9th Standard Maths Annual Exam 2026 — Question Paper with Answers | Kalvi Mini

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9th Standard Mathematics
Common Annual Examination — 2026

9th Standard Maths Annual Exam 2026

📘 Standard IX ⏱ 2.30 Hours 📊 100 Marks Tamil Nadu State Board

Dear Students, This post contains the complete 9th Standard Maths Annual Exam 2026 question paper with full answer key. All correct answers are shown in green boxes. Study well and score full marks! — Kalvi Mini 🌟

📝 PART I — Multiple Choice Questions 14 × 1 = 14 Marks

Answer ALL the questions. ✅ Correct answer is highlighted in green.

1If A = {x : x ∈ ℤ, −1 < x < 1}, then A is a:

a) Singleton set ✅

b) Power set

c) Null set

d) Subset

✅ Answer

a) Singleton set — The only integer between −1 and 1 is 0. So A = {0} → one element → Singleton set.

2Which one of the following is an irrational number?

a) √25

b) √(9/4)

c) 7/11

d) π ✅

✅ Answer

d) π — π = 3.14159… is non-terminating and non-repeating → Irrational number.

3The degree of the constant polynomial is:

a) 3

b) 2

c) 1

d) 0 ✅

✅ Answer

d) 0 — A constant polynomial like f(x) = 5 has degree 0.

4If the y-coordinate of a point is zero, then the point always lies:

a) In the I quadrant

b) In the II quadrant

c) On x-axis ✅

d) On y-axis

✅ Answer

c) On x-axis — Any point with y = 0 lies on the x-axis.

5The value of tan 1° × tan 2° × tan 3° × … × tan 89° is:

a) 0

b) 1 ✅

c) 2

d) √3/2

✅ Answer

b) 1 — tan θ × tan(90°−θ) = 1 for each pair. tan 45° = 1. So entire product = 1.

6The capacity of a water tank of dimensions 10m × 5m × 1.5m is:

a) 75 litres

b) 750 litres

c) 7500 litres

d) 75000 litres ✅

✅ Answer

d) 75000 litres — Volume = 10 × 5 × 1.5 = 75 m³. 1 m³ = 1000 litres → 75 × 1000 = 75,000 litres.

7The distance between the point (5, −1) and the origin is:

a) √24

b) √37

c) √26 ✅

d) √17

✅ Answer

c) √26 — Distance = √(5² + (−1)²) = √(25 + 1) = √26

8The exterior angle of a triangle is equal to the sum of two:

a) Exterior angles

b) Interior opposite angles ✅

c) Alternate angles

d) Interior angles

✅ Answer

b) Interior opposite angles — By Exterior Angle Theorem, exterior angle = sum of two non-adjacent interior angles.

9If the ratio of sides of two cubes is 2:3, then the ratio of their surface areas will be:

a) 4:6

b) 4:9 ✅

c) 6:9

d) 16:36

✅ Answer

b) 4:9 — Surface area ∝ side². Ratio = 2² : 3² = 4 : 9

10Which of the following is a linear equation in two variables?

a) x + 1/x = 2

b) x(x−1) = 2

c) 3x + 5 = 2/3

d) x − y = 5 ✅

✅ Answer

d) x − y = 5 — Two variables (x, y), degree 1 → Linear equation in two variables.

11If P(E) = 0.4, then the probability of failure P(E') is:

a) 0.4

b) 0.5

c) 0.6 ✅

d) 1

✅ Answer

c) 0.6 — P(E') = 1 − P(E) = 1 − 0.4 = 0.6

12The mean of the first 11 natural numbers is:

a) 6 ✅

b) 46

c) 48

d) 52

✅ Answer

a) 6 — Sum = 11×12/2 = 66. Mean = 66/11 = 6

13If n(A) = 4, then n[P(A)] is:

a) 8

b) 16 ✅

c) 32

d) 64

✅ Answer

b) 16 — n[P(A)] = 2^n(A) = 2⁴ = 16

14The zero of the polynomial 2x + 5 is:

a) 5/2

b) −5/2 ✅

c) 2/5

d) −2/5

✅ Answer

b) −5/2 — Set 2x + 5 = 0 → x = −5/2

✏️ PART II — Short Answer Questions 10 × 2 = 20 Marks

Answer ANY TEN questions. Question No. 28 is compulsory.

15If A = {2,3,5,7} and B = {1,3,5,8,10}, find A∪B and A∩B.

✅ Answer

A∪B = {1, 2, 3, 5, 7, 8, 10}
A∩B = {3, 5}

16Express 0.35 in the form of p/q.

✅ Answer

0.35 = 35/100 = 7/20

17Rationalize the denominator of 7/√14.

✅ Answer

7/√14 × √14/√14 = 7√14/14 = √14/2

18Find the value of f(y) = 6y − 3y² + 3 at y = −1.

✅ Answer

f(−1) = 6(−1) − 3(1) + 3 = −6 − 3 + 3 = −6

19Find the remainder when f(x) = x³ + 3x² + 3x + 1 is divided by (x + 1).

