SEE Mathematics Sample Paper (Class IX)

Class 9 Maths Sample Paper 2025–26

Here are the **detailed answers with explanations** for all the questions from the sample paper:

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## **SECTION A (MCQs)**

**1. Zero of the zero polynomial is**  

✅ **c) Not defined**  

*Explanation:* Zero polynomial is \( p(x) = 0 \). Every real number is a zero, so the zero is not uniquely defined.

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**2. If one zero of \( kx^2 + 4x + k \) is 1, then \( k = ? \)**  

✅ **c) -2**  

*Explanation:* Put \( x = 1 \): \( k(1)^2 + 4(1) + k = 0 \) → \( k + 4 + k = 0 \) → \( 2k = -4 \) → \( k = -2 \)

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**3. \( x = 2, y = -1 \) is a solution of**  

✅ **b) \( x + 2y = 0 \)**  

*Explanation:* \( 2 + 2(-1) = 2 - 2 = 0 \) ✅

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**4. Which pair is a solution of \( 3x - 2y = 7 \)?**  

✅ **b) (1, -2)**  

*Explanation:* \( 3(1) - 2(-2) = 3 + 4 = 7 \) ✅

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**5. How many lines pass through two points?**  

✅ **d) only one**  

*Explanation:* Unique line through two distinct points.

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**6. If \( \angle BOD = 63^\circ \), then \( \angle BOC = ? \)**  

✅ **a) 117°**  

*Explanation:* Linear pair: \( 180^\circ - 63^\circ = 117^\circ \)

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**7. If \( AB \parallel CD \), then \( x \) and \( y \) are?**  

✅ **d) 50°, 77°**  

*Explanation:* Using alternate interior angles and angle sum property.

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**8. If \( \triangle ABC \cong \triangle PQR \) by SSS, then**  

✅ **d) \( BC = QR \)**  

*Explanation:* Corresponding sides are equal.

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**9. In \( \triangle ABC \), \( BC = AB \) and \( \angle B = 80^\circ \). Then \( \angle A = ? \)**  

✅ **a) 50°**  

*Explanation:* Isosceles triangle: \( \angle A = \angle C \), sum = \( 180^\circ \) → \( 80 + 2A = 180 \) → \( A = 50^\circ \)

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**10. Which is a binomial?**  

✅ **b) \( x^2 + 4 \)**  

*Explanation:* Two terms.

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**11. In parallelogram ABCD, \( \angle ADC = 85^\circ \), find \( x + y \)**  

✅ **c) 95°**  

*Explanation:* Adjacent angles supplementary: \( x + y = 180 - 85 = 95^\circ \)

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**12. Two adjacent angles of a parallelogram in ratio 4:5 are**  

✅ **b) 80°, 100°**  

*Explanation:* \( 4k + 5k = 180 \) → \( 9k = 180 \) → \( k = 20 \) → angles \( 80^\circ, 100^\circ \)

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**13. Diagonals are equal in**  

✅ **b) Rectangle**  

*Explanation:* Only rectangle has equal diagonals.

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**14. In given figure, if \( \angle DAB = 62^\circ \), \( \angle ABD = 58^\circ \), then \( \angle ACB = ? \)**  

✅ **c) 60°**  

*Explanation:* Use angle sum property in triangles and cyclic properties.

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**15. Angle in a semicircle =**  

✅ **d) 90°**  

*Explanation:* Thales' theorem.

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**16. Three or more lines intersecting at same point are called**  

✅ **c) Concurrent**

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**17. Surface area of sphere of radius 3.5 cm =**  

✅ **b) 154 cm²**  

*Explanation:* \( 4\pi r^2 = 4 \times \frac{22}{7} \times (3.5)^2 = 154 \)

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**18. 1 cm = 30 km, length for 75 km =**  

✅ **b) 2.5 cm**  

*Explanation:* \( \frac{75}{30} = 2.5 \) cm

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**19. Assertion-Reason**  

✅ **c) A is true but R is false**  

*Explanation:* Opposite angles sum to \( 180^\circ \), not \( 360^\circ \).

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**20. Assertion-Reason**  

✅ **a) Both A and R are true and R is correct explanation**  

*Explanation:* \( 120^\circ + 60^\circ = 180^\circ \) → supplementary.

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## **SECTION B**

**21. \( 0.6 \) in \( \frac{p}{q} \) form**  

\( 0.6 = \frac{6}{10} = \frac{3}{5} \)

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**22. C lies between A and B, AC = BC, prove AC = \( \frac{1}{2} AB \)**  

*Proof:* \( AC + CB = AB \), \( AC = BC \) ⇒ \( 2AC = AB \) ⇒ \( AC = \frac{1}{2}AB \)

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**23. Verify \( x = -\frac{1}{3} \) is zero of \( p(x) = 3x + 1 \)**  

\( p\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1 = -1 + 1 = 0 \) ✅

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**24. Express \( 3x + 2 = 0 \) in \( ax + by + c = 0 \)**  

\( 3x + 0y + 2 = 0 \) ⇒ \( a = 3, b = 0, c = 2 \)  

**OR** Check (4, 0) in \( x - 2y = 4 \): \( 4 - 0 = 4 \) ✅ Yes.

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**25. Prove AB = CD**  

Perpendicular from center to chord bisects it. Since both chords are in same circle, AB = CD.  

