·f(y) c) f(xy) ≤ f(x)·f(y) d) None of these

Answer: c) f(xy) ≤ f(x)·f(y)
Solution: [f(x)·f(y)]² = (1 + x²)(1 + y²) = 1 + x² + y² + x²y². Since x² + y² ≥ 0, 1 + x²y² ≤ (1 + x²)(1 + y²), which gives f(xy) ≤ f(x)·f(y).

4. Using Euclid's division lemma, if the cube of any positive integer is divided by 9 the possible remainders are
a) 1, 4, 8 b) 0, 1, 8 c) 0, 1, 3 d) 1, 3, 5

Answer: b) 0, 1, 8
Solution: Any positive integer can be written as 3k, 3k + 1, or 3k + 2. Cubing each gives: (3k)³ = 27k³ = 9(3k³) (rem = 0); (3k + 1)³ = 9q + 1 (rem = 1); (3k + 2)³ = 9q + 8 (rem = 8).

5. An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P is
a) 16m b) 62m c) 31m d) (31/2)m

Answer: c) 31m
Solution: S₃₁ = (31/2)[2a + (31 − 1)d] = (31/2)[2a + 30d] = 31(a + 15d) = 31 × t₁₆ = 31m.

6. If the sequence t₁, t₂, t₃, ... are in A.P then the sequence t₆, t₁₂, t₁₈, ... is
a) a Geometric Progression b) An Arithmetic progression
c) neither an Arithmetic Progression nor a Geometric Progression d) a constant sequence

Answer: b) An Arithmetic progression
Solution: Let the common difference of t_n be d. Then t₁₂ − t₆ = 6d and t₁₈ − t₁₂ = 6d. Since the difference between successive terms is constant (6d), it forms an Arithmetic Progression.

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: From 3z = 9 ⇒ z = 3. Substituting into −7y + 7(3) = 7 ⇒ −7y = −14 ⇒ y = 2. Then x + 2 − 3(3) = −6 ⇒ x − 7 = −6 ⇒ x = 1.

8. y² + 1/y² is not equal to
a) (y⁴ + 1) / y² b) [y + 1/y]² c) [y − 1/y]² + 2 d) [y + 1/y]² − 2

Answer: b) [y + 1/y]²
Solution: [y + 1/y]² = y² + 1/y² + 2 ≠ y² + 1/y².

9. 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 z⁴)| c) ¹⁶/₅ |y / (x z²)| d) ¹⁶/₅ |(x z²) / y|

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

10. In ΔLMN, ∠L = 60°, ∠M = 50°. If ΔLMN ~ ΔPQR then the value of ∠R is
a) 40° b) 70° c) 30° d) 110°

Answer: b) 70°
Solution: ∠N = 180° − (60° + 50°) = 70°. Since ΔLMN ~ ΔPQR, corresponding angles are equal, so ∠R = ∠N = 70°.

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

Answer: b) 25 sq. units
Solution: Area = ¹/₂ | −5(−5 − 0) + 0(0 − 0) + 5(0 − (−5)) | = ¹/₂ | 25 + 0 + 25 | = 25 sq. units.

12. 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.

13. 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 represents a vertical line, which is always parallel to the y-axis.

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

15. Let A = {1, 2, 3} and B = {x | x is a prime number < 10}. Find A × B and B × A.

Solution:
Prime numbers less than 10 are B = {2, 3, 5, 7}.
A × B = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7)}.
B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (5, 1), (5, 2), (5, 3), (7, 1), (7, 2), (7, 3)}.

16. 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.

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

Solution:
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.

18. In GP 729, 243, 81, ... find t₇.

Solution:
First term a = 729, common ratio r = 243 / 729 = ¹/₃.
t_n = a rⁿ⁻¹ ⇒ t₇ = 729 × (¹/₃)⁷⁻¹ = 729 × (¹/₃)⁶ = 729 × (1 / 729) = 1.

19. Find the square root of the following expression: (144 a⁸ b¹² c¹⁶) / (81 f¹² g⁴ h¹⁴).

Solution:
√[ (144 a⁸ b¹² c¹⁶) / (81 f¹² g⁴ h¹⁴) ] = (12/9) |(a⁴ b⁶ c⁸) / (f⁶ g² h⁷)| = ⁴/₃ |(a⁴ b⁶ c⁸) / (f⁶ g² h⁷)|.

20. Find the excluded values of the following: (7p + 2) / (8p² + 13p + 5).

Solution:
Set denominator equal to zero: 8p² + 13p + 5 = 0.
8p² + 8p + 5p + 5 = 0 ⇒ (p + 1)(8p + 5) = 0.
The excluded values are −1 and −⁵/₈.

21. Find LCM of 5x − 10, 5x² − 20.

Solution:
5x − 10 = 5(x − 2).
5x² − 20 = 5(x² − 4) = 5(x − 2)(x + 2).
LCM = 5(x − 2)(x + 2) = 5(x² − 4).

22. 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 ≈ 6.43 cm.

23. 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 is 0, the points P, Q, R are collinear.