✅ Answer

By Remainder Theorem, put x = −1:
f(−1) = (−1)³ + 3(1) + 3(−1) + 1 = −1 + 3 − 3 + 1 = 0

20In cyclic quadrilateral ABCD, ∠A = 72°. Find supplementary angle ∠C.

✅ Answer

Opposite angles in a cyclic quadrilateral are supplementary.
∠A + ∠C = 180° → ∠C = 180° − 72° = 108°

21Find the midpoint of the line segment joining (3, 0) and (−5, 4).

✅ Answer

Midpoint = ((3−5)/2, (0+4)/2) = (−2/2, 4/2) = (−1, 2)

22If tan A = 2/3, find sin A.

✅ Answer

tan A = opp/adj = 2/3 → hyp = √(4+9) = √13
sin A = opp/hyp = 2/√13

23Evaluate: sin 30° + cos 30°

✅ Answer

sin 30° = 1/2, cos 30° = √3/2
= 1/2 + √3/2 = (1 + √3) / 2

24Find the area of an equilateral triangle whose side is 10 cm.

✅ Answer

Area = (√3/4) × a² = (√3/4) × 100 = 25√3 cm²

25Find the volume of a cube whose side is 10 cm.

✅ Answer

Volume = a³ = 10³ = 1000 cm³

26The average marks of 10 students is 48. If 7 is subtracted from each score, what is the new mean?

✅ Answer

When a constant is subtracted from each value, mean also reduces by that constant.
New Mean = 48 − 7 = 41

27A coin is tossed twice. What is the probability of getting two heads?

✅ Answer

Sample space = {HH, HT, TH, TT} → n(S) = 4
Favourable = {HH} → n(E) = 1
P(two heads) = 1/4 = 0.25

28Find the value of k if (x−2) is a factor of 2x³ − 6x² + mx + 4.COMPULSORY

✅ Answer

If (x−2) is a factor, f(2) = 0
2(8) − 6(4) + m(2) + 4 = 0
16 − 24 + 2m + 4 = 0
2m − 4 = 0 → m = 2

📖 PART III — Long Answer Questions 10 × 5 = 50 Marks

Answer ANY TEN questions. Question No. 42 is compulsory.

29Verify (A∩B)' = A'∪B' using a Venn diagram (De Morgan's Law).

✅ Answer

Let U = {1,2,3,4,5,6,7,8}, A = {1,2,3,4}, B = {3,4,5,6}

A∩B = {3,4} → (A∩B)' = {1,2,5,6,7,8}

A' = {5,6,7,8}, B' = {1,2,7,8}
A'∪B' = {1,2,5,6,7,8}

∴ (A∩B)' = A'∪B' ✅ Verified (De Morgan's Second Law)

30In a school, 300 play Hockey, 250 play Cricket, 110 play both. Find: (i) Only Hockey (ii) Only Cricket (iii) Total students.

✅ Answer

n(H) = 300, n(C) = 250, n(H∩C) = 110

(i) Only Hockey = 300 − 110 = 190
(ii) Only Cricket = 250 − 110 = 140
(iii) Total = n(H∪C) = 300 + 250 − 110 = 440 students

31Arrange the surds in descending order: ³⁰√5, ³¹√4, ³²√3

✅ Answer

LCM of 30, 31, 32 = 14880
³⁰√5 = 5^(496/14880), ³¹√4 = 4^(480/14880), ³²√3 = 3^(465/14880)
Comparing: 5^496 > 4^480 > 3^465
Descending order: ³⁰√5 > ³¹√4 > ³²√3

32Factorize: 2x³ − 3x² − 3x + 2

✅ Answer

f(2) = 16−12−6+2 = 0 → (x−2) is a factor
f(−1) = −2−3+3+2 = 0 → (x+1) is a factor
f(1/2) = 0 → (2x−1) is a factor

∴ 2x³ − 3x² − 3x + 2 = (x−2)(x+1)(2x−1)

33Dividing (8x⁴ − 2x² + 6x − 7) by (2x+1), quotient is (4x³ + px² − qx + 3). Find p, q and remainder.

✅ Answer

Perform long division of 8x⁴ + 0x³ − 2x² + 6x − 7 by (2x+1):

Step 1: 8x⁴ ÷ 2x = 4x³ → multiply: 8x⁴ + 4x³ → subtract → −4x³
Step 2: −4x³ ÷ 2x = −2x² → but quotient shows +px², so p = −2
Step 3: Continuing → q = −3
Remainder = −10

34Find all angles of cyclic quadrilateral ABCD where ∠A = (2y+4)°, ∠B = (6x−4)°, ∠C = (4y−4)°, ∠D = (7x+2)°.

✅ Answer

Opposite angles are supplementary:
∠A + ∠C = 180° → (2y+4) + (4y−4) = 180 → 6y = 180 → y = 30
∠B + ∠D = 180° → (6x−4) + (7x+2) = 180 → 13x − 2 = 180 → 13x = 182 → x = 14

∠A = 2(30)+4 = 64°, ∠B = 6(14)−4 = 80°, ∠C = 4(30)−4 = 116°, ∠D = 7(14)+2 = 100°

35Show that A(5,4), B(2,0), C(−2,3) form an isosceles triangle.