**OR** Equal angles subtend equal chords.

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## **SECTION C**

**26. Locate \( \sqrt{3} \) on number line**  

Draw right triangle with legs 1 and 1 → hypotenuse \( \sqrt{2} \), then using \( \sqrt{2} \) and 1 → \( \sqrt{3} \).

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**27. Hemisphere radius 21 cm**  

CSA = \( 2\pi r^2 = 2 \times \frac{22}{7} \times 441 = 2772 \) cm²  

TSA = \( 3\pi r^2 = 3 \times \frac{22}{7} \times 441 = 4158 \) cm²  

**OR** Inner dia 10.5 cm → r = 5.25 cm  

CSA = \( 2\pi r^2 = 2 \times \frac{22}{7} \times 27.5625 = 173.25 \) cm²  

Cost = \( \frac{173.25}{100} \times 16 = 27.72 \) rupees

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**28. \( \angle POY = 90^\circ \), a:b = 2:3, find \( \angle c \)**  

Let a = 2k, b = 3k, a + b = 90° ⇒ 5k = 90 ⇒ k = 18  

a = 36°, b = 54°, c = 180° - (a + b) = 90°

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**29. Factorise \( x^3 - 3x^2 - 9x - 5 \)**  

Try x = 5: 125 - 75 - 45 - 5 = 0 ⇒ factor (x - 5)  

Divide → \( x^2 + 2x + 1 = (x + 1)^2 \)  

So \( (x - 5)(x + 1)^2 \)

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**30. 3 solutions of \( 2x + 5y = 13 \)**  

(0, 13/5), (1, 11/5), (4, 1)

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**31. AD altitude in isosceles triangle ABC, AB = AC**  

1. In right triangles ADB and ADC, AB = AC, AD common ⇒ RHS ⇒ BD = DC ⇒ AD bisects BC.  

2. ∠BAD = ∠CAD ⇒ AD bisects ∠A.  

**OR** BE = CF, RHS ⇒ ∆ABE ≅ ∆ACF ⇒ AB = AC.

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## **SECTION D**

**32. \( x = 2 - \sqrt{3} \), find \( \left(x - \frac{1}{x}\right)^3 \)**  

\( \frac{1}{x} = 2 + \sqrt{3} \), so \( x - \frac{1}{x} = -2\sqrt{3} \)  

Cube: \( (-2\sqrt{3})^3 = -8 \times 3\sqrt{3} = -24\sqrt{3} \)  

**OR** Simplify: \( \frac{7+3\sqrt{5}}{3+\sqrt{5}} - \frac{7-3\sqrt{5}}{3-\sqrt{5}} \)  

Multiply numerator and denominator by conjugates:  

First term = \( \frac{(7+3\sqrt{5})(3-\sqrt{5})}{4} = \frac{21 - 7\sqrt{5} + 9\sqrt{5} - 15}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} \)  

Second term = \( \frac{(7-3\sqrt{5})(3+\sqrt{5})}{4} = \frac{21 + 7\sqrt{5} - 9\sqrt{5} - 15}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \)  

Difference = \( \frac{3 + \sqrt{5} - 3 + \sqrt{5}}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5} \)  

So a = 0, b = 1.

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**33. Runs scored by teams**  

Bar graph or histogram with class intervals 1-6, 7-12, etc., showing runs.

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**34. Sides 5:12:13, perimeter 150 m**  

5k + 12k + 13k = 150 ⇒ 30k = 150 ⇒ k = 5  

Sides: 25, 60, 65 m  

Area = \( \sqrt{s(s-a)(s-b)(s-c)} \), s = 75  

= \( \sqrt{75 \times 50 \times 15 \times 10} = \sqrt{562500} = 750 \) m²  

**OR** Sides 42, 34, 20 cm, s = 48  

Area = \( \sqrt{48 \times 6 \times 14 \times 28} = \sqrt{112896} = 336 \) cm²  

Height corresponding to longest side (42 cm) = \( \frac{2 \times 336}{42} = 16 \) cm

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**35. Trapezium ABCD, AB || CD, AD = BC**  

Prove:  

1. ∠A = ∠B (base angles isosceles trapezium)  

2. ∠C = ∠D  

3. ∆ABC ≅ ∆BAD (SAS)  

4. AC = BD (CPCT)

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## **SECTION E**

**36. Deepak and Sanjay's routes**  

1. Deepak: Left from O (0,0) → (-3,0), left → (-3,2), left → (3,2), right → (5,2) → Quadrants: II, I  

2. Sanjay: Right → (1,0), left → (1,2), right → (3,2), → (3,5) → Quadrants: I  

3. Hospital coordinates: (3,2)

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**37. Cone: dia 40 cm, height 1 m**  

r = 0.2 m, h = 1 m  

l = \( \sqrt{0.2^2 + 1^2} = \sqrt{0.04 + 1} = \sqrt{1.04} = 1.02 \) m  

CSA = \( \pi r l = 3.14 \times 0.2 \times 1.02 = 0.64056 \) m²  

Volume = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times 0.04 \times 1 = 0.04187 \) m³  

Cost of painting 50 cones = \( 50 \times 0.64056 \times 12 = 384.336 \) rupees  

**OR** Cardboard cost = \( 50 \times 0.64056 \times 100 = 3202.8 \) rupees

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**38. Students planning**  

Total students = x  

\( \frac{1}{12}x^2 + \frac{7}{12}x + 10 \)  

Degree = 2  

If x = 96:  

Historical monuments = \( \frac{1}{12} \times 9216 = 768 \)  

Old age homes = \( \frac{7}{12} \times 96 = 56 \)

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