24. Find the equation of a straight line which has slope −⁵/₄ and passing through the point (−1, 2).

Solution:
Using point-slope form y − y₁ = m(x − x₁):
y − 2 = −⁵/₄ (x − (−1)) ⇒ 4(y − 2) = −5(x + 1) ⇒ 4y − 8 = −5x − 5 ⇒ 5x + 4y − 3 = 0.

25. Show that the straight lines x − 2y + 3 = 0 and 6x + 3y + 8 = 0 are perpendicular.

Solution:
Slope of x − 2y + 3 = 0 is m₁ = −1 / (−2) = ¹/₂.
Slope of 6x + 3y + 8 = 0 is m₂ = −6 / 3 = −2.
m₁ × m₂ = (¹/₂) × (−2) = −1.
Since product of slopes is −1, the lines are perpendicular.

26. Prove that tan²θ − sin²θ = tan²θ sin²θ.

Solution:
LHS = tan²θ − sin²θ = (sin²θ / cos²θ) − sin²θ = sin²θ [ (1 / cos²θ) − 1 ]
= sin²θ (sec²θ − 1) = sin²θ · tan²θ = tan²θ sin²θ = RHS. Hence proved.

27. Prove that (1 − tan²θ) / (cot²θ − 1) = tan²θ.

Solution:
LHS = (1 − tan²θ) / [ (1 / tan²θ) − 1 ] = (1 − tan²θ) / [ (1 − tan²θ) / tan²θ ]
= (1 − tan²θ) × [ tan²θ / (1 − tan²θ) ] = tan²θ = RHS. Hence proved.

28. [Compulsory] If 3 + k, 18 − k, 5k + 1 are in A.P. then find k.

Solution:
In an A.P., t₂ − t₁ = t₃ − t₂ ⇒ 2t₂ = t₁ + t₃.
2(18 − k) = (3 + k) + (5k + 1) ⇒ 36 − 2k = 6k + 4 ⇒ 8k = 32 ⇒ k = 4.

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

29. 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. A = {0, 1, 2, 3} and B = {1, 3, 5, 7, 9} be two sets. f: A → B be a function given by f(x) = 2x + 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(0) = 1, f(1) = 3, f(2) = 5, f(3) = 7.
i) Arrow diagram: Domain A = {0, 1, 2, 3} mapped to Co-domain B = {1, 3, 5, 7, 9} with arrows: 0 → 1, 1 → 3, 2 → 5, 3 → 7.
ii) Table form:

x	0	1	2	3
f(x)	1	3	5	7

iii) Set of ordered pairs: f = {(0, 1), (1, 3), (2, 5), (3, 7)}.
iv) Graphical form: Points (0, 1), (1, 3), (2, 5), (3, 7) plotted on Cartesian coordinate plane.

31. 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 finding 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.

32. Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.

Solution:
Subtract 12 from each number: 1230 − 12 = 1218, 1926 − 12 = 1914.
Required number = HCF(1218, 1914).
1914 = 1218 × 1 + 696
1218 = 696 × 1 + 522
696 = 522 × 1 + 174
522 = 174 × 3 + 0.
Hence, the largest number is 174.

33. 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 = Σ_k=1²⁴ k² − Σ_k=1⁹ k².
= [(24 × 25 × 49) / 6] − [(9 × 10 × 19) / 6] = 4900 − 285 = 4615 cm².

34. Solve: x + y + z = 5, 2x − y + z = 9, x − 2y + 3z = 16.

Solution:
x + y + z = 5 — (1)
2x − y + z = 9 — (2)
x − 2y + 3z = 16 — (3)
(1) + (2) ⇒ 3x + 2z = 14 — (4)
2(1) + (3) ⇒ 3x + 5z = 26 — (5)
(5) − (4) ⇒ 3z = 12 ⇒ z = 4.
Sub z = 4 into (4): 3x + 8 = 14 ⇒ 3x = 6 ⇒ x = 2.
Sub x = 2, z = 4 into (1): 2 + y + 4 = 5 ⇒ y = −1.
Solution: x = 2, y = −1, z = 4.

35. Find the GCD of following polynomials: x⁴ + 3x³ − x − 3 and x³ + x² − 5x + 3.

Solution:
Let f(x) = x⁴ + 3x³ − x − 3 and g(x) = x³ + x² − 5x + 3.
Dividing f(x) by g(x) gives quotient (x + 2) and remainder 3x² + 6x − 9 = 3(x² + 2x − 3).
Dividing g(x) by x² + 2x − 3 gives remainder 0.
Therefore, GCD = x² + 2x − 3.

36. Simplify: [(b² + 3b − 28) / (b² + 4b + 4)] ÷ [(b² − 49) / (b² − 5b − 14)].

Solution:
Factorising polynomials:
b² + 3b − 28 = (b + 7)(b − 4)
b² + 4b + 4 = (b + 2)²
b² − 49 = (b − 7)(b + 7)
b² − 5b − 14 = (b − 7)(b + 2)
Now, [ (b + 7)(b − 4) / (b + 2)² ] × [ (b − 7)(b + 2) / ((b − 7)(b + 7)) ] = (b − 4) / (b + 2).

37. 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 a line parallel to BC intersect AB at D and AC at E, so DE || BC.
In ΔADE and ΔABC:
1. ∠ADE = ∠ABC (corresponding angles)
2. ∠AED = ∠ACB (corresponding angles)
3. ∠DAE = ∠BAC (common angle)
By AAA similarity, ΔADE ~ ΔABC.
Therefore, corresponding sides are proportional: 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.

38. Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices taken in order are (−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).

39. Find the equation of a straight line parallel to x-axis and passing through the point of intersection of the lines 7x − 3y = −12 and 2y = x + 3.

Solution:
From the second equation: x = 2y − 3. Substitute into 7x − 3y = −12:
7(2y − 3) − 3y = −12 ⇒ 14y − 21 − 3y = −12 ⇒ 11y = 9 ⇒ y = ⁹/₁₁.
A line parallel to x-axis has the equation y = c.
Therefore, the required equation is y = ⁹/₁₁ ⇒ 11y − 9 = 0.

40. A quadrilateral has vertices at A(−4, −2), B(5, −1), C(6, 5) and D(−7, 6). Show that the mid-points of its sides form a parallelogram.

Solution:
Mid-point of AB, E = (¹/₂, −³/₂).
Mid-point of BC, F = (¹¹/₂, 2).
Mid-point of CD, G = (−¹/₂, ¹¹/₂).
Mid-point of DA, H = (−¹¹/₂, 2).
Slope of EF = ⁷/₁₀, Slope of GH = ⁷/₁₀ ⇒ EF || GH.
Slope of FG = −⁷/₁₂, Slope of HE = −⁷/₁₂ ⇒ FG || HE.
Since both pairs of opposite sides are parallel, EFGH forms a parallelogram.

41. Prove that √[(1 + sin θ) / (1 − sin θ)] + √[(1 − sin θ) / (1 + sin θ)] = 2 sec θ.

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

42. [Compulsory] If the function f is defined by
f(x) = x + 2, if x > 1
f(x) = 2, if −1 ≤ x ≤ 1
f(x) = x − 1, if −3 < x < −1
Find the values of: i) f(3) ii) f(0) iii) f(−1.5) iv) f(2) + f(−2).

Solution:
i) For x = 3 > 1: f(3) = 3 + 2 = 5.
ii) For x = 0 ∈ [−1, 1]: f(0) = 2.
iii) For x = −1.5 ∈ (−3, −1): f(−1.5) = −1.5 − 1 = −2.5.
iv) f(2) = 2 + 2 = 4 (since 2 > 1). f(−2) = −2 − 1 = −3 (since −2 ∈ (−3, −1)).
Therefore, f(2) + f(−2) = 4 + (−3) = 1.

Part - IV • 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 ΔPQR in which PQ = 8 cm, ∠R = 60° and the median RG from R to PQ is 5.8 cm. Find the length of the altitude from R to PQ.

Solution Summary:
a) 1. Construct arbitrary ΔPQR. 2. Draw ray QX making an acute angle with QR. 3. Mark 7 equidistant points Q₁, Q₂, ..., Q₇ on QX. 4. Join Q₃R. 5. Through Q₇, draw line parallel to Q₃R intersecting extended QR at R'. 6. Through R', draw line parallel to RP intersecting extended QP at P'. ΔP'QR' is the required similar triangle.
b) 1. Draw base PQ = 8 cm. 2. At P, draw ∠QPE = 60° downwards. Draw ray PF ⊥ PE. 3. Draw perpendicular bisector of PQ intersecting PQ at G and ray PF at O. 4. Draw circle with centre O and radius OP. 5. With G as centre and radius 5.8 cm, draw an arc cutting circle at R. Join PR and QR to form ΔPQR. 6. Draw perpendicular from R to PQ. The measured length of the altitude is approximately 3.8 cm.

44. a) 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.
(OR) b) The following table shows the data about the number of pipes and the time taken to fill the same tank:

No. of pipes (x)	2	3	6	9
Time taken (in min) (y)	45	30	15	10

Draw the graph for the above data and hence:
i) find the time taken to fill the tank when five pipes are used.
ii) Find the number of pipes when the time is 9 minutes.

Solution Summary:
a) Let time be x (hours) and distance be y (km). Direct variation: y = kx.
i) Constant of variation: k = y / x = 50 / 1 = 50 (or ⁵/₆ km/min).
ii) In 90 minutes (1.5 hours): y = 50 × 1.5 = 75 km.
iii) For distance y = 300 km: x = 300 / 50 = 6 hours (360 minutes).
b) As x increases, y decreases ⇒ Inverse variation: xy = k.
k = 2 × 45 = 3 × 30 = 6 × 15 = 9 × 10 = 90. Equation: xy = 90.
i) When x = 5 pipes: y = 90 / 5 = 18 minutes.
ii) When y = 9 minutes: x = 90 / 9 = 10 pipes.