✅ Answer

AB = √((5−2)²+(4−0)²) = √(9+16) = √25 = 5
BC = √((2+2)²+(0−3)²) = √(16+9) = √25 = 5
CA = √((−2−5)²+(3−4)²) = √(49+1) = √50

AB = BC = 5 ∴ Two sides are equal → Isosceles Triangle ✅

36Using section formula, find the point dividing A(−3,6) and B(1,−2) in ratio m:n internally.

✅ Answer

Section formula: P = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))

P = ((m(1) + n(−3))/(m+n), (m(−2) + n(6))/(m+n))
P = ((m − 3n)/(m+n), (−2m + 6n)/(m+n))

For ratio 1:3 (example): P = ((1−9)/4, (−2+18)/4) = (−2, 4)

37Evaluate: tan 7° × tan 23° × tan 60° × tan 67° × tan 83°

✅ Answer

Using identity: tan θ × tan(90°−θ) = 1
tan 7° × tan 83° = 1 (since 7+83=90)
tan 23° × tan 67° = 1 (since 23+67=90)
tan 60° = √3

Result = 1 × 1 × √3 = √3

38Sides of a triangular ground are 22m, 120m, 122m. Find area and cost of levelling at ₹20 per m².

✅ Answer

s = (22+120+122)/2 = 264/2 = 132
s−a = 110, s−b = 12, s−c = 10
Area = √(132×110×12×10) = √(1742400) = 1320 m²

Cost = 1320 × 20 = ₹26,400

39A cubical tank holds 64,000 litres of water. Find the length of its side in metres.

✅ Answer

64,000 litres = 64 m³ (since 1000 litres = 1 m³)
Volume = a³ = 64 → a = ∛64 = 4 metres

40Find the mean for the frequency table: Classes 100–120, 120–140, 140–160, 160–180, 180–200, 200–220, 220–240 | Frequencies: 10, 8, 4, 4, 3, 1, 2

✅ Answer

Midpoints (x): 110, 130, 150, 170, 190, 210, 230
Σf = 10+8+4+4+3+1+2 = 32
Σfx = 1100+1040+600+680+570+210+460 = 4660
Mean = Σfx/Σf = 4660/32 = 145.625

41Find the mode: Marks (0–10, 10–20, 20–30, 30–40, 40–50) | Students (22, 38, 46, 34, 20)

✅ Answer

Modal class = 20–30 (highest frequency = 46)
l = 20, f₁ = 46, f₀ = 38, f₂ = 34, h = 10
Mode = l + ((f₁−f₀)/(2f₁−f₀−f₂)) × h
= 20 + (8/(92−38−34)) × 10
= 20 + (8/20) × 10 = 20 + 4 = 24

42If (x+2) and (x−4) are sides of a rectangle with area x²−2x−8, verify using factor theorem.COMPULSORY

✅ Answer

Let f(x) = x² − 2x − 8

Check (x+2): f(−2) = 4+4−8 = 0 ✅ → (x+2) is a factor
Check (x−4): f(4) = 16−8−8 = 0 ✅ → (x−4) is a factor

Also: (x+2)(x−4) = x²−4x+2x−8 = x²−2x−8 ✅
Verified: (x+2) and (x−4) are indeed the sides of the rectangle.

📐 PART IV — Graph & Construction Questions 2 × 8 = 16 Marks

Answer ALL the questions.

43a) Draw the graph for the linear equation y = 3x − 1.

✅ Answer — Graph Points

Prepare a table of values:

x = −1 → y = −4 → Point: (−1, −4)
x = 0 → y = −1 → Point: (0, −1)
x = 1 → y = 2 → Point: (1, 2)
x = 2 → y = 5 → Point: (2, 5)

Plot these points on graph paper and join them to get a straight line.

— OR —

b) Solve graphically: x + y = 7 and x − y = 3.

✅ Answer

For x + y = 7: (0,7), (7,0), (3,4)
For x − y = 3: (0,−3), (3,0), (5,2)

Plot both lines → Intersection point = (5, 2)
∴ x = 5, y = 2

44a) Construct △ABC with AB = 5cm, ∠B = 100°, BC = 6cm. Locate circumcentre and draw circumcircle.

✅ Steps

1. Draw BC = 6 cm
2. At B, construct ∠B = 100° and mark A such that AB = 5 cm
3. Join AC to complete △ABC
4. Draw perpendicular bisectors of any two sides (AB and BC)
5. The intersection point is the Circumcentre (O)
6. With O as centre and OA as radius, draw the Circumcircle

— OR —

b) Draw equilateral triangle of side 6.5cm. Locate incentre and draw incircle.

✅ Steps

1. Draw AB = 6.5 cm
2. With A and B as centres, draw arcs of radius 6.5 cm to get point C
3. Join AC and BC to complete equilateral △ABC
4. Draw angle bisectors of any two angles
5. Intersection point = Incentre (I)
6. Draw perpendicular from I to any side → that length is radius
7. Draw Incircle with centre I